A buffer is 0.25 M in ${CH}_{3} COOH$ and $0.56 M$ in ${CH}_{3} COONa$ . What is the value of $p H i f 0.006 m o l o f H C l$ is added to 0.300 L of a buffer solution? $(p k_{w} = 14)$ $[p K_{b} (C H_{3} C O O^{-}) = 9.26, \log 2 = 0.3010]$

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answer is 5.03.

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Detailed Solution

$\begin{array}{l} C H_{3} C O O H = 0.25 M = 0.25 m o l L^{- 1} \\ = 0.25 \times 0.3 m o l i n 0.3 L = 0.075 m o l \\ C H_{3} C O O N a = 0.56 M = 0.56 m o l L^{- 1} \\ = 0.56 \times 0.3 m o l i n 0.3 L = 0.168 m o l \\ H C I = 0.006 m o l \end{array}$
$\begin{array}{l} p H = p K_{a} + \log \frac{[C H_{3} C O O N a]}{[C H_{3} C O O H]} \\ = 4.74 + \log (\frac{0.162}{0.081}) \\ = 4.74 + \log 2 \\ = 4.74 + 0.3 = 5.04 \end{array}$