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The ratio of volumes of $\mathrm{CH}_{3} \mathrm{COOH } \text{ 0.1}$ (N) to CH COONa 0.1 (N) required to prepare a buffer solution of $\mathrm{pH} 5.74$ is (Given, $p K_{a}$ of $C H_{3} C O O H$ is $\left.4.74\right)$
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The correct answer is:
$1 : 10$
Given, that $\mathrm{pH}=5.74, \mathrm{pK}_{a}=4.74$
Suppose that volume of acid solution $=x \mathrm{L}$ Volume of salt solution $=y \mathrm{L}$
From Henderson equation,
$$
\mathrm{pH}=\mathrm{p} K_{\mathrm{a}}+\log \frac{[\mathrm{Salt}]}{[\mathrm{Acid}]}
$$
or, $\mathrm{pH}-\mathrm{p} K_{\mathrm{a}}=\log \frac{\left[\mathrm{CH}_{3} \mathrm{COONa}\right]}{\left[\mathrm{CH}_{3} \mathrm{COOH}\right]}$
or, $5.74-4.74=1=\log \frac{\left[\mathrm{CH}_{3} \mathrm{COONa}\right]}{\left[\mathrm{CH}_{3} \mathrm{COOH}\right]}$
or $\frac{\left.\mathrm{CH}_{3} \mathrm{COONa}\right]}{\left[\mathrm{CH}_{3} \mathrm{COOH}\right]}=10$
or $\frac{\left[\mathrm{CH}_{3} \mathrm{COOH}\right]}{\left[\mathrm{CH}_{3} \mathrm{COONa}\right]}=\frac{1}{10}=\frac{\frac{x+y}{0.1 \mathrm{y}}}{\mathrm{x}+\mathrm{y}}$
Thus, $\frac{x}{y}=\frac{1}{10}$
Suppose that volume of acid solution $=x \mathrm{L}$ Volume of salt solution $=y \mathrm{L}$
From Henderson equation,
$$
\mathrm{pH}=\mathrm{p} K_{\mathrm{a}}+\log \frac{[\mathrm{Salt}]}{[\mathrm{Acid}]}
$$
or, $\mathrm{pH}-\mathrm{p} K_{\mathrm{a}}=\log \frac{\left[\mathrm{CH}_{3} \mathrm{COONa}\right]}{\left[\mathrm{CH}_{3} \mathrm{COOH}\right]}$
or, $5.74-4.74=1=\log \frac{\left[\mathrm{CH}_{3} \mathrm{COONa}\right]}{\left[\mathrm{CH}_{3} \mathrm{COOH}\right]}$
or $\frac{\left.\mathrm{CH}_{3} \mathrm{COONa}\right]}{\left[\mathrm{CH}_{3} \mathrm{COOH}\right]}=10$
or $\frac{\left[\mathrm{CH}_{3} \mathrm{COOH}\right]}{\left[\mathrm{CH}_{3} \mathrm{COONa}\right]}=\frac{1}{10}=\frac{\frac{x+y}{0.1 \mathrm{y}}}{\mathrm{x}+\mathrm{y}}$
Thus, $\frac{x}{y}=\frac{1}{10}$
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