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A buffer solution is prepared by mixing $10 \mathrm{~mL}$ of $1.0 \mathrm{M}$ acetic acid and $20 \mathrm{~mL}$ of $0.5 \mathrm{M}$ sodium acetate and then diluted to $100 \mathrm{~mL}$ with distilled water, The $\mathrm{pH}$ of the buffer solution is ( $\mathrm{p} K_a$ of acetic acid is 4.76 )
Options:
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Verified Answer
The correct answer is:
4.76
Milliequivalents of acetic acid $=10 \mathrm{~mL} \times 1.0 \mathrm{M}$
Milliequivalents of sodium acetate $=2.0 \mathrm{~mL} \times 0.5 \mathrm{M}$ From Henderson equation,
$$
\mathrm{pH}=\mathrm{p} K_a+\log \left[\frac{\text { salt }}{\text { acid }}\right]
$$
Here, $\quad \mathrm{p} K_a=4.76$
[For acetic acid]
$$
\begin{aligned}
\because \quad \mathrm{pH} & =4.76+\log \left(\frac{0.5 \times 20}{10 \times 1.0}\right) \\
\mathrm{pH} & =4.76
\end{aligned}
$$
Hence, option (d) is correct.
Milliequivalents of sodium acetate $=2.0 \mathrm{~mL} \times 0.5 \mathrm{M}$ From Henderson equation,
$$
\mathrm{pH}=\mathrm{p} K_a+\log \left[\frac{\text { salt }}{\text { acid }}\right]
$$
Here, $\quad \mathrm{p} K_a=4.76$
[For acetic acid]
$$
\begin{aligned}
\because \quad \mathrm{pH} & =4.76+\log \left(\frac{0.5 \times 20}{10 \times 1.0}\right) \\
\mathrm{pH} & =4.76
\end{aligned}
$$
Hence, option (d) is correct.
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