A student needs to prepare a buffer solution of propanoic acid and its sodium salt with pH 4 . The ratio of $\frac{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COO}^{-}\right]}{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}\right]}$ required to make buffer is \_\_\_\_ Given : $\mathrm{K}_{\mathrm{a}}\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}\right)=1.3 \times 10^{-5}$

JEE Main · 2022 · 28 Jun 2022 · Shift 2 · Equilibrium

A student needs to prepare a buffer solution of propanoic acid and its sodium salt with pH 4 . The ratio of

\frac{[ CH _{3} CH _{2} COO ^{-} ]}{[ CH _{3} CH _{2} COOH ]}

required to make buffer is ____

Given :

K_{a} (CH_{3} CH_{2} COOH) = 1.3 \times 1 0^{- 5}

Answer: (B)

Use the Henderson-Hasselbalch equation:

pH = pK_{a} + lo g \frac{[ Salt ]}{[ Acid ]}

. Given

K_{a} = 1.3 \times 1 0^{- 5}

, so

pK_{a} = 5 - lo g 1.3

. Substituting pH = 4:

4 = 5 - lo g 1.3 + lo g \frac{[ CH _{3} CH _{2} COO ^{-} ]}{[ CH _{3} CH _{2} COOH ]}

. Rearranging:

lo g \frac{[ CH _{3} CH _{2} COO ^{-} ]}{[ CH _{3} CH _{2} COOH ]} = lo g 1.3 - 1 = lo g \frac{1.3}{10}

. Therefore,

\frac{[ CH _{3} CH _{2} COO ^{-} ]}{[ CH _{3} CH _{2} COOH ]} = 0.13

JEE Main · 2022 · 28 Jun 2022 · Shift 2 · Single correct option