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Equilibrium

Question
CBSEENCH11006549
Wired Faculty App

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

Solution
left parenthesis straight a right parenthesis space Volume space of space 0.2 space straight M thin space Ca left parenthesis OH right parenthesis subscript 2 space equals space 10 space mL space left parenthesis given right parenthesis Volume space of space 0.1 space straight M space HCl space equals space 25 space mL space space space left parenthesis given right parenthesis therefore space space space space straight M left parenthesis OH to the power of minus right parenthesis space equals space fraction numerator straight M subscript 1 straight V subscript 1 left parenthesis Base right parenthesis space minus space straight M subscript 2 straight V subscript 2 left parenthesis Acid right parenthesis over denominator straight V subscript 1 plus straight V subscript 2 end fraction space space space space space space space space space space space space space space space space space space space space equals space fraction numerator 0.2 space cross times space 2 space cross times space 10 minus space 0.1 space cross times space 2 over denominator 10 plus 25 end fraction straight M left parenthesis OH to the power of minus right parenthesis space equals space fraction numerator 1.5 over denominator 25 end fraction space equals space 0.06 space equals space 6 cross times 10 to the power of negative 2 end exponent space space space space space space space space space
But space space space space space space straight p subscript OH space equals space minus log space open square brackets OH to the power of minus close square brackets space equals space minus log space left parenthesis 6 space cross times space 10 to the power of negative 2 end exponent right parenthesis space space space space space space space space space space space space space space space space space equals negative log space 6 space minus space log space 10 to the power of negative 2 end exponent space space space space space space space space space space space space space space space space space equals space minus 0.779 space plus space 2 space space space space space space space space space space space space space space space space space equals space 1.221 space space space space space space space space space space space space space space space space space space space space space
because space space straight p subscript straight H space plus space straight p subscript OH space equals space 14 therefore space space space space space straight p subscript straight H space equals space 14 minus straight p subscript OH space space space space space space space space space space space space space equals 14 minus 1.221 space equals space 12.78

left parenthesis straight b right parenthesis space Volume space of space 0.01 space MH subscript 2 SO subscript 4 space equals space 10 space mL space left parenthesis given right parenthesis Volume space of space 0.01 space straight M space Ca left parenthesis OH right parenthesis subscript 2 space equals space 10 space mL space left parenthesis given right parenthesis
Since two solutions have equal moles of H+ and OH- ions, therefore the two solutions get completely neutralised and the pH = 7.0

left parenthesis straight c right parenthesis space Volume space of space 0.1 space MH subscript 2 SO subscript 4 space equals space 10 space mL left parenthesis given right square bracket Volume space of space 0.1 space straight M space KOH space equals space 10 space mL space left parenthesis given right square bracket straight M left parenthesis straight H to the power of plus right parenthesis space equals space fraction numerator straight M subscript 1 straight V subscript 1 left parenthesis Acid right parenthesis space minus space straight M subscript 2 straight V subscript 2 left parenthesis Base right parenthesis over denominator straight V subscript 1 space plus space straight V subscript 2 end fraction space space space space space space space space space equals space fraction numerator 0.1 space cross times space 2 space cross times space 10 space minus space 0.1 space cross times space 10 space over denominator 10 space plus space 10 end fraction equals space fraction numerator 2 minus 1 over denominator 20 end fraction space equals 1 over 20 space space space space space space space space space space space 0.05 space equals space 5 space cross times 10 to the power of negative 2 end exponent straight M
 
But space pH space equals space minus log space open square brackets straight H to the power of plus close square brackets space equals space minus log space left parenthesis 5 space cross times space 10 to the power of negative 2 end exponent right parenthesis space space space space space space space space space space equals negative log space 5 space minus space log space 10 to the power of negative 2 end exponent space space space space space space space space space space equals negative 0.6990 space plus 2 space equals space 1.30

Some More Questions From Equilibrium Chapter

What is the role of forward and backward reactions at equilibrium of a reversible reaction?

What is a forward reaction?

What condition favours a reversible reaction?

What is a backward reaction?

What type of equilibrium is attained in a reversibe chemical change?

What is meant by a system at dynamic equilibrium?

discuss the effect of a catalyst on equilibrium ? 

At equilibrium, the mass of each of the reactants and products remains constant. Does it mean that the reaction has stopped? Explain.

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