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Equilibrium

Question
CBSEENCH11006587

Determine the degree of hydrolysis, hydrolysis constant and pH of 0·02N of sodium acetate. The dissociation constant of acetic acid is 1·8×10–5, Kw = 10–14.

Solution
HereKh = 10-14 
straight K subscript straight a space equals space 1.8 space cross times space 10 to the power of negative 5 end exponent comma space space space straight c space equals space 0.02 space straight N space equals space 0.02 space straight M
For salts of weak acid and strong base,
straight K subscript straight h space equals space straight K subscript straight w over straight K subscript straight a space equals fraction numerator 10 to the power of negative 14 end exponent over denominator 1.8 space cross times space 10 to the power of negative 5 end exponent end fraction space equals space 5.55 space cross times space 10 to the power of negative 10 end exponent
Degree space of space hydrolysis space left parenthesis straight alpha right parenthesis space equals square root of fraction numerator straight K subscript straight w over denominator straight K subscript straight a space cross times space straight c end fraction end root
space space space space space space space space space space space space space space space space equals square root of fraction numerator 10 to the power of negative 14 end exponent over denominator 1.8 space cross times space 10 to the power of negative 5 end exponent space cross times space 10 to the power of negative 2 end exponent end fraction end root
space space space space space space space space space space equals space 2.35 space cross times space 10 to the power of 14

pH space equals space 1 half open square brackets pK subscript straight w space plus space pK subscript straight a space plus space log space straight c close square brackets
space space space space equals space minus 1 half open square brackets log space 10 to the power of negative 14 end exponent space plus space log space left parenthesis 1.8 space cross times space 10 to the power of negative 5 end exponent right parenthesis space minus space log space 10 space minus space 2 close square brackets
space space space space space equals space minus 1 half left square bracket negative 14 minus 5 plus 0 plus 0.2553 space plus 2 right square bracket space equals space 8.37