Question

Calculate the pH of a solution obtained by mixing 6·0g of acetic acid and 12·30g of sodium acetate and making the volume of solution to 500 ml. K_a for acetic acid is 1·8 × 10^–5

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Solution

Molecular space mass space of space acetic space acid comma space space left parenthesis CH subscript 3 COOH right parenthesis space equals 60 Molecular space mass space of space sodium space acetate comma space left parenthesis CH subscript 3 COONa right parenthesis space equals space 82 open square brackets CH subscript 3 COOH close square brackets space equals fraction numerator 6.0 over denominator 60 end fraction cross times 1000 over 50 space equals 0.2 space straight M open square brackets CH subscript 3 COONa close square brackets space equals space fraction numerator 12.30 over denominator 82 end fraction cross times 1000 over 500 space equals 0.3 space straight M

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 space... left parenthesis 1 right parenthesis Now 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 equals negative log space left parenthesis 1.8 space cross times space 10 to the power of negative 5 end exponent right parenthesis space space space space space space space space space space space space space space equals negative left square bracket log space 1.8 space plus space log space 10 to the power of negative 5 end exponent right square bracket space space space space space space space space space space space space space space equals negative open square brackets log space 1.8 space minus 5 close square brackets space space space space space space space space space space space space space space equals plus 5 space minus space log space 1.8 therefore space space space space pK subscript straight a space space equals space 4.7417 space space space space space space space space space

Substituting the values in equation (1), we get,

pH space equals space 4.7447 space plus space log space fraction numerator open square brackets 0.3 close square brackets over denominator open square brackets 0.2 close square brackets end fraction space equals space 4.9248

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