JEE MAIN - Chemistry (2022 - 28th June Evening Shift - No. 4)
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 : $${K_a}(C{H_3}C{H_2}COOH) = 1.3 \times {10^{ - 5}}$$
0.03
0.13
0.23
0.33
Explanation
$\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH} \rightleftharpoons \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COO}^{-}+\mathrm{H}^{+}$
From Henderson equation
$\mathrm{pH}=\mathrm{pK}_{\mathrm{a}}+\log \frac{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COO}^{-}\right]}{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}\right]}$
$4=-\log 1.3 \times 10^{-5}+\log \frac{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COO}^{-}\right]}{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}\right]}$
$-\log 10^{-4}=-\log 1.3 \times 10^{-5}+\log \frac{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COO}^{-}\right]}{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}\right]}$
$-\log 10^{-4}=-\log 1.3 \times 10^{-5} \frac{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}\right]}{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COO}^{-}\right]}$
$10^{-4}=1.3 \times 10^{-5} \frac{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}\right]}{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COO}^{-}\right]}$
$\frac{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COO}^{-}\right]}{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}\right]}=0.13$
From Henderson equation
$\mathrm{pH}=\mathrm{pK}_{\mathrm{a}}+\log \frac{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COO}^{-}\right]}{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}\right]}$
$4=-\log 1.3 \times 10^{-5}+\log \frac{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COO}^{-}\right]}{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}\right]}$
$-\log 10^{-4}=-\log 1.3 \times 10^{-5}+\log \frac{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COO}^{-}\right]}{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}\right]}$
$-\log 10^{-4}=-\log 1.3 \times 10^{-5} \frac{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}\right]}{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COO}^{-}\right]}$
$10^{-4}=1.3 \times 10^{-5} \frac{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}\right]}{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COO}^{-}\right]}$
$\frac{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COO}^{-}\right]}{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}\right]}=0.13$
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