A student needs to prepare a buffer solution of propanoic acid and its sodium salt with pH 4.The ratio of CH3CH2COO−CH3CH2COOH required to make buffer is _____________.Given : KaCH3CH2COOH=1.3×10−5

pH=pKa+log⁡[ Salt ][ Acid ]4=5−log⁡1.3+log⁡CH3CH2COO−CH3CH2COOHlog⁡CH3CH2COO−CH3CH2COOH−=log⁡1.3−1=log⁡1.310CH3CH2COO−CH3CH2COOH−=0.13The problem can be solved by Henderson equation.

A student needs to prepare a buffer solution of propanoic acid and its sodium salt with pH 4.The ratio of CH3CH2COO−CH3CH2COOH required to make buffer is _____________.Given : KaCH3CH2COOH=1.3×10−5

pH=pKa+log⁡[ Salt ][ Acid ]4=5−log⁡1.3+log⁡CH3CH2COO−CH3CH2COOHlog⁡CH3CH2COO−CH3CH2COOH−=log⁡1.3−1=log⁡1.310CH3CH2COO−CH3CH2COOH−=0.13The problem can be solved by Henderson equation.

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A student needs to prepare a buffer solution of propanoic acid and its sodium salt with pH 4.
The ratio of $\frac{[{CH}_{3} {CH}_{2} {COO}^{-}]}{[{CH}_{3} {CH}_{2} COOH]}$ required to make buffer is _____________.
Given : $K_{a} ({CH}_{3} {CH}_{2} COOH) = 1 .3 \times 10^{- 5}$

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0.03

0.13

0.23

0.33

answer is B.

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Detailed Solution

$\begin{array}{l} pH = {pK}_{a} + \log \frac{[Salt]}{[Acid]} \\ 4 = 5 - \log 1 .3 + \log \frac{[{CH}_{3} {CH}_{2} {COO}^{-}]}{[{CH}_{3} {CH}_{2} COOH]} \\ \log \frac{[{CH}_{3} {CH}_{2} {COO}^{-}]}{[{CH}_{3} {CH}_{2} {COOH}^{-}]} = \log 1 .3 - 1 = \log \frac{1 .3}{10} \\ \frac{[{CH}_{3} {CH}_{2} {COO}^{-}]}{[{CH}_{3} {CH}_{2} {COOH}^{-}]} = 0 .13 \end{array}$