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Calculate the molar ratio of a weak acid \(\mathrm{H} A\left(K_a=10^{-6}\right)\) and its salt with strong base, so that the \(\mathrm{pH}\) of buffer solution is 6 .
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Verified Answer
The correct answer is:
0.1
\(\begin{aligned}
& \mathrm{pH}=\mathrm{pK}+\log \frac{[\text { Salt }]}{[\text { Acid }]} \log \frac{[\text { Salt }]}{[\text { Acid }]} \\
& =\frac{\mathrm{pH}}{\mathrm{p} K_a}=\frac{6}{6}=1 \\
& {\left[K_a=10^{-6} \therefore \mathrm{p} K_a=-\log K_a=\log \left(10^{-6}\right) \Rightarrow 6\right]} \\
& \log \frac{[\text { Salt }]}{[\text { Acid }]}=1 \\
& \frac{[\text { Acid }]}{[\text { Salt }]}=10^{-1} \text { or } 0.1
\end{aligned}\)
Hence, the correct option is (d).
& \mathrm{pH}=\mathrm{pK}+\log \frac{[\text { Salt }]}{[\text { Acid }]} \log \frac{[\text { Salt }]}{[\text { Acid }]} \\
& =\frac{\mathrm{pH}}{\mathrm{p} K_a}=\frac{6}{6}=1 \\
& {\left[K_a=10^{-6} \therefore \mathrm{p} K_a=-\log K_a=\log \left(10^{-6}\right) \Rightarrow 6\right]} \\
& \log \frac{[\text { Salt }]}{[\text { Acid }]}=1 \\
& \frac{[\text { Acid }]}{[\text { Salt }]}=10^{-1} \text { or } 0.1
\end{aligned}\)
Hence, the correct option is (d).
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