A student needs to prepare a buffer solution of propanoic acid and its sodium salt with pH 4. The ratio of ${{[C{H_3}C{H_2}CO{O^ - }]} \over {[C{H_3}C{H_2}COOH]}}$ required to make buffer is ___________.
Given : ${K_a}(C{H_3}C{H_2}COOH) = 1.3 \times {10^{ - 5}}$
Solution
$\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH} \rightleftharpoons \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COO}^{-}+\mathrm{H}^{+}$
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From Henderson equation
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$\mathrm{pH}=\mathrm{pK}_{\mathrm{a}}+\log \frac{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COO}^{-}\right]}{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}\right]}$
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$4=-\log 1.3 \times 10^{-5}+\log \frac{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COO}^{-}\right]}{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}\right]}$
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$-\log 10^{-4}=-\log 1.3 \times 10^{-5}+\log \frac{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COO}^{-}\right]}{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}\right]}$
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$-\log 10^{-4}=-\log 1.3 \times 10^{-5} \frac{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}\right]}{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COO}^{-}\right]}$
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$10^{-4}=1.3 \times 10^{-5} \frac{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}\right]}{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COO}^{-}\right]}$
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$\frac{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COO}^{-}\right]}{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}\right]}=0.13$
About this question
Subject: Chemistry · Chapter: Equilibrium · Topic: Chemical Equilibrium and Kc, Kp
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