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In a mixture of acetic acid and sodium acetate the ratio of concentrations of the salt to the acid is increased ten times. Then the pH of the solution
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increases by one
The $\mathrm{pH}$ of buffer solution is given as,
$\mathrm{pH}_{1}=\mathrm{p} K_{a}+\log \frac{\left[\mathrm{CH}_{3} \mathrm{COONa}\right]}{\left[\mathrm{CH}_{3} \mathrm{COOH}\right]}...(i)$
Given, $\left[\mathrm{CH}_{3} \mathrm{COONa}\right]_{2} /\left[\mathrm{CH}_{3} \mathrm{COOH}\right]_{2}$ $=10\left[\mathrm{CH}_{3} \mathrm{COONa}\right] /\left[\mathrm{CH}_{3} \mathrm{COOH}\right]$ $\mathrm{pH}_{2}=\mathrm{pK}_{a}+\log 10\left[\mathrm{CH}_{3} \mathrm{COONa}\right] /\left[\mathrm{CH}_{3} \mathrm{COOH}\right]$ $\mathrm{pH}_{2}=\mathrm{pK}_{a}+\log \left[\mathrm{CH}_{3} \mathrm{COONa}\right] /\left[\mathrm{CH}_{3} \mathrm{COOH}\right]+\log 10$ $\therefore \mathrm{pH}_{2}=\mathrm{pH}$ [From Eq. (i) $]$
$\mathrm{pH}_{1}=\mathrm{p} K_{a}+\log \frac{\left[\mathrm{CH}_{3} \mathrm{COONa}\right]}{\left[\mathrm{CH}_{3} \mathrm{COOH}\right]}...(i)$
Given, $\left[\mathrm{CH}_{3} \mathrm{COONa}\right]_{2} /\left[\mathrm{CH}_{3} \mathrm{COOH}\right]_{2}$ $=10\left[\mathrm{CH}_{3} \mathrm{COONa}\right] /\left[\mathrm{CH}_{3} \mathrm{COOH}\right]$ $\mathrm{pH}_{2}=\mathrm{pK}_{a}+\log 10\left[\mathrm{CH}_{3} \mathrm{COONa}\right] /\left[\mathrm{CH}_{3} \mathrm{COOH}\right]$ $\mathrm{pH}_{2}=\mathrm{pK}_{a}+\log \left[\mathrm{CH}_{3} \mathrm{COONa}\right] /\left[\mathrm{CH}_{3} \mathrm{COOH}\right]+\log 10$ $\therefore \mathrm{pH}_{2}=\mathrm{pH}$ [From Eq. (i) $]$
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