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(i) Evaporation of water in a closed vessel.
(ii) Dissolution of a gas in a liquid under pressure in a closed vessel.
What kind of molecules in a liquid can evaporate?
Name the factors on which vapour pressure of any liquid depends.
Under what condition solid 
liquid, equilibrium can exist?
The following property becomes constant in the equilibria.
(i) A solubility of the solid
(ii) Mass of the gas dissolved.
Which measurable property becomes constant in water vapour equilibrium?
What is the role of forward and backward reactions at equilibrium of a reversible reaction?
What is a forward reaction?
A forward reaction is a reversible reaction in which reactants produce products, and a reverse reaction turns those products into their original reactants.
What condition favours a reversible reaction?
What is a backward reaction?
What is meant by a system at dynamic equilibrium?
It is a system, in which the forward and backward reactions proceed at the same rate and the respective quantities remain unaltered.
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discuss the effect of a catalyst on equilibrium ?
Characteristics of equilibrium constant:
(i) It changes with temperature.
(ii) It does not change with concentration.
Equilibrium constant in terms of partial pressures:
Kp = Kc (RT)∆n, where Kp is the equilibrium constant in terms of partial pressure.
Kc is the equilibrium constant in terms of concentration, R is gas constant, T is kelvin temperature and ∆n = No. of moles of the gaseous products – No. of moles of the gaseous reactants.
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#13 {main}</pre>](/application/zrc/images/qvar/CHEN11088872.png)
We have given,
N2 = 3·0 × 10–3 M,
O2 = 4·2 × 10–3 M
NO = 2·8 × 10–3 M
Applying the law of chemical equilibrium. We have
Putting the values in expression (1), we have
What is the effect of temperature on the equilibrium state of a system at equilibrium?
In exothermic reactions, increase in temperature decreases the equilibrium constant, K, whereas, in endothermic reactions, increase in temperature increases the K value.
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The reaction
is at equilibrium, 
formed is absorbed in sulphuric acid, what happens to the equilibrium?
Define Le-Chatelier's principle ?
In a gaseous system at equilibrium 
3 mole of inert gas is pumped into the vessel at constant pressure. What will be its effect on equilibrium?
What is the effect of pressure on the solubility of a gas in a liquid?
Or
State Henry's Law.
A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature. The volume of the container is suddenly increased.
(i) What is the initial effect of the change on vapour pressure?
(ii) How do rates of evaporation and condensation change initially?
(iii) What happens when equilibrium is restored finally and what will be the final vapour pressure?
(i) Vapour pressure will decrease initially.
(ii) The rate of evaporation remains constant at a temperature in a closed vessel. However, the rate of condensation will be low initially due to the presence of lesser molecules per unit volume in the vapour phase and hence the number of collisions per unit time with the liquid surface decreases.
(iii) When equilibrium is restored finally, then the rate of evaporation becomes equal to the rate of condensation. The final vapour pressure will be the same as it was initial.
(i) In case of liquid 
gas equilibrium, the vapour pressure of the liquid becomes constant at a given temperature.
(ii) In case of solid 
solution equilibrium, the concentration of solute in solution becomes constant at a given temperature.
(iii) In the case of gas 
solution equilibrium, the pressure of the gas above liquid becomes constant at a given temperature.
(iv) In the case of solid 
liquid equilibrium, there is only one temperature (melting point) at which two phases can co-exist i.e. temperature remains constant at a given pressure.
Hence the common characteristics of physical equilibrium are:
1. Equilibrium can be established only in case of a closed system.
2. The equilibrium is dynamic in nature i.e. the process does not stop after the establishment of equilibrium but the rate of the forward reaction becomes equal to the rate of backward reaction.
3. The measurable properties of the system such as melting point, boiling point, vapour pressure and solubility remain constant since the concentration of the substances remains constant.
4. When equilibrium is attained there exists an expression involving the concentration of reacting substances which acquire a constant value at a given temperature.
5. In the case of a gas dissolving in a liquid, the increase of pressure always increases the solubility of the gas in the liquid.
(i) The initial effect on the vapour pressure of increasing the volume of the container will be that the vapour pressure is lowered. This is because the same amount of vapours now occupy large space.
(ii) Due to a sudden increase in volume, the pressure inside the sealed container suddenly decreases. According to Le-Chatelier’s principle for the equilibrium,
A decrease in pressure will shift the equilibrium in the forward direction. Thus the rate of evaporation will increase initially. The rate of condensation decreases initially because vapour pressure per unit volume decreases. However, due to increase in the rate of evaporation, the amount of vapour begins to increase and so the rate of condensation also begins to increase.
(iii) When the equilibrium is again reached, the rate of evaporation becomes equal to the rate of condensation. The final vapour pressure will remain the same as vapour pressure before increasing the volume of the container. This is because the vapour pressure of a liquid does not depend on upon the amount of the liquid or the space above it but depends on only upon temperature.
Reversible reactions: Reactions which proceed in both directions are called reversible reactions. In a reversible reaction, the reactants are changed into products and simultaneously the products are changed into reactants. A reversible reaction is shown by using two half arrows in opposite direction 
. In a reversible reaction, the reaction proceeding from left to right (reactants that give the products) is called forward reaction. The reaction proceeding from right to left (products that give back the reactants) is called a backwards reaction. The reversible reaction generally proceeds in a closed vessel. This prevents the escape of the products. Example of reversible reaction are:
Irreversible reactions: A chemical reaction is said to be irreversible if reactants are changed into products (proceeds only in the forward direction) but the products do not combine to form the reactants. These reactions are indicated by a single arrow 
in a chemical equation. Examples of irreversible reactions are:
(i) Burning of magnesium
(ii) Decomposition of potassium chlorate.
(iii) The reaction between hydrogen and oxygen to form water vapours. 
A reversible reaction is said to be in a state of equilibrium when the rate of the. forward reaction is equal to the rate of backward reaction. At equilibrium, the concentrations of the reactants and products remain constant provided the conditions under which the reaction is performed are not changed. Some observable properties (colour, concentration, partial pressure or density) of the reactants or products may be used to indicate the concentration of the reaction at equilibrium.
For example, the colour of nitrogen dioxide (NO2) may be used in recognising the equilibrium state when the reaction takes place between carbon monoxide and nitrogen dioxide. During the reaction, the reddish brown colour of nitrogen dioxide slowly fades as it is changed to colourless nitric oxide. After some time, the intensity of the colour decreases and there will be no further change in colour. Then we can say that the equilibrium state is reached.
In the case of dissociation of calcium carbonate, the attainment of equilibrium is recognised by observing the constant partial pressure of carbon dioxide in the manometer.
The important characteristics of chemical equilibrium are as follows:
(i) At the equilibrium state, all the measurable properties such as concentration, colour, density, partial pressure become constant.
(ii) Chemical equilibrium is dynamic in nature i.e. reaction does not stop but proceeds in both the directions with the same rate.
(iii) Free energy change ∆G = 0.
(iv) Chemical equilibrium can be approached from either direction (either from the reactants or products side).
(v) Equilibrium can be reached only in a closed system.
(vi) The equilibrium state is not affected by the presence of a catalyst as it lowers the same magnitude of activation energy by speeding up both forward and backward reactions to the same extent. However, a catalyst helps to reach the equilibrium state in a shorter period of time.
We know that when the temperature increases, the vapour pressure of a liquid also increases and at the boiling point, vapour pressure becomes equal to the atmospheric pressure. A liquid with lower vapour pressure will require a higher temperature for it to attain its vapour pressure equal to the atmospheric pressure. Therefore, the liquid with the highest vapour pressure will have the lowest boiling point. Acetone has the lowest boiling point and water has the highest boiling point.
At 293K, water evaporates least in the sealed container before equilibrium is established.
The law of mass action (Guldberg and Waage) may be stated as:
The rate at which a substance reacts is directly proportional to its active mass (molar concentration) and the rate of chemical reaction is proportional to the product of active masses of the reacting substances.
By the term active mass, we mean molar concentration or number of moles per litre. The active mass of A is denoted as [A].
Consider a simple reaction A 
Product
According to the law of mass action,
Rate of reaction 
where [A] represents the active mass of A, k is proportionality constant known as velocity constant. The velocity constant is also called velocity co-efficient or rate constant or specific reaction rate.
According to the law of mass action,
Rate of the reaction 
For a general reaction,
According to the law of mass action,
Rate of the reaction = 
Hence the law of mass action may also be stated as the rate of the reaction is proportional to the product of the active masses (or molar concentration) of reactants with each active mass raised to the power equal to the number of molecules of the respective reactants participating in the reaction.
is called equilibrium constant and has a constant value at a given temperature. Similarly, for a reaction 
Balanced chemical equation
How is equilibrium constant expressed when the reaction is carried in the gaseous phase ?
...(2)
Write the expressions for equilibrium constant for the reactions:
Expression for equilibrium constant for the reactions:
The concentration of pure liquids and solids in the equilibrium expression for heterogeneous reactions are ignored because their concentrations remain constant. By convention, the concentration of all pure solids and pure liquids is taken as unity, i.e.
[solid] = 1, [liquid] = 1.
(i) The equilibrium constant has a definite value for every chemical reaction at a given temperature. However, it varies with the change in temperature.
(ii) Its value is not influenced by the change in the concentration of reactants and products.
(iii) It is not affected by the presence of a catalyst.
(iv) The equilibrium constant for the forward reaction is the inverse of the equilibrium constant for the backward reaction.
For 
and for
Clearly
(v) The value of K tells us the extent to which the forward or backward reaction has taken place. The Greater value of Kc and Kp means that the reaction has proceeded to a greater extent in the forward direction.
(vi) The value of K changes if the coefficient of various species in the equation representing equilibrium are multiplied by the same number.
For example,
However, if we write 
Then 
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will have the units 
What is the significance of equilibrium constant?
(i) The numerical value of equilibrium constant is a measure of the extent to which reactants have been converted into the products. A large value of K indicates that reactants have been converted to the products to a large extent. The lower value of K indicates that only small amounts of reactants have been converted into products.
(ii) If we know the initial concentration of reactants and equilibrium constant of the reaction, we can calculate the equilibrium concentration of various reactants and products.
A mixture of 
with molar concentration 
and 
respectively was prepared at 500 K. At this temperature the value of 
for reaction 
Predict whether at this state the concentration of 
will increases or decrease.
Since Qc ≠ Kc, reaction is not in equilibrium.
Since Qc < Kc reaction will proceed in the forward direction i.e. towards the formation of ammonia.
The value of 
for the reaction 
At a given time the composition of reaction mixture is 
In which direction the reaction will proceed?
For the above reaction
...(1)
Here,
Putting the values in expression (1), we have
As Qc > Kc, so the reaction will proceed in the reverse direction.
A mixutre of 1.57 mol of N2, 1.92 mol of H2 and 8.13 mol of NH3 is introduced into a 20L reaction vessel at 500K. At this temperature, the equilibrium constant, Kc, for the reaction
is 
. Is the reaction mixture at equilibrium? If not, what is the direction of the net reaction?
The reversible reaction is
Concentration ratio,
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the reaction mixture is not in equilibrium.
As Qc > Kc, the net reaction will be in the backward direction.
is at equilibrium at 1300K in a 1 L flask. It also contains 0.30 mol of CO, 0.10 mol of H2 and 0.02 mol of H2O and an unknown amount of CH4 in the flask. Determine the concentration of CH4 in the mixture. The equilibrium constant, Kc, for the reaction at the given temperature is 3.90.
What is Kc for the following equilibrium when the equilibrium concentration of each substance
is: 
The reversible reaction is
The equilibrium constant Kc is given by
Putting the values in expression (1), we have
The given reversible reaction is
(Decrease in the pressure of HI = 0.2 - 0.04 = 0.16 atm).
Applying the law of chemical equilibrium
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Putting the values in expression (1), we have
for the equlibrium.
The reversible reaction is,
Applying the law of chemical equilibrium
Putting the values, we have,
The value of Kc for the reaction 
is 
at 
If the equilibrium concentration of 
in air at 
is 
what is the concentration of O3?v
The given reversible reaction is
Applying the law of chemical equilibrium
...(1)
Putting this value in expression (1), we have
Calculate the value of equilibrium constant for the reaction:
There is 10.0 mol of N2, 14·0 mol of O2 and 0·2 mol of NO2 present at equilibrium in a 3·0L vessel at 298K.What will be the effect of increased temperature on the equilibrium constant?
the partial pressure of SO2, NO2, SO3 and NO at equilibrium are 0.5, 0.8, 0.7 and 1.2 bar respectively. Calculate Kp for the reaction. 
Write the equilibrium constant expression for the following reactions:
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State and explain Le-Chatelier’s principle.
Le-Chatelier’s principle. This principle may be stated as if a stress (such as a change in concentration, temperature or pressure) is applied to a system in equilibrium, the equilibrium shifts in a way to undo or nullify the effect of the imposed stress.
(i) Effect of change of concentration on equilibrium. If the concentration of any one or all the reactants is increased, the equilibrium shifts towards right hand side to form more products whereas increase in the concentration of any one or all the products shifts the equilibrium towards left hand side to form more reactants in order to nullify the effect of increase in the concentration of reactants or products respectively. For example, consider the reaction,
Increase in concentration of reactants (N2, H2) will shift the equilibrium in the forward direction in order to decrease their concentration. The addition of extra NH3 from outside the equilibrium mixture will shift the equilibrium in the backward direction.
(ii) Effect of temperature on equilibrium. According to Le-Chatelier’s principle of increasing the temperature, the equilibrium shifts towards that direction where absorption of heat (endothermic change) takes place in order to nullify the effect of the rise in temperature. On the other hand, on decreasing the temperature, the equilibrium shifts towards that direction where the evolution of heat (exothermic change) takes place in order to nullify the effect of a decrease in temperature. For example.
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On decreasing the temperature, the equilibrium shifts in the forward direction i.e. towards the exothermic reaction (evolution of heat). Thus, a decrease in temperature favours the formation of sulphur trioxide.
On increasing the temperature, the equilibrium shifts in the forward direction i.e. towards the endothermic reaction (absorption of heat). Thus, an increase in the temperature favours the formation of nitric oxide.
(iii) Effect of change in pressure on equilibrium. On increasing the pressure, the number of moles per unit volume increases and thus according to Le-Chatelier’s principle the equilibrium shifts towards the side where the number of moles per unit volume decreases in order to nullify the effect of an increase in pressure. For example.
On increasing the pressure, the number of moles per unit volume increases and thus according to Le-Chaterlier's principle the equilibrium shifts towards the right-hand side (i.e. towards the formation of ammonia) where the number of moles per unit volume decreases.
(iv) Effect of the catalyst. A catalyst has no effect on equilibrium point. This is because it increases the rate of the forward as well as backward reaction to the same extent. Thus, a catalyst does not affect the position of equilibrium, but simply helps to achieve the equilibrium in a shorter time i.e. quickly.
(i) Effect of concentration. Increase in concentration of reactants (N2, H2) will shift the equilibrium in the forward direction to form more ammonia in order to decrease their concentrations. The addition of extra NH3 from outside to the equilibrium mixture will shift the equilibrium in the backward direction. Thus, the addition of N2 and H2 favours the formation of ammonia.
(ii) Effect of temperature. The forward reaction is exothermic in nature while the backward reaction is endothermic in nature. According to Le-Chatelier’s principle, on decreasing the temperature, the equilibrium shifts towards that direction where the evolution of heat takes place in order to nullify the effect of decreasing temperature. Thus, a decrease in temperature favours the formation of ammonia.
On the other hand, on increasing the temperature the equilibrium shifts towards the backward direction where absorption of heat takes place in order to nullify the effect of the rise in temperature. Thus, lower temperature favours the formation of ammonia.
(iii) Effect of pressure. On increasing the pressure, the number of moles per unit volume increases and thus according to Le-Chatelier’s principle, the equilibrium shifts towards that side where the number of moles per unit volume decreases in order to nullify the effect of an increase in pressure. On the other hand, on decreasing the pressure, the equilibrium shifts towards the backwards direction i.e. ammonia decomposes to give N2 and H2. Thus, an increase in pressure favours the formation of ammonia.
Hence favourable conditions in the formation of ammonia are:
(i) The addition of N2 and H2
(ii) Lower temperature
(iii) Higher pressure.
What is the effect of adding an inert gas(say He or N2):
(i) at constant volume and
(ii) at the constant pressure on the following equilibrium:
The given dissociation equilibrium is
(i) Adding inert gas at constant volume. If an inert gas like helium or nitrogen etc. is added to a system at equilibrium at constant volume, then the total pressure increases. However, there will be no change in the position of equilibrium, even though the pressure has changed. This is because the concentrations of reactants and products (number of moles/volume) will not change. Hence the values of concentrations will continue to satisfy the equilibrium law. Hence the state of equilibrium will remain unaffected.
(ii) Adding inert gas at constant pressure. When an inert gas is added to a system at equilibrium keeping the pressure constant, the volume of the system increases. This results in a decrease in a number of moles of reactants per unit volume. According to Le-Chatelier’s principle, the equilibrium shifts in a direction in which there is an increase in the number of moles of gases. Hence the equilibrium shifts towards the forward direction where the number of moles of gases increases. In other words, more PCl5 dissociates to give PCl3 and Cl2. Hence dissociation of PCl5 increases with the addition of an inert gas.
According to Le Chatelier’s principle, on decreasing the pressure, moles of reaction products will:
(a) increase
(b) decrease
(c) remain the same.
[∵ np = nr gaseous)
Which of the following reactions will get affected by increasing the pressure? Also, mention whether the change will cause the reaction to go into forward or backward direction?
(i) In this case np (=2) is greater than nr (=1), the reaction will go in the backward direction.
(ii) In this case np ( = 3) is equal to nr ( = 3), reaction will not be affected by the pressure.
(iii) In this case np (= 2) is greater than nr (=1), the reaction will go in the backward direction.
(iv) In this case np (=1) is less than nr (= 3), the reaction will go in a forward direction.
(v) In this case np (=1) is greater than nr (= 0) reaction will go in the backward direction.
(vi) In this case np (= 10) is greater than nr (= 9), the reaction will go in the backward direction.
The change of ice into water is a reversible endothermic (i.e. accompanied by absorption of heat) process. The reaction involves a decrease in volume. Hence according to Le-Chatelier’s principle:
(i) With the increase in pressure: The equilibrium tends to shift in that direction in which there occurs a decrease in the volume. So in this case, an increase in pressure favours the conversion of ice into water.
(ii) With the increase in temperature: The equilibrium shifts to the right i.e. towards the direction in which heat is absorbed. Thus, increase in temperature favours melting of ice.
Solution; 
(a) 
(b)
(i) The value of Kp remains unchanged on increasing the pressure. According to Le Chatelier’s principle, equilibrium will shift in the backward direction.
(ii) In case of endothermic reactions the value of increases with the increase in temperature. According to Le Chatelier principle, equilibrium will shift in the forward direction.
(iii) Kp will remain undisturbed i.e. equilibrium composition will not be disturbed but equilibrium will be attained equally.However, in the presence of a catalyst, the equilibrium would be attained quickly.
Describe the effect of:
(a) addition of H2
(b) addition of CH3OH
(c) removal of CO
(d) removal of CH3OH
on the equilibrium of the reaction:
(a) According to Le-Chatelier’s principle, equilibrium will shift in the forward direction.
(b) According to Le-Chatelier’s principle, equilibrium will shift in the backward direction.
(c) According to Le-Chatelier’s principle, equilibrium will shift in the backward direction.
(d) According to Le-Chatelier’s principle, equilibrium will shift in the forward direction.
(c)
(i) Kc remains unchanged when more PCl5 is added.
(ii) When pressure is increased, Kc remains unchanged.
(iii) As given reaction is endothermic, on increasing the temperature, Af will increase.
As 
will increase with the increase of temperature.
ions (from NaOH) combine with 
ions to form colourless water.
(i) write the expression for the equilibrium constant for the reaction;
(ii) at room temperature (300 K) will Kp be greater than, less than or equal to Kp at 900 K;
(iii) how will the equilibrium be affected if the volume of the vessel containing the three gases is reduced, keeping the temperature constant; what happens;
(iv) what is the effect of adding 1 mole He(g) to a flask containing SO2,O2 and SO3 at equilibrium at constant temperature?
The dissociation of phosgene is represented as follows:
When the mixture of these three gases is compressed at constant temperature, what happens to:
(i) the amount of CO in the mixture.
(ii) the partial pressure of COCl2
(iii) the equilibrium constant for the reaction?
For the equilibrium 
the value of the equilibrium constant KC is 
Calculate Kp for the reaction at this temperature.
...(1)
For the equilibrium
the value of the constant, Kc is 
Calculate 
for the reaction at this temperature.
We know
...(1)
Putting the values in expression (1), we have
The value of Kc for the reaction 
is found to be 186.9 at 750K. Calculate the value of 
if partial pressures are expressed in bars.
...(1)
...(1)
...(1)
is 
What is the chemical value in moles per litre of Kc for this reaction at the same temperature?
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...(1) 
Sponsor Area
The equilibrium constant at 278K for 
is 
. In a solution in which copper has displaced some silver ions from the solution, concentration of Cu2+ ions is 1.8 x 10-2 mol L-1 and the concentration of Ag+ ions is 3·0 ×10–9 mol L–1. Is the system at equilibrium?
which is same as for the reaction in equilibrium.
Hence the given system is in equilibrium.
One mole of 
and one mole of CO are taken in a 10-litre vessel and heated to 725K. At equilibrium, 40% of water (by mass) reacts with carbon monoxide according to the equation
Calculate the equilibrium constant for the reaction.
40% by mass of water = 40% by mole of water
40% of one mole of water = 
Initial mole 1 1 0 0
Final mole 1-0.4 1-0.4 0.4 0.4
= 0.6 = 0.6
According to the law of chemical equilibrium,
Substituting the values, we have,
2N2O(g)
x
Nitric oxide reacts with Br2 and gives nitrosyl bromide as per reaction given below:
When 0.087 mol of NO and 0.0437 mol of Br2 are mixed in a closed container at constant temperature, 0.0518 mol of NOBr is obtained at equilibrium. Calculate equilibrium amount of NO and Br2.
At 700K, equilibrium constant for the reaction:
is 54.8. If 0.5 mole/litre of HI(g) is present at equilibrium at 700K, what are the concentration of H2(g) and I2(g) assuming that we initially started with HI(g) and allowed it to reach equilibrium at 700 K?
Ethyl acetate, is formed by the reaction of ethanol and acetic acid and equilibrium is represented as
(i) Write the concentration ratio (reaction quotient), Qc, for this reaction (note:water is not in excess and is not a solvent in this reaction).
(ii) At 293K, if one starts with 1·00 mol of acetic acid and 0·18 mol of ethanol, there is 0·171 mol of ethyl acetate in the final equilibrium mixture. Calculate the equilibrium constant.
(iii) Starting with 0·5 mole of ethanol and 1·0 mole of acetic acid and maintaining it at 293K, 0·214 mole of ethyl acetate is formed after sometime. Has equilibrium been reached ?
(ii) The various molar concentrations before the chemical reaction and at equilibrium point may be represented as
Initialconc.
1 mole 0.180 ml 0 0
Equilibrium conc.
1 - 0.171 0.180 - 0.171 = 0.171 0.171
(0.829) (0.009) = 0.171 0.171
Applying the law of chemical equilibrium,
(iii) The various molar concentrations before the reaction and at equilibrium point may be represented as
Initial conc.
1.00 mole 0.5 mole 0 0
Equilibrium conc.
1 - 0.214 0.5- 0.214 = 0.214 0.214
(0.786) (0.286) = 0.214 0.214
Concentration ratio,
As the value of Qc is less than the value of equilibrium constant, therefore, the equilibrium has not reached.
(i) Why some concentrated sulphuric acid is usually added to the reaction mixture in a laboratory preparation of ethyl acetate?
(ii) Since the heat of reaction is nearly zero for this reaction, how will the equilibrium constant depend on upon the temperature?
(i) Concentrated sulphuric acid is a dehydrating agent. It can absorb water. According to Le-Chatelier’s principle, the removal of water from the reaction mixture, equilibrium will shift in the forward direction producing more of ethyl acetate.
(ii) Since the heat of reaction is nearly zero, the equilibrium constant is almost not affected by temperature.
Since at equilibrium 0·33 mole of acetic acid is left, we say that 1 – 0·333 = 0·667 mole of acetic acid is used up.
Now according to the equation, it is clear that 0·667 mole of acetic acid combines with a 0·667 mole of ethyl alcohol to form a 0·667 mole of ethyl acetate and a 0·667 mole of water.
At equilibrium [CH3COOH] = 0·333 mole ; [C2H5OH] = 0·333 mole [CH3COOC2H5] = 0·667 mole;
[H2O] = 0·667 mole
Applying the law of chemical equilibrium, we have
Let at equilibrium,
Then 
Initial conc. 0.78 M 0 0
At eqm. 0.78 - 2x x x
Applying the law of chemical equilibrium
Putting the values in expression (1), we have
Hence, at equilibrium 
and
Let at equilibrium,
The reaction is
Initial pressure 4.0 atm 0 0
At equm. 4 - p p p
Applying the law of chemical equilibrium
Putting the reaction in expression (1),
we have
Taking positive value,
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Bromine monochloride (BrCl) decomposes into bromine and chlorine and attains the equilibrium:
for which Kc = 32 at 500 K. If initially pure BrCl is present at a concentration of 3·30×10–3 mol L–1 what is its molar concentration in the mixture at equilibrium?
At equilibrium
Suppose the partial pressure of H2 at equilibrium = p bar
Initial pressure 4.0 bar 4.0bar = - -
At eqm.
(4 -p) (4-p) = p p
Applying the law of chemical equilibrium,
Putting the values in expression (1), we have
The value of 
for the phoshorylation of glucose in glycolysis is 13.8 kJ/mol. Find the value of Kc at 298 K.
...(1)
Hydrolysis of sucrose gives,
Equilibrium constant Kc for the reaction is 
Calculate 
at 300K.
We know,
Putting the values in expression (1), we have
At 1127 K and 1 atm pressure, a gaseous mixture of CO and CO2 in equilibrium with solid carbon has 90.55% CO by mass
Calculate Kc for this reaction at the above example.
If the total mass of the mixture of CO and CO2 is 100g. then the mass of CO=90.55g and mass of
CO2= 100-90.55=9-45g
therefore,
Number of moles of CO= 90.55/28=3.23
Number of moles of CO= 9.45/44 =0.22
Neutralisation is the process of formation of a conjugate base and a conjugate acid of the original acid and base which react respectively. For example,
is the conjugate base of HCl and 
is the conjugate acid of NH3.
is the conjugate base of HCl.
is the conjugate acid of NH3.
as a Bronsted base:
are conjugate acid base pair.
are conjugate acid base pair.
Lewis bases : NH3, OH– ;
Lewis acids : AlCl3, Ag+
A Lewis base is a substance that can donate a pair of electrons to form a new bond.Neutralisation is the sharing of an electron pair between an acid and base.
How does degree of dissociation of a weak acid (HA) vary with concentration?
Or
What is Ostwald’s dilution law?
The pH of a solution is a measure of the molar concentration of hydrogen ions in the solution and as such is a measure of the acidity or basicity of the solution.
Hydronium ion concentration in molarity is more conveniently expressed on a logarithmic scale known as the pH scale.
Acids: pH ranges 1 to 7.
Bases: pH ranges 7 to 14.
What happens to the pH if a few drops of acid are added to CH3 COONH4 solution ?
What change is observed in the H+ ion concentration of water on adding on acid to it?
Give the conjugate acid - base pairs in dilute sulphuric acid.
Out of CH3COO– and OH– which is stronger base and why?
What is the difference between solubility product and ionic product?
What is the maximum value of the ionic product of a salt at a particular temperature ?
The maximum value of the ionic product is equal to the solubility product constant, Ksp.
How does the degree of ionisation of a weak acid change with the increase in concentration ?
Under what condition a precipitate is formed if two solutions are mixed ?
What is common ion effect?
The phenomenon of suppression of the degree of dissociation of a weak acid or a weak base by the addition of a strong electrolyte containing a common ion is known as common ion effect.
How does the ionisation of H2S change in the presence of HCl?
Due to common-ion effect (H+ ions), ionisation of H2S is suppressed.
During qualitative analysis, why NH4Cl and NH4OH are added in the III group?
What happens when sufficient quantities of NH4Cl, NH4OH and (NH4)2CO3 are added to an aqueous solution containing chlorides of magnesium and calcium?
What are electrolytes and non-electrolytes?
Generally,the substances which can conduct electric current by the migration of ions through their molten states or aqueous solutions.
Acids: HCl, HNO3, H2SO4, CH3COOH etc.
Bases: NaOH, KOH, NH4OH, Ca(OH)2 etc.
Salts: NaCl, NH4Cl, CH3COONH4, CaCl2 etc.
Non-electrolytes: The substances which do not conduct electric current through their molten states or aqueous solutions are known as non-electrolytes. For example, glucose, fructose, cane sugar, glycerol.
Discuss the ionisation of weak electrolytes (Ostwald’s dilution law).
H+(aq) + A-(aq)
Thus, the degree of ionisation of a weak electrolyte is inversely proportional to the square root of the molar concentration of the solution of the electrolyte.
Hence the degree of ionisation of weak electrolyte goes on increasing with the decrease in molar concentration i.e. with dilution and it reaches the maximum value (unity) in very dilute solution. Thus, all weak electrolytes undergo almost complete ionisation at infinite dilution.
For a weak base BOH:
Let us consider a weak base BOH dissolved in water, a be the degree of ionisation and C be the molar concentration of BOH.
Initial conc. C 0 0
Equilibrium 
conc.
Applying law of chemical equilibrium,
CH3COO-(aq) + H3O+
Initial concentration of acetic acid = 0·1 M
As the degree of ionisation is 1·34% thus amount of acetic acid ionised
Thus we have,
CH3COOH 
CH3COO-(aq) + H3O+
Initial conc
0.1 M 0 0
Equilibrium conc
0.1 - 0.00134 0.00134 M 0.00134M
= 0.9866 M
Applying the law of chemical equilibrium,
Substituting the values, we have,
What is the fate of proton in aqueous medium?
(i) Limited scope: Arrhenius concept is limited to aqueous medium only. It fails to explain the behaviour of acids and bases in a non-aqueous solvent like ammonia,sulphur dioxide,alcohol, etc.
(ii) According to Arrhenius, the acidic and basic properties are due to H+ and OH–ions respectively. However, these ions do not exist as such that readily get hydrated and represented as H+ (aq) and OH– (aq).
(iii) It fails to explain the acidic nature of CO2, SO2, SO3, N2O5 which do not contain any hydrogen.
(iv) It fails to explain the basic nature of CaO, NH3,Na2CO3 which do not contain any hydroxyl group.
Discuss the Lowry Bronsted concept of acids and bases.
Or
Discuss the protonic concept of acids and bases.
A species formed by adding a proton to Bronsted base is called conjugate acid and a species formed by losing a proton from Bronsted acid is called conjugate base. 
Thus, every acid has its conjugate base and every base has its conjugate acid. Consider a reaction between HCl and H2O.
In the forward reaction, HCl donates a proton to water.
∴ HCl is an acid and H2O is a base.
In the backward reaction, H3O+ ion donates a proton to Cl– ion.
∴ H3O+ is an acid and Cl– is a base.
Thus Cl– ion is a conjugate base of HCl and H3O+ is the conjugate acid of base H2O. Such pairs of substances which differ from one another by a proton are known as conjugate acid-base pairs. Thus any acid-base reaction really involves two acids and two bases, forming conjugate pairs.
Given species Conjugate base Name
Given species Conjugate acid Name 
State the formula and name of conjugate acid of the following bases:
(i) 
(ii) NH3
(iii) CH3COO– (iv) HS–
Given Species Conjugate acid Name
Identify the acids and bases in the following equilibrium reactions:
ions are bases as both have the tendency to accept a proton.
According to Bronsted Lowry concept:
(i) the strength of an acid depends on upon its tendency to donate proton and
(ii) the strength of base depends on upon its tendency to accept a proton.
Consider a reaction between HNO3 and water.
HNO3 has a greater tendency to donate a proton than H3O+ ion. Therefore. HNO3 is a strong acid, NO3– ion has a very little tendency to accept a proton, therefore, NO3– ion is a weak base.
Thus strong acid has a weak conjugate base.
Now consider the reaction between acetic acid and water.
Acetic acid has a small tendency to donate a proton, therefore, it is a weak acid. CH3COO– ion has strong tendency to accept a proton, therefore, CH3COO– ion is a strong base.
Thus weak acid has a strong conjugate base.
All Arrhenius acids are Bronsted acids while all Arrhenius bases arc not Bronsted bases. Discuss.
All Arrhenius acids are Bronsted acids: According to Arrhenius, an acid is a substance that yields hydrogen ions in an aqueous solution. According to Bronsted concept, an acid is a substance that can donate a proton (H+ ion). Both definitions require an acid to be a source of protons. Thus HCl is an acid according to both the theories. Therefore, all Arrhenius acids are also Bronsted acids.
All Arrhenius bases are not Bronsted bases: According to Arrhenius concept, a base is a substance that dissociates in aqueous solution to give hydroxyl ions. According to Bronsted concept, a base is a substance that accepts a proton, e.g. NaOH is a base according to Arrhenius concept because it yields OH– ions in aqueous solution.
But NaOH does not accept a proton as such, thus it is not a base according to Bronsted concept. Therefore, all Arrhenius bases are not Bronsted bases.
What are the limitations (drawbacks) of Bronsted-Lowry concept of acids and bases?
(i) It fails to explain the reaction between acidic oxides (SO2, SO3 etc.) and basic oxides (BaO, CaO, Na2O etc.) because, in such reactions, no proton transfer occurs.
(ii) It fails to recognise the acidic character of FeCl3, AlCl3 BF3 etc.
Discuss in brief Lewis concept of acids and bases.
Or
Discuss the electronic concept of acids and bases.
According to Lewis concept, an acid is any substance (atom, molecule or ion) which is capable of accepting a pair of electrons from another substance to form a co-ordinate (dative) bond.
A base is any substance (atom, molecule or ion) which is capable of donating a pair of electrons to form a co-ordinate (dative) bond. Hence,
In terms of electronic structure.
(i) an acid must have a vacant orbital into which an electron pair donated by a base can be accommodated e.g. AlCl3, BF3
(ii) a base is a substance which has at least one lone pair of electrons e.g
(a) OH– can donate electron pair. Hence it is a Lewis base.
(b) F– can also donate electron pair. Hence it is a Lewis base.
(c) H+ can accept electron pair. Hence it is a Lewis acid.
(d) BCl3 is deficient in electrons. Hence it can accept electron pair and is, therefore, a Lewis acid.
How does Lewis concept account for the acidic character of carbon dioxide?
What are the limitations of Lewis concept of acids and bases ?
(i) The concept fails to recognise the relative strength of acids and bases.
(ii) Formation of a coordination compound is a slow process but acid-base reactions are fast. This concept could not explain this behaviour.
(iii) Protonic acids (H2SO4, HCl, HNO3) do not form a coordinate bond with bases and therefore are not included in the Lewis acids.
(iv) The catalytic activity of many acids is due to H+ ion. Since a Lewis acid need not contain hydrogen hence many Lewis acids will not have catalytic activity.
Justify the statement: All Bronsted bases are Lewis bases, but all Bronsted acids are not Lewis acids.
Out of Lowry Bronsted concept and Lewis concept, which is regarded as better and why?
Lewis concept of acids and basis is regarded as better than Bronsted concept due to the following reasons:
(i) Many species which are acids and bases according to Bronsted and Arrhenius concept are also covered by Lewis concept.
(ii) The acidic and basic character of certain species can be explained only with the help of Lewis concept. For example, acidic nature of CO2 and basic nature of CaO can not be explained by any other concept except the Lewis concept.
(iii) According to Lewis concept, an acid need not contain the element hydrogen but it must be electron deficient in nature. For example AlCl3 is an acid though it has no H+ion present.
What do you understand by ionic product of water?
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The ionic product of water is 0·11 × 10–14 at 273 K; 1·0 × 10–14 at 298K and 51 × 10–14 at 373K. Deduce from this data whether the ionisation of water into hydrogen and hydroxide ion is exothermic or endothermic.
Comment on the statement: An acidic solution contains OH– ions and even a basic solution contains H3O+ ions.
Or
How the values of Kw, [H3O+] and [OH–] are affected if acid or base is added to pure water at 298K?
We know, [H3O+][OH–] = Kw
For pure water, [H3O+] =[OH–]
= 1 × 10–7 mol L–1
On adding few drops of an acid (say HCl) to water, the concentration of H3O+ increases and thus concentration OH– decreases accordingly in order to maintain Kw constant.![<pre>uncaught exception: <b>Http Error #404</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/HttpImpl.class.php line 61<br />#0 [internal function]: com_wiris_plugin_impl_HttpImpl_0(Object(com_wiris_plugin_impl_HttpImpl), NULL, 'http://www.wiri...', 'Http Error #404')
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Hence acidic solution contains both hydronium and hydroxyl ions. But 
But on adding.
But on adding few drops of a base (say NaOH) to water, the concentration of OH–increases and thus the concentration of H3O+ decreases accordingly in order to maintain Kw constant.
Hence basic solution contains both hydronium and hydroxyl ions.
But [H3O+] < [OH–]
In general, in neutral solution,
[H3O+] = [OH–]
In acidic solution, [H3O+] > [OH–]
In basic solution, [H3O+] < [OH–]
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#13 {main}</pre>](/application/zrc/images/qvar/CHEN11090169-5.png)
As pH of a solution goes on decreasing below 7, the acidic nature of solution goes on increasing.
If pH lies between:
(a) 0 and 2, the solution is strongly acidic,
(b) 2 and 4, the solution is moderately acidic,
(c) 4 and 7, the solution is weakly acidic.
(iii) In basic solution:
As pH of a solution goes on increasing above 7, the basic nature of the solution goes on increasing.
If pH lies between:
(a) 7 and 10, the solution is weakly basic,
(b) 10 and 12, the solution is moderately basic,
(c) 12 and 14, the solution is strongly basic.
What is the effect of temperature on pH?
We know that ionic product of water (Kw.) increases with the rise of temperature. Therefore, the concentration of H3O+ ions increases with temperature.
Therefore, pH value is inversely proportional to 
ion concentration
Hence pH value decreases with the rise in temperature.
What is the significance (or importance) of pH value?
The significance of the pH value. pH is a measure of the acidic or basic (alkaline) nature of a solution (concentration of the hydrogen ion [H+] activity in a solution determines the pH).
For neutral solution: pH = 7
For acidic solution: pH < 7
For basic solution: pH > 7
Certain chemical reactions take place at a fixed pH value e.g. fresh cow’s milk (pH = 6·4) turns to the presence of acids produced by the bacteria. The sour milk has a pH 3 to 4.
What is pOH of a solution? How is it related to its pH value?
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The concentration of hydrogen ion in a sample of soft drink is 3·8 × 10-3 M. What is its pH value?
0.005 M NaOH:
Since NaOH is completely dissociated,
0·002M HBr
Since HBr is completely dissociated,
0.002M KOH
As KOH is completely dissociated, therefore, 
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What is the pH of an aqueous solution with hydrogen ion concentration equal to 3 × 10–5.mol L–l ? Is the solution acidic, basic or neutral?
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Calculate the pH of 1 ml of 13.6 MHCl diluted with water to give 1 litre of the solution.
If 0.561 KOH is dissolved in water to give 200 mL, of the solution at 298 K, calculate the concentration of potassium, hydrogen and hydroxyl ions. What is its pH ?
Calculate the pH of the resultant mixtures:
(a) 10 mL of 0·2 M Ca(OH)2 + 25 mL of 0·1M HCl
(b) 10 mL of 0·01 M H2SO4 + 10mLof 0·01 IM Ca(OH)2
(c) 10mL of 0·1 M H2SO4 + 10mL of 0·1 M KOH
How many grammes of NaOH must be dissolved in one litre of the solution to give it a pH value of 12?
Calculate the pH of a solution obtained by mixing 150 mL of 0.1 N - NaOH and 150 ml of 0.2 N - HCl.
are neutralised by 15 mL of N-HCl
How does Arrhenius theory help in comparing the relative strengths of weak acids and weak bases?
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#13 {main}</pre>](/application/zrc/images/qvar/CHEN11090273.png)
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#6 [internal function]: _hx_lambda->execute('Http Error #404')
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#8 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/haxe/Http.class.php(444): com_wiris_plugin_impl_HttpImpl->onError('Http Error #404')
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#11 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(80): com_wiris_plugin_impl_HttpImpl->request(true)
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#13 {main}</pre>](/application/zrc/images/qvar/CHEN11090273-1.png)
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#13 {main}</pre>](/application/zrc/images/qvar/CHEN11090273-2.png)
Hence the relative strength of two acids having the same molar concentration in aqueous solution may be compared in terms of the square root of their dissociation constants.
Relative strength of weak base:
The dissociation of weak base, say BOH may be represented as,
According to law of chemical equilibrium
Arrange the following in decreasing order of acidic strength:
(i) HOCl (Ka = 3·0 × 10–8)
(ii) HCN (Ka = 4·0 × 10–10)
(iii) HNO3 (Ka = 4·5 × 10–4)
(iv) HF (Ka = 6·7 × 10–4)
Greater the Ka value, stronger will be the acid.
∴
Decreasing order of acidic strength is
HF > HNO3 > HOCl > HCN.
H3O+ + ClO- 
be its degree of dissociation, then
H3O+ + X-
It has been found that the pH of a 0.01 M solution of an organic acid is 4.15. Calculate the concentration of the anion, the ionisation constant of the acid and its pKa.
We know,
The concentration of codeine ion is also same as that of hydroxyl ion. The concentration of the ions is very small and hence the concentration of undissociated base can be taken as equal to 0·005 M.
We know, 
Suppose x is the amount of dimethylamine dissociated in the presence of 0.1 M NaOH.
(CH3)2NH + H2O 
(CH3)2NH+OH- + OH-
Initial conc
0.02 M 0 0
After
dissociation
0.02 - x x 0.1 + x
= 0.02 = 0.1
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#6 [internal function]: _hx_lambda->execute('Http Error #404')
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#13 {main}</pre>](/application/zrc/images/qvar/CHEN11090625.png)
The reaction is
At 298 K, calculate the pH of 0.200 M solution of methylamine, CH3NH2(ionisation constant = 4·4×105).
The ionisation constant of propanoic acid is 1·32 × 10–5. Calculate the degree of ionisation of the acid in its 0·05M solution and also it's pH. What will be its degree of ionisation if the solution is 0·01M in HCl also?
Explain common in effect.
Discuss the salt hydrolysis in the following cases:
(i) Salts of strong acid and strong base
(ii) Salts of strong acid and weak base
(iii) Salts of weak acid and strong base
(iv) Salts of weak acid and weak base.
(i) Salts of strong acid and strong base:
When a salt of the strong acid and a strong base (say NaCl) is dissolved in water, the resulting solution is neutral having pH equal to 7. The hydrolysis of the salt can be represented as:
i.e. there is the only dissociation of sodium chloride and no hydrolysis.
(ii) Salts of strong acid and weak base:
When a salt of the strong acid and a weak base is dissolved in water, it gives an acidic solution having pH value less than 7. For example hydrolysis of NH4Cl can be represented as:
The overall reaction will be
i.e. it is a cationic hydrolysis.
(iii) Salts of weak acid and strong base:
When a salt of weak acid and a strong base is dissolved in water, it gives a basic solution having pH greater than 7. For example hydrolysis of sodium acetate can be represented as:
The overall reaction will be
i.e. it is an anionic hydrolysis.
(iv) Salts of weak acid and weak base:
When a salt of weak acid and a weak base is dissolved in water, it forms an almost neutral aqueous solution having value almost equal to 7. For example, hydrolysis of ammonium acetate can be represented as:
The overall reaction will be
i.e. it involves both anionic as well as cationic hydrolysis.
(i) NH4NO3 solution is acidic, as it is a salt of the strong acid and a weak base, having pH < 7.
(ii) NaCN, NaNO2, KF solutions are basic, as they are salts of a strong base and a weak acid, having pH > 7.
(iii) NaCl, KBr solutions are neutral, as they are the salts of a strong acid and strong base, having pH = 7.
Explain the following terms:
(i) Hydrolysis constant
(ii) Degree of hydrolysis.
(i) Hydrolysis constant:
Hydrolysis of a salt (BA) may be represented as
Applying the law of chemical equilibrium, we get,
Since water is present in very large excess in aqueous solution, its concentration 
may be regarded as constant.
Since [HA] and [BOH] represent the molar concentration of the free acid and the free base produced and [BA] represents the molar concentration of the unhydrolyzed salt, the expression for the hydrolysis constant may be written as,
(ii) The degree of hydrolysis: The degree of hydrolysis (α) of a salt is defined as the fraction (or percentage) of the total salt which is hydrolysed. For example, if in a solution of aniline hydrochloride, 90% of the salt is hydrolyzed into aniline and hydrochloric acid, the degree of hydrolysis is 0·90 Or may be expressed as percentage i.e. 90%.
Derive the relation,
in the case of hydrolysis of salts of strong acids and weak acids.
Show that for an aqueous solution of a salt of weak base and strong acid 
where 
is the degree of hydrolysis.
XOH + H+
is very small as compared to unity. 
Derive an expression for the degree of hydrolysis of NH4Cl.
+ H2O 
NH4OH + H+
is very small as compared to unity,
.Let the salt of strong acid and weak base be XY and hydrolysis reaction be represented as:
Let c be the initial concentration of the salt and α be the degree of hydrolysis. At equilibrium, the concentration of various species is represented as:
X+ + H2O 
XOH + H+
Initial conc.
c c 0 0
Equilibrium conc.
Since 
is very small as compared to unity,
Derive the relations:
for a solution of a salt of weak acid and strong base.
OH- + HA
is very small as compared to unity,
Show that degree of hydrolysis of a salt of weak acid and weak base is independent of concentration of the solution.
Or
Derive the relation, 
in case of hydrolysis of a salt of weak acid and weak base.
is very small as compared to unity,
The ionisation constant of nitrous acid is 4·5 × 10–4. Calculate the pH of 0·04M sodium nitrite solution and also its degree of hydrolysis.
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#13 {main}</pre>](/application/zrc/images/qvar/CHEN11090708-1.png)
Calculate the pH of 0.1M solution of sodium acetate (pKa = 4·74).
Calculate the degree of hydrolysis of 0.015 M solution of NH4Cl. Given Kb for NH4OH is 1·8 × 10–5, Kw = 10–14at 25°C.
Explain the terms: buffer solution and buffer capacity.
Buffer solution: A buffer solution may be defined as a solution which resists change in its pH on the addition of water or small amount of acid or base. The ability of the buffer solution to resist change in its pH on the addition of acid or base is called buffer action.
Types of buffer solutions : Buffer solutions are of two types:
(i) Acidic buffer. An acidic buffer is a mixture (equimolar) of a weak acid and its salt with strong base. For example CH3COOH and CH3COONa ; H2CO3 ; H3PO4 and K3PO4.
(ii) Basic buffer. A basic buffer is a mixture (equimolar) of a weak base and its salt with strong acid. For example NH4OH and NH4Cl ; Ca(OH)2 and CaSO4.
Buffer capacity or Buffer index: The buffer capacity is defined as the amount (number of moles) of the acid or base which when added to 1 litre of the given buffer solution change its pH by unity i.e.
where d[B] = change in the concentration of base
d[A] = change in the concentration of acid
or dpH = change in pH of the buffer solution
Buffer capacity is always positive, because
(i) On adding acid to buffer, its pH is decreased i.e. d[A] is positive but dpH is negative so that 
(ii) On adding base to a buffer solution, its pH increases, i.e. d[B] and dpH are both positive and hence 
Show with an example how buffer solution resists the action of acid or base towards change in pH.
Or
Discuss the buffer action of:
(i) acidic buffer
(ii) basic buffer.
Ionisation of NH4OH is suppressed by NH4 + ions [common ion effect].
There will be a large concentration of NH4+ ions, Cl– ions and NH4OH molecules.
(i) When a few drops of acid (say HCl) are added, the H+ ions of the added acid combine with OH- ions of-NH4OH (of buffer) to form feebly ionised water. Thus no change in pH occurs.
According to Le Chatelier’s principle, the above reaction results in the greater dissociation of NH4OH to restore the original concentration of OH– ions.
(ii) When a few drops of a base (say NaOH) are added to the buffer, the OH– ions of added base combine with NH4+ ions to form unionised NH4OH.![<pre>uncaught exception: <b>Http Error #404</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/HttpImpl.class.php line 61<br />#0 [internal function]: com_wiris_plugin_impl_HttpImpl_0(Object(com_wiris_plugin_impl_HttpImpl), NULL, 'http://www.wiri...', 'Http Error #404')
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#13 {main}</pre>](/application/zrc/images/qvar/CHEN11090745-6.png)
As the concentration of OH– ions does not increase, the pH value remains unchanged.
In this buffer, reserve acidity is due to NH4+ ions and reserve alkalinity is due to NH4OH molecules.
ions to form very slightly dissociated NH4OH. 
ions, hence there is very little or no change in the pH value of ammonium acetate.
ions and reserve alkalinity due to CH3COO– ions.Calculate the pH of:
(i) an acidic buffer mixture
(ii) a basic buffer mixture.
Or
Derive Henderson’s equation for an acidic and basic buffer mixture.
Or
Derive the following equation for the pH of an acidic buffer:
Describe Ostwald’s theory of acid-base indicators.
Or
How does Ostwald’s theory explain the colour change of:
(i) Phenolphthalein
(ii) Methyl orange in acid-base titrations?
(i) Ostwald’s theory: According to this theory, an acid-base indicator is either a weak organic acid or a weak organic base whose undissociated molecule has a. colour different from the ions furnished by it. These different coloured ions are produced in the solution under the influence of a strong acid or a strong base.
To understand this theory, consider the case of phenolphthalein which is written as HPh. It is a weak acid. Its undissociated molecules are colourless. When its molecules are dissociated, they give colourless H+ ions and deep pink coloured Ph+ ions.![<pre>uncaught exception: <b>Http Error #404</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/HttpImpl.class.php line 61<br />#0 [internal function]: com_wiris_plugin_impl_HttpImpl_0(Object(com_wiris_plugin_impl_HttpImpl), NULL, 'http://www.wiri...', 'Http Error #404')
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#13 {main}</pre>](/application/zrc/images/qvar/CHEN11090781.png)
When phenolphthalein is added to the acidic solution, the dissociation of phenolphthalein is practically nil due to increasing the concentration of hydrogen ions. It means that the solution remains colourless.
When strong alkali is added to the phenolphthalein, OH– ions furnished by alkali combine with H+ ions furnished by phenolphthalein to form feebly ionised water and sodium salt NaPh. The NaPh remains in ionic state and imparts a pink colour to it. This is explained as follows:
Similarly, this theory also explains the action of methyl orange. It is a weak base and may be, represented as MeOH. Its undissociated molecules are yellow while its ions (Me+) are red. thus,
When a base is added to MeOH. the OH– ions of alkali suppress the ionisation of MeOH so that the solution of MeOH is yellow in colour. When an acid is added to MeOH. the H+ ions furnished by acid combine with MeOH to form feebly ionised water along with the formation of MeCl. The McCl. is strongly ionised to give a large concentration of Me+ ions, thereby producing a pink-red colour. Thus, in the acid solution, methyl orange gives a pink-red colour.
Ostwald’s theory does not explain the cause of colour changes in the indicator.
How does the concept of solubility product help in finding out the solubility of sparingly soluble salts?
How does the concept of solubility product help in predicting the ionising and precipitating nature of a salt?
Knowing the solubility product of a salt, it is possible to predict that on mixing the solutions of its ions, a precipitate will be formed or not. For the precipitation to occur, its ionic product should exceed its solubility product. Therefore, to predict the precipitation reaction, we just calculate the ionic product of the ions and find out whether it is greater than Ksp or not. If
Ionic product > Ksp, precipitation occurs;
Ionic product < Ksp, no precipitation occurs.
Ionic product = Ksp, the solution is just saturated and no precipitation occurs.
|
Solubility product |
Ionic product |
|
1. It is applicable to saturated solution. |
1. It is applicable to any kind of solution. |
|
2. It is the product of the concentrations of the ions of the electrolyte each concentration raised to the power equal to the number of times each ion occurs in the equation in a saturated solution. |
2. It is the product of the concentrations of the ions of the electrolyte. Each concentration raised to the power equal to the number of times each ion occurs in the equation in solution at any concentration. |
|
3. It is constant for a given electrolyte at a given temperature. |
3. Its value is not constant and varies with change in concentration of the ions. |
Give the points of difference betw een solubility and solubility product.
|
Solubility |
Solubility product |
|
1. It is applicable to electrolytes such as NaCl as well as non-electrolytes such as urea. |
1. It can be increased or decreased in case of electrolytes by common ions even at constant temperature. |
|
2. It is applicable only to electrolytes like urea. |
2. Ksp remains constant at a particular temperature. |
The solubility of Ag2CrO4 at 298K is 6·6 × 10–5 moles/litre. Find its solubility product.
on dissociation gives 2 moles of Ag+ ions and 1 mole of 
ions. 
The solubility of CaF2 in water at 298K is 1·7×10–3 gram per 100 cm3. Calculate the solubility product of CaF2 at 298K.
Calculate the molar solubility of Ni(OH)2 in 0·10M NaOH. The ionic product of Ni(OH)2 is 2·0 × 10–15.
At a certain temperature, the solubility product of AgCl is 1 × 10–10, calculate the solubility of AgCl in g/litre.
Let the solubility of AgCl be S mole litre
Applying the law of solubility product, 
Molecular mass of AgCl
= 108 + 35.5 = 143.5
Hence solubility of AgCl
= 143.5 x 1 x 10-5 g L-1
= 14.35 x 10-4 g L-1
Calculate the solubility of lead iodide in water at 298 K. (Ksp of PbI2 = 7·1×10–9).
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#13 {main}</pre>](/application/zrc/images/qvar/CHEN11090950.png)
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#13 {main}</pre>](/application/zrc/images/qvar/CHEN11090950-2.png)
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#11 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(80): com_wiris_plugin_impl_HttpImpl->request(true)
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#13 {main}</pre>](/application/zrc/images/qvar/CHEN11090960-2.png)
Determine the solubility of silver chromate at 298K. Given that Ksp for Ag2CrO4 = 1·1 × 10–12. Also determine the molarities of each ion.
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#13 {main}</pre>](/application/zrc/images/qvar/CHEN11090972-2.png)
The solubility of Sr(OH)2 at 298K is 19·23g/L of the solution. Calculate the concentration of strontium and hydroxyl ions and pH of the solution.
in the solution gives 1 mole of 
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#13 {main}</pre>](/application/zrc/images/qvar/CHEN11090978-5.png)
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#13 {main}</pre>](/application/zrc/images/qvar/CHEN11091000.png)
How many moles of AgBr (Ksp = 5 × 10–13 mol2 L–2) will dissolve in a 0·01 M NaBr solution (NaBr is completely dissociated into Na+ and Br- ion).
can not exceed 5 x 10-13 mol2 L-2 
If 20 ml of 1·5 × 10–5 M BaCl2 solution is mixed with 40 ml of 0·9 × 10–5MNa2SO4 solution, will a precipitate get formed? Ksp for BaSO4 = 0·1 × 10–10.
Solubility product of BaSO4 = 0·1 × 10–10
Total volume of the mixture solution = 20 + 40 = 60 ml
(i) Determining Ba2+ ions concentration ; (Before mixing) M1 V1 = M2V2 (After mixing)
As sodium sulphate is fully ionised
(iii) Determining ionic product;
As the ionic product of BaSO4 (0·30 × 10–10) is more than the Ksp value of the salt, therefore, a precipitate of barium sulphate will be formed.
If 25.0 cm3 of 0.50 M - Ba(NO3)2 are mixed with 25·0 cm3 of 0·0220M-NaF, will any BaF2 precipitate [Ksp of Ba F2 is 1·7 × 10–6 at 298 K.]
Equal volumes of 0.002M solutions of sodium iodate and cupric chlorate are mixed together. Will it lead precipitation of copper iodate? (For cupric iodate Ksp = 7·4 × 10–8).
ions in the solution after mixing will be reduced to one-half.![<pre>uncaught exception: <b>Http Error #404</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/HttpImpl.class.php line 61<br />#0 [internal function]: com_wiris_plugin_impl_HttpImpl_0(Object(com_wiris_plugin_impl_HttpImpl), NULL, 'http://www.wiri...', 'Http Error #404')
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The concentration of sulphide ion in 0·1M HCl solution saturated with hydrogen sulphide is 1·0 × 10–19M. If 10 mL of this solution is added to 5 mL of 0·04M solution of FeSO4, MnCl2, ZnCl2 and CaCl2, in which solutions precipitation will take place? Given Ksp for FeS = 6·3 × 10–18, MnS = 2·5 × 10–13, ZnS =1·6 × 10–24and CdS = 8·0 × 10–27.
The values of Ksp of two sparingly soluble salts Ni(OH)2 and AgCN are 2·0 × 10–15 and 6 × 10–17 respectively. Which salt is more soluble ? Explain
For AgCN
Let the solubility of AgCN be S1 mol L–1
Applying the law of solubility product, 
Applying the law of solubility product
Since the solubility of Ni(OH) is more than that of AgCN, so Ni(OH)2 is more soluble than AgCN.
The solubility product constant of Ag2CrO4 and AgBr are 1·1 × 10–12 and 5·0 × 10–13 respectively. Calculate the ratio of the molarities of their saturated solutions.
Depending upon the ionic product, the solution of the salt may be unsaturated, saturated or super-saturated.
(i) If the ionic product < solubility product, the solution is unsaturated.
(ii) If the ionic product solubility product, the solution is saturated.
(iii) If the ionic product > solubility product, the solution is super-saturated.
Flow will you purify an impure sample of sodium chloride containing water soluble impurities?
Due to common ion effect of H+ ions, the degree of dissociation of H2S is suppressed, so the concentration of sulphide ions decreases. Thus the ionic product of sulphides of group II cations (Hg2+, Pb2+, Cu2+, Cd2+, Bi3+) decreases. But this is large enough to exceed the value of their solubility products and radicals of group II get precipitated.
Other groups cations are not precipitated because their Ksp are comparatively higher.
ions, the degree of dissociation of NH4OH is suppressed, so the concentration of hydroxide ions decreases and thereby the ionic product of the hydroxides of group III cations decreases. Even with their suppressed concentration of OH– ions, their ionic products exceed the solubility product and hence radicals of group III get precipitated. The solubility products of hydroxides of higher group cations are relatively high and thus these cations are not precipitated in group III.What is the function of ammonium hydroxide in group IV?
What is the function of NH4Cl in group V analysis?
Due to common ion effect of NH4+ ions, the degree of dissociation of (NH4)2CO3 is suppressed, so the concentration of 
ions decreases. Thus. ionic product of carbonates of group V cations decreases. Even the low concentration of 
ions is sufficient for the ionic product of these ions and cations of the fifth group to exceed the solubility product of corresponding carbonates. Hence, the carbonates of the fifth group get precipitated.
Other group cations (Na+, K+ and NH4+ ) are either soluble or their solubility product (in the case of Mg2+) is higher.
What is the necessity of adding NH4OH in group V analysis?
Give reason: The precipitation of Mg(OH)2 is prevented by the addition of NH4Cl prior to the addition of NH4OH. But its precipitation by NaOH is not prevented by the prior addition of NaCl.
Due to common ion effect, NH4ions suppress the ionisation of NH 4OH. The low concentration of OH– ions is not sufficient for the ionic product [Mg2+][OH-]2 to exceed the solubility product of Mg(OH–)2. Hence it is not precipitated.
On the other hand, NaOH is a strong base and ionises almost completely. Its dissociation is not much affected by the common ion (Na+) from NaCl. Thus the ionic product [Mg2+][OH–]2 exceeds the solubility product value of Mg(OH)2. Hence Mg(OH)2 gets precipitated.
In which of the following, silver chloride will dissolve more:
(i) Pure water
(ii) 0·1M AgNO3 solution?
For the reaction,
if Kp = Kc (RT)x where the symbol has usual meaning then the value of x is (assuming ideality)
-1
-1/2
1/2
1
B.
-1/2
For the given reaction, ∆ng = np-nR
where np = number of moles of products
nR = number of moles of reactants
Kp = Kc (RT)∆ng
∆ng = -1/2
The species which can best serve as an initiator for the cationic polymerization is
LiAlH4
HNO3
AlCl3
BaLi
C.
AlCl3
Electron-deficient species (Lewis acid) like AlCl3, BF3, etc, are is used as an initiator for cationic polymerisation.
Consider the reaction:
Cl2(aq) + H2S(aq) → S (s) + 2H+ (aq) + 2Cl- (aq)
The rate equation for this reaction is
I. Cl2 + H2S → H+ +Cl- + Cl+ +HS-
II. H2S ⇌ H+ + HS- (fast equilibrium)
Cl2 + HS- → 2Cl- + H+ + S (slow)
II only
Both (I) and (II)
Neither (I) nor (II)
(I) only
D.
(I) only
For (A)
rate = K[Cl2] [H2S]
For (B)
rate = K[Cl2] [HS-] … (i)
In aqueous solution the ionization constants for carbonic acid are
K1 = 4.2 x 10-7 and K2 = 4.8 x 10-11.
Select the correct statement for a saturated 0.034 M solution of the carbonic acid.
The concentration of CO32- is 0.034 M
The concentration of CO32- is greater than that of HCO3-
The concentration of H+ and HCO3- are approximately equal
The concentration of H+ is double that of CO32-
C.
The concentration of H+ and HCO3- are approximately equal
The correct order of increasing basicity of the given conjugate bases (R = CH3) is
A.
Stronger acid has a weaker conjugate base.
Order of acidity : RCOOH > CH≡CH > NH3 > RH
Order of basicity : 
pKa of a weak acid (HA) and pKb of a weak base (BOH) are 3.2 and 3.4, respectively. The pH of their salt (AB) solution is
7.2
6.9
7.0
1.0
B.
6.9
Given
pKa (HA) = 3.2
pKb (BOH) = 3.4
As given salt is of weak acid and weak base
therefore,
pH = 7 + (1/2)pKa - (1/2)pKb
= 7 + (1/2)(3.2) - (1/2)(3.4)
= 6.9
In the following reactions, ZnO is respectively acting as a/an :
(a) ZnO + Na2O → Na2ZnO2
(b) ZnO + CO2 → ZnCO3
base and acid
base and base
acid and acid
acid and base
D.
acid and base
ZnO is amphoteric oxide,in given reaction,
(a) ZnO + Na2O → Na2ZnO2
In this reaction ZnO act as Acid.
(b) ZnO + CO2 → ZnCO3
In this reaction ZnO act as Base.
For the following three reactions a, b and c, equilibrium constants are given
a. CO (g) + H2O (g) ⇌ CO2 (g)+ H2 (g); K1
b. CH4(g) + H2O (g) ⇌ CO (g) + 3H2 (g); K2
c. CH4(g) + 2H2O (g) ⇌ CO2 (g) + 4H2 (g); K2
Which of the following relations is correct?
K1√K2 = K3
K2K3 = K1
K3 = K1K2
K3.K32 = K12
C.
K3 = K1K2
Equation (c) = equation (a) + equation (b) Thus K3 = K1.K2
The pKa of a weak acid, HA, is 4.80. The pKb of a weak base, BOH, is 4.78. The pH of an aqueous solution of the corresponding salt, BA, will be
9.58
4.79
7.01
9.22
C.
7.01
It is a salt of weak acid and weak base
The pKa of a weak acid (HA) is 4.5. The pOH of an aqueous buffered solution of HA in which 50% of the acid is ionized is
4.5
2.5
9.5
7.0
C.
9.5
For buffer solution
as HA is 50% ionized so [Salt] = [Acid]
pH = 4.5 pH + pOH = 14
⇒ pOH = 14 – 4.5 = 9.5
Regular use of which of the following fertilizers increases the acidity of soil ?
Potassium nitrate
Urea
Superphosphate of lime
Ammonium sulphate
D.
Ammonium sulphate
(NH4)2 SO4 is a salt of the strong acid and weak base, on hydrolysis it ill produce H+ ion. This will increase the acidity of the soil.
For the reaction
When Kp and Kc are compared at 184oC, it is found that
Kp is greater than Kc
Kp is less than Kc
Kp = Kc
Whether Kp is greater than, less than or equal to Kc depends upon the total gas pressure
A.
Kp is greater than Kc
Kp = Kc RT∆n n =1
Kp > Kc
Hydrogen ion concentration in mol / L in a solution of pH = 5.4 will be
3.98 x 108
3.88 x 106
3.68 x 10-6
3.98 x 10-6
D.
3.98 x 10-6
pH = - log (H+ )
The conjugate base of H2PO4- is
PO43-
HPO42-
H3PO4
P2O5
B.
HPO42-
H3PO4 is a tribasic acid, thus ionising in three steps.
What is the equilibrium expression for the reaction P4(s) +5O2(g)⇌ P4O10(s)?
Kc = [P4O10] / P4] [O2]5
Kc = 1/[O2]5
Kc = [O2] 5
Kc = [P4O10] / 5[P4][O2]
B.
Kc = 1/[O2]5
In the expression for equilibrium constant (Kp or Kc) species state are not written (i.e., their molar concentrations are not taken as 1)
P4 (s) + 5O2 (g) ⇌ P4O10 (s)
Thus, Kc = 1/[O2]5
For the reaction, CO(g) + Cl2(g) ⇌ COCl2(g) the Kp/Kc is equal to
1/RT
1.0
√RT
RT
A.
1/RT
Kp =Kc (RT)Δn
Δn = sum of coefficients of gaseous products
Sum of coefficients of gaseous reactants.
CO (g) + Cl2 (g) → COCl2 (g)
Δn = 1-2 = -1
Kp = Kc (RT)-1
Kp/Kc = (RT)-1
1/RT
The predominant form of histamine present in human blood is (pKa, Histidine = 6.0)
A.
The N-atoms present in the ring will have the same pKa values (6.0), while N atom outside the ring will have different pKa value (pKa > 7.4)
Therefore, two N-atoms inside the ring will remain in the unprotonated form in human blood because of their pKa(6.0) < pH of blood (7.4), while the N-atom outside the ring will remain in protonated form because of its pKa > pH of blood (7.4).
Which of the following are Lewis acids?
BCl3 and AlCl3
PH3 and BCl3
AlCl3 and SiCl4
PH3 and SiCl4
A.
BCl3 and AlCl3
Both BCl3 and AlCl3 are Lewis acids as both ‘B’ and ‘Al’ has vacant p-orbitals. SiCl4 is also a Lewis acid as silicon atom has vacant 3d-orbital.
The addition of a catalyst during a chemical reaction alters which of the following quantities ?
Internal energy
Enthalpy
Activation energy
Entropy
C.
Activation energy
A catalyst is a substance which alters the reaction but itself remains unchanged in the chemical reaction. In a chemical reaction, it provides a new reaction path by the lowering the activation energy barrier.
Which one of the following characteristics is associated with adsorption?
ΔG, ΔH and ΔS all are negative
ΔG and ΔH are negative but ΔS is positive
ΔG and ΔS are negative but ΔH is positive
ΔG is negative but ΔH and ΔS are positive
A.
ΔG, ΔH and ΔS all are negative
Adsorption is a spontaneous process that occurs with the release of energy and decrease in the entropy of the substance. For a spontaneous process, ΔG must be negative,
ΔG = ΔH - T ΔS
As the process is exothermic and randomness of the molecule (entropy) decreases hence both ΔH and ΔS will be negative as well.
MY and NY3, two nearly insoluble salts. have the same Ksp values of 6.2 x 10-13 at room temperature. Which statement would be true in regard to MY and NY3 ?
The molar solubility of MY in water is less than that of NY3
The salts MY and NY3 are more soluble in 0.5 M KY than in pure water
The addition of the Salt of KY to the solution of MY and NY3 will have no effect on their solubilities.
The molar solubilities of MY and NY3 in water identical.
A.
The molar solubility of MY in water is less than that of NY3
For MY,
Consider the following liquid-vapour equilibrium
Which of the following relations is correct?
C.
The given phase equilibria is 
This equilibrium states that, when liquid is heated, it converts into vapour but on cooling, it further converts into liquid, which is derived by Clausius clapeyron and the relationship is written as,
Which of these is least likely to act as a lewis base?
CO
F-
BF3
PF3
C.
BF3
Electron rich species are called lewis base. Among the given, BF3 is an electron deficient species, so have a capacity of electrons accepting instead of donating that's why it is least likely to act as a lewis base. It is a lewis acid.
pH of a saturated solution of Ba(OH)2 is 12. The value of solubility product Ksp of Ba(OH)2 is
3.3 x 10-7
5.0 x 10-7
4.0 x 10-6
5.0 x 10-6
B.
5.0 x 10-7
Given, pH of Ba(OH)2 = 12
pOH = 14-pH
= 14-12 = 2
We know that,
pOH = -log [OH-]
2 =-log [OH-]
[OH-] = antilog (-2)
[OH-] = 1 x 10-2
Ba(OH)2dissolves in water as 
Equimolar solutions of the following substances were prepared separately, which one of these will record the highest pH value?
BaCl2
AlCl3
LiCl
BeCl2
A.
BaCl2
BaCl2 is a salt of strong acid HCl and strong base Ba(OH)2. So, its aqueous solution is neutral with pH 7. All other salts give acidic solution due to cationic hydrolysis, so their pH is less than 7. Thus, pH value is highest for the solution of BaCl2.
Buffer solutions have constant acidity and alkalinity because
these give unionised acid or base on reaction with added acid or alkali
acids and alkalies in these solutions are shielded from attack by other ions.
they have a large excess of H+ or OH- ions
they have fixed value of pH
A.
these give unionised acid or base on reaction with added acid or alkali
If a small amount of an acid or alkali is added to a buffer solution, it converts them into unionised acid or base. Thus, remains unaffected or in other words its acidity/alkalinity remains constant. e.g.,
H3O+ + A- ⇌ H2O +HA
-OH +HA → H2O +A-
If acid is added, it reacts with A- to form undissociated HA. Similarly, if base/alkali is added, O H- combines with HA to give H2O and A- and thus, maintains the acidity/alkalinity of buffer solution.
Give that the equilibrium constant for the reaction,
2SO2 (g) + O2 (g) ⇌ 2SO3 (g)
has a value of 278 at a particular temperature. What is the value of the equilibrium constant for the following reaction t the same temperature?
SO3 (g) ⇌ SO2 (g) +1/2 O2 (g)
1.8 x 10-3
3.6 x 10-3
6.0 x 10-2
1.3 x 10-5
C.
6.0 x 10-2
2SO2 (g) +O2 (g) ⇌ 2SO3 (g)
Equilibrium constant for this reaction,
The value of ΔH for the reaction
X2 (g) + 4Y2 (g) ⇌ 2XY4 (g) is less than zero. Formation of XY4 (g) will be favoured at
Low pressure and low temperature
high temperature and low pressure
high pressure and low temperature
high temperature and high pressure
C.
high pressure and low temperature
X2 (g) + 4Y2 (g) ⇌ 2XY4 (g);
For the reaction N2 (g) + O2 (g) ⇌ 2NO (g), the equilibrium constant is K1. The equilibrium constant is K2 for the reaction 2NO (g) + O2(g) ⇌ 2NO2 (g). What is K for the reaction 
1/(4K1K2)
[1/K1K2]1/2
1/(K1K2)
1/(2K1K2)
B.
[1/K1K2]1/2
N2 (g) + O2 (g) ⇌ 2NO (g) ; K1 (i)
2NO (g) + O2 (g) ⇌ 2NO2 (g) ; K2 (ii)
______________________________
N2 (g) + O2(g) ⇌ 2NO2 (g) ; K = K1 x K2
therefore, For NO2 (g) = N2/2 (g) +O2 (g);
K = [1/K1K2]1/2
What is the pH of the resulting solution when equal volumes of 0.1 M NaOH and 0.01 M HCl are mixed?
12.65
2.0
7.0
1.04
A.
12.65
When equal volumes of acid and base are mixed then resulting solution become alkaline if the concentration of base is taken high.
Let normality of the solution after mixing 0.1 M
NaOH and 0.01 M HCl is N.
therefore, N1V1 - N2V2 = NV
or 0.1 x 1 - 0.01 x 1 = N x 2
Since normality, of NaOH, is more than that of HCl. Hence, the resulting solution is alkaline.
Or
[OH-] = N = 0.09/2 = 0.045 N
Or pOH = - log (0.045) = 1.35
therefore, pH = 14-pOH
= 14 - 1.35 = 12.65
Which one of the following pairs of the solution is not an acidic buffer?
HClO4 and NaClO4
CH3COOH and CH3COONa
H2CO3 and Na2PO4
H2PO4 and Na3PO4
A.
HClO4 and NaClO4
A buffer solution is one which resists changes in pH when small quantities of an acid or an alkali are added to it.
HClO4 and NaClO4 are not an acidic buffer because strong acid with its salt cannot form buffer solution.If the equilibrium constant for 
K1/2
K/2
K
K2
A.
K1/2
As we can see the reaction for which we have to find out equilibrium constant is different only in stoichiometric coefficient as compared to the given reaction. Hence, we can find equilibrium constant as,
An aqueous solution of which of the following compounds is the best conductor of electric current?
Acetic acid C2H4O2
Hydrochloric acid, HCl
Ammonia, NH3
Fructose, C6H12O6
B.
Hydrochloric acid, HCl
HCl is strong acid and dissociates completely. Hence, it conducts electricity best in its aqueous solution.
If pH ofa saturated solution of Ba(OH)2 is 12, the value of its Ksp is
4.00 x 10-6 M3
4.00 x 10-7 M3
5.00 x 10-6 M3
5.00 x 10-7 M3
C.
5.00 x 10-6 M3
Given pH of Ba(OH)2 = 12
therefore, [H+] = [1 x 10-12]
and [OH-] = 1 x 10-14 / 1 x 10-12 {[H+][OH-] = 1 x 10-14}
What is [H+] in mol/L of a solution that is 0.20 M in CH3COONa and 0.10 M in CH3COOH?
(Ka for CH3COOH = 1.8 x 10-5 )
3.5 x 10-4
1.1 x 10-5
1.8 x 10-5
9.0 x 10-6
D.
9.0 x 10-6
CH3COOH (weaK acid) and CH3COONa (conjugated salt) form acidic buffer and for acidic buffer,
In which of the following equilibrium Kc and Kp are not equal?
2NO (g) ⇌ N2 (g) + O2 (g)
SO2 (g) + NO2 (g) ⇌ SO3 (g) + NO (g)
H2 (g) + I2 (g) ⇌ 2 HI (g)
2C (s) + O2 (g) ⇌ 2CO2 (g)
D.
2C (s) + O2 (g) ⇌ 2CO2 (g)
The reaction for which the number of moles of gaseous products (np) is not equal to the number of moles of gaseous reactants (nR), has a different value of Kc and Kp.
a) np = nR = 2 thus, Kp = Kc
b) np = nR = 2 thus, Kp = Kc
c) np = nR = 2 thus, Kp = Kc
d) np = 2, nR = 1 thus, Kp not equal to Kc
Which of the following molecules acts as a Lewis acid?
(CH)3B
(CH3)O
(CH3)P
(CH3)2)O
A.
(CH)3B
According to Lewis concept " Acids are electron acceptor and bases are electron donor" Since, electron deficient compounds also have an ability to accept electrons, these are regarded as acids.
Trimethylborane (CH)3B have incomplete octet thus, act as Lewis acid.
The ionisation constant of ammonium hydroxide is 1.77 x 10-5 at 298 K. Hydrolysis constant of ammonium chloride is
5.65 x 10-10
6.50 x 10-12
5.65 x 10-13
5.65 x 10-12
A.
5.65 x 10-10
Nitrobenzene can be prepared from benzene by using a mixture of conc. HNO3 and Conc. H2SO4.In the mixture, nitric acid acts as a/an:
reducing agent
acid
base
catalyst
C.
base
Proton donor is acids and proton acceptor is bases.
Conc. H2SO4 and conc. HNO3 react in the following manner:
HNO3 + H2SO4 → H2NO3+ +HSO4-
H2NO3+ → NO2+ +H2O
Hence, in this reaction HNO3 acts as a base and H2SO4 as an acid.
Equal volumes of three acid solutions of pH 3.4 and 5 are mixed in a vessel. What will be the H+ ion concentration in the mixture?
1.11 x 10-4 M
3.7 x 10-4 M
3.7 x 10-3 M
1.11 x 10-3 M
B.
3.7 x 10-4 M
[H]+ in mixture
[H]+ of 1st acid x its volume + [H+]
of IInd acid x its volume + [H+] of 
Assume the volume of each solution is 1 L.
[H3O+] in solution of pH = 3 is 10-3M
[H3O+] in solution of pH = 4 is 10-4 M
[H3O+] in solution is pH is 5 is 10-5 M
If the concentration of OH- ions in the reaction Fe(OH)3 (s) ⇌ Fe3+ (aq) + 3OH-(aq) is decreased by 1/4 times, then equilibrium concentration of Fe3+ will increase by
8 times
16 times
64 times
4 times
C.
64 times
The concentration of solids taken to be unity.
Fe(OH)3 (s) ⇌ Fe3+ (aq) + 3OH- (aq)
therefore,
K = [Fe3+][OH-]3
Hence, if OH- ion concentration is decreased by 1/4 times. then equilibrium concentration of Fe3+ will increase by 64 times.
Base strength of 
is in the order of
(2) > (1) > (3)
(3) > (2) > (1)
(1) > (3)> (2)
(1) > (2) > (3)
D.
(1) > (2) > (3)
Stronger the acid, weaker is its conjugate base.
The strength of their conjugate acids are in the order: 
Therefore, the correct order of strength of their conjugate base is: 
Equimolar solutions of the following were prepared in water separately which one of the solutions will record the highest pH?
SrCl2
BaCl2
MgCl2
CaCl2
B.
BaCl2
As the basic nature increases
pH of base > 7
pH of acid < 7
In alkaline earth metals on moving downward the size of cation increases, thus basicity increases. Hence, the increasing order or basicity is as:
MgCl2 < CaCl2 < SnCl2< BaCl2
Therefore, the solution of BaCl2 will record the highest pH.
The dissociation equilibrium of gas AB2 can be represented as
2AB2 (g) ⇌ 2AB (g) + B2 (g)
The degree of dissociation is 'x' and is small compared to 1. The expression relating the degree of dissociation (x) with equilibrium constant Kp and total pressure p is
(2Kp / p)
(2Kp/p)1/3
(2Kp/p)1/2
(Kp/p)
D.
(Kp/p)
Initial 1 0 0
at equ 2(1-x) 2x x
where, x = degree of dissociation
Total moles at equilibrium = 2-2x + 2x + x
= ( 2 + x)
The value of KP1 and Kp2 for the reactions
X ⇌ Y +Z ... (i)
A ⇌ 2B ... (ii)
are in the ratio of 9:1. if the degree of dissociation of X and A be equal, then the total pressure at equilibrium (i) and (ii) are in the ratio
3:1
1:9
36:1
1:1
C.
36:1
From equation.
X ⇌ Y + Z
1 0 0 initial mole
(1-α) α α mole at equilibrium
The value of equilibrium constant of the reaction HI (g) ⇌ H2(g)/2 + I2/2 is 8.0
The equilibrium constant of the reaction H2(g) + I2(g) ⇌ 2 HI (g) will be
1/16
1/64
16
1/8
B.
1/64
HI (g) ⇌ H2/2(g) + I2/2(g)
Calculate the pOH of a solution at 250C that contains 1 x 10-10 M of hydronium ions.
7
4
9
1
B.
4
[H3O+] = [H+] = 10-10
pH + pOH = [14]
pH = - log [ H+]
pH = - log [10-10]
pH = 10
⇒ pOH + 10 = 14
⇒ pOH = 14-10 = 4
The following equilibrium constants are given:
N2 + 3H2 ⇌ 2NH3; K1
N2 +O2 ⇌ 2NO; K2
H2 + 1/2O2 ⇌ H2O' K3
The equilibrium constants for the oxidation of NH3 by oxygen to give NO is:
K2K33 /K1
KK2K32 /K1
K22K3 /K1
K1K2 /K3
A.
K2K33 /K1
The required equation for the oxidation of NH3 by oxygen to give NO is:
Identify the correct statement for the change of Gibbs energy for a system (ΔGsystem) at constant temperature and pressure
If ΔGsystem > 0, the process is spontaneous
If ΔGsystem =0, the system has attained equilibrium
If ΔGsystem < 0, the system is still moving in a particular direction
If ΔGsystem < 0, the process is not spontaneous
B.
If ΔGsystem =0, the system has attained equilibrium
If the Gibbs free energy for a system (ΔGsystem) is equal to zero, then system is present in equilibrium at a constant temperature and pressure.
The enthalpy and entropy change for the reaction:
Br2 (l) + Cl2 (g)→ 2BrCl (g)
are 30 kJ mol-1 and 105 JK-1 mol-1 respectively. The temperature at which the reaction will be in equilibrium is:
285.7 K
273 K
450 K
300 K
A.
285.7 K
At equilibrium, Gibbs free energy change (ΔGo) is equal to zero. The following thermodynamic relation is used to show the relation of ΔGo with the enthalpy change (ΔHo) and entropy change (ΔSo)
ΔGo = ΔHo - ΔSo
0 = 30 x 103 (J mol-1) - T x 105 (J K-1) mol-1
For the reaction,
CH4 (g) + 2 O2 (g) ⇌ CO2 (g) + 2H2O (l),
ΔrH = - 170. 8 kJ mol-1
Which of the following statements is not true?
At equilibrium, the concentrations of CO2 (g) and H2O (l) are not equal
The equilibrium constant for the reaction is given by Kp = 
Addition of CH4 (g) or O2 (g) at equilibrium will cause a shift to the right
The reaction is exothermic
B.
The equilibrium constant for the reaction is given by Kp =
For the reaction,
CH4 (g) + 2 O2 (g) ⇌ CO2 (g) + 2H2O (l),
ΔrH = - 170. 8 kJ mol-1
This equilibrium is an example of heterogeneous chemical equilibrium.Hence, for it
not correct expression.
In addition of CH4 (g) or O2 (g) at equilibrium Kc = will be decreased according to expression (i) but Kc remains constant at constant at constant temperature fro a reaction, so for maintaining the constant value of Kc, the concentration of CO2 will increase in same order. Hence, on the addition of CH4 or O2 equilibrium will cause to the right.
This reaction is an example of an exothermic reaction.
Which of the following pairs consitutes a buffer?
HNO2 and NaNO2
NaOH and NaCl
HNO3 and NH4NO3
HCl and KCl
A.
HNO2 and NaNO2
A pair constituent with an HNO2 and NaNO2 because HNO2 is weak acid and NaNO2 is a salt of the weak acid (HNO2) with a strong base (NaOH). Hence, it is an example of acidic buffer solution.
The hydrogen ion concentration of a 10-8 M HCL aqueous solution at 298 K (Kw = 10-14) is:
1.0 x 10-6
1.0525 x 10-7 M
9.525 x 10-8 M
1.0 x 10-8 M
B.
1.0525 x 10-7 M
In aqueous solution of 10-8 M HCl, [H+] is based upon the concentration of H+ ion of 10-8 M HCl and concentration of H+ ion of water Kw of H2O = 10-14 = [H+][OH-] or [H+] = 10-7 M (due to neutral behaviour) os, in aqueous solution of 10-8 M HCl,
[H+] = [H+] of HCl +[H+] of water
= 10-8 + 10-7 = 11 x 10-8 M
1.10 x 10-7 M
Which of the following, not a correct statement?
The electron -deficient molecules can act as Lewis acids
The canonical structures have no real existence
Every AB5 molecules does, in fact, have square pyramid structure
Multiple bonds are always shorter than corresponding single bonds
C.
Every AB5 molecules does, in fact, have square pyramid structure
Every AB5 molecule does not, in fact, have square pyramidal structure but AB5 molecules have trigonal bipyramidal structures due to sp3d hybridisation.
HgCl2 and I2 both when dissolved in water containing I– ions the pair of species formed is
HgI2 , I3-
Hgl2, I-
HgI43-, I3-
Hg2I2, I-
C.
HgI43-, I3-
In a solution containing HgCl2, I2 and I–, both HgCl2 and I2
compete for I–.
Since formation constant of [HgI4]2– is 1.9 × 1030
which is very large as compared with I3– (Kf= 700)
therefore, I– will preferentially combine with HgCl2
HgCl2 + 2l- → Hgl2↓ + 2Cl-
Hgl2 + 2l- → [Hgl4]2-
When 750 mL of 0.5 M HCl is mixed with 250 mL of 2 M NaOH solution, the value of pH will be
pH = 7
pH > 7
pH < 7
pH = 0
B.
pH > 7
nHCl = MHCl x VHCl and nNaOH = MNaOH x VNaOH
where n = number of moles
M = molarity, V = volume
∴ nHCl = 0.5 x 750 = 375 milimoles
nNaOH = 2.0 x 250 = 500 milimoles
∵ nNaOH > nHCl
The solution is basic and has pH > 7
The most stable carbonium ion among the following is
C.
Due to more possibility for delocalisation in
Sponsor Area
Sponsor Area
At equilibrium the vessel contains 0.1 mole of 
0.20 mole of 
and 0.20 mole of Cl2. What is the value of equilibrium constant of the reaction?
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2NO(g) + Cl2(g); Kp = 1.8 x 10-2 at 500 K
CaO(s) + CO2(g); Kp = 167 at 1073 K
If initially pure BrCl is present at a concentration of 
what is its molar concentration in the mixture at equilibrium?
?
(Sulphate ion) is the conjugate base.
?
is the conjugate base.
can act both as Bronsted acids and bases. For each case give the corresponding conjugate acid and base.
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at the same temperature is
