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Equilibrium

Question
CBSEENCH11006568
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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? 

Solution
Applying the Ostwald dilution law,
straight alpha space equals square root of straight K subscript straight a over straight C end root space equals square root of fraction numerator 1.32 space cross times space 10 to the power of negative 5 end exponent over denominator 0.05 end fraction end root space equals space 1.62 space cross times space 10 to the power of negative 2 end exponent Also space space open square brackets straight H to the power of plus close square brackets space equals space square root of straight K subscript straight a space cross times space straight C end root space equals space square root of 1.32 space cross times space 10 to the power of negative 5 end exponent space cross times space 0.05 end root space space space space space space space space equals space 8.124 space cross times space 10 to the power of negative 4 end exponent therefore space space 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 left parenthesis 8.124 space cross times space 10 to the power of negative 4 end exponent right parenthesis space space space space space space space space space space space equals negative left parenthesis log space 8.124 space minus space 4 space log space 10 right parenthesis space space space space space space space space space space space equals left parenthesis 4 minus log space 8.124 right parenthesis space equals space left parenthesis 4 minus 0.909 right parenthesis space equals space 3.09
Calculation of a for propanoic acid in (0·01M )HCl solution:
space CH subscript 3 CH subscript 2 COOH space space rightwards harpoon over leftwards harpoon space space CH subscript 3 CH subscript 2 COO to the power of minus space plus space straight H to the power of plus
In the presence of HCl, the ionisation of CH3CH3COOH will decrease (common ion effect). If C is the initial concentration of acid and x is the amount dissociated at equilibrium, then
open square brackets CH subscript 3 CH subscript 2 COOH close square brackets space equals space straight C minus straight x semicolon space space space open square brackets CH subscript 3 CH subscript 2 COO to the power of minus close square brackets space equals straight x semicolon space space space open square brackets straight H to the power of plus close square brackets space equals space 0.01 space plus space straight x therefore space space space straight K subscript straight a space equals space fraction numerator open square brackets CH subscript 3 CH subscript 2 COO to the power of minus close square brackets space open square brackets straight H to the power of plus close square brackets over denominator open square brackets CH subscript 3 CH subscript 2 COOH close square brackets end fraction space space space space space space space space space space space space equals fraction numerator left parenthesis straight x right parenthesis space cross times space left parenthesis 0.01 space plus space straight x right parenthesis over denominator straight C minus straight x end fraction space equals space fraction numerator straight x left parenthesis 0.01 right parenthesis over denominator straight C end fraction space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space open square brackets because space straight x space is space very minus very space small close square brackets space space space space space space space or space space straight x over straight C space equals space fraction numerator straight K subscript straight a over denominator 0.01 end fraction space equals space fraction numerator 1.32 space cross times space 10 to the power of negative 3 end exponent over denominator 10 to the power of negative 2 end exponent end fraction or space space space space space straight alpha space equals space straight x over straight C space equals space 1.32 space cross times space 10 to the power of negative 3 end exponent

Some More Questions From Equilibrium Chapter

A crystal of common salt of a given mass is kept in its aqueous solution. After 24 hours, its mass remains the same. Is the crystal in equilibrium with the solution?

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 ? 

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