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The $p K_a$ of a weak acid is 4.8 . What should be the ratio of $\frac{\text { [acid] }}{\text { [salt }]}$, if a buffer of $\mathrm{pH}=5.8$ is required?
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$0.1$
$\begin{aligned} & \mathrm{pH}=\mathrm{p} K_{\mathrm{a}}+\log \frac{[\text { salt }]}{[\text { acid }]} \\ & 5.8=4.8+\log \frac{\text { [salt }]}{\text { [acid] }} \\ & \therefore \quad \log \frac{\text { [salt }]}{\text { [acid] }}=5.8-4.8=1.0 \\ & \frac{\text { [ salt }]}{\text { [acid }]}=\text { antilog of }[1.0]=\frac{10}{1} \\ & \frac{[\text { acid }]}{[\text { salt }]}=\frac{1}{10}=0.1 \\ & \end{aligned}$
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