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Equilibrium

Question
CBSEENCH11006595

Calculate the pH value of a solution obtained by mixing 0·083 moles of acetic acid and 0·091 moles of sodium acetate and making the volume 500 ml. Ka for acetic acid is 1·75 × 10–5.

Solution
For acidic buffer, Henderson equation is
pH space equals space pK subscript straight a space plus space log space fraction numerator open square brackets Salt close square brackets over denominator open square brackets Acid close square brackets end fraction space space... left parenthesis 1 right parenthesis
Here space open square brackets Acid close square brackets space equals space open square brackets CH subscript 3 COOH close square brackets
space space space space space space space space space space space space space space space equals space fraction numerator 0.083 over denominator 500 end fraction space cross times space 1000 space equals space 0.166 space mol space straight L to the power of negative 1 end exponent
space space space space space space space space space space space open square brackets Salt close square brackets space equals space open square brackets CH subscript 3 COONa close square brackets
space space space space space space space space space space space space space space space space space space space space equals fraction numerator 0.091 over denominator 500 end fraction cross times 1000 space equals space 0.182 space mol space straight L to the power of negative 1 end exponent
space space space space space space space space space pK subscript straight a space equals space minus log space straight K subscript straight a
space space space space space space space space space space space space space space space space equals negative log space open square brackets 1.75 space space cross times 10 to the power of negative 5 end exponent close square brackets space equals space 4.757 space space space space space space space space space space space space space space space space space space space space space
Substituting the values in equation (1), we get, 
pH space equals space 4.757 space plus space log space fraction numerator 0.182 over denominator 0.166 end fraction
space space space space space equals space 4.757 space plus space 0.040 space equals space 4.797