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$\mathrm{pH}$ of a buffer solution decreases by 0.02 units when $0.12 \mathrm{~g}$ of acetic acid is added to $250 \mathrm{~mL}$ of a buffer solution of acetic acid and potassium acetate at $27^{\circ} \mathrm{C}$. The buffer capacity of the solution is
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The correct answer is:
0.4
Buffer capacity, $\beta=\frac{d C_{\mathrm{HA}}}{d_{\mathrm{pH}}}$,
where, $d C_{\mathrm{HA}}=$ no. of moles of acid added per litre
$\begin{aligned}
d_{\mathrm{pH}} & =\text { change in } \mathrm{pH} \\
d C_{\mathrm{HA}} & =\frac{\text { moles of acetic acid }}{\text { volume }} \\
& =\frac{0.12 / 60}{250 / 1000}=\frac{1}{125} \\
\therefore \quad \beta & =\frac{1 / 125}{0.02}=\frac{1}{2.5}=0.4
\end{aligned}$
where, $d C_{\mathrm{HA}}=$ no. of moles of acid added per litre
$\begin{aligned}
d_{\mathrm{pH}} & =\text { change in } \mathrm{pH} \\
d C_{\mathrm{HA}} & =\frac{\text { moles of acetic acid }}{\text { volume }} \\
& =\frac{0.12 / 60}{250 / 1000}=\frac{1}{125} \\
\therefore \quad \beta & =\frac{1 / 125}{0.02}=\frac{1}{2.5}=0.4
\end{aligned}$
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