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Assuming that the degree of hydrolysis is small, the $\mathrm{pH}$ of $0.1 \mathrm{M}$ solution of sodium acetate $\left(\mathrm{K}_{\mathrm{a}}=1.0 \times 10^{-5}\right)$ will be:
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The correct answer is:
$9.0$
$9.0$
Sodium acetate is a salt of strong base and weak acid.
$$
\begin{aligned}
&\therefore \mathrm{pH}=7+\frac{1}{2} \mathrm{p} K_a+\frac{1}{2} \log c \quad \text { where } \\
&\mathrm{p} k_a=-\log K_a=-\log 10^{-5}=5 \\
&\log \mathrm{c}=\log 10^{-1}=-1 \\
&\mathrm{pH}=7+\frac{5}{2}-\frac{1}{2}=9.0
\end{aligned}
$$
$$
\begin{aligned}
&\therefore \mathrm{pH}=7+\frac{1}{2} \mathrm{p} K_a+\frac{1}{2} \log c \quad \text { where } \\
&\mathrm{p} k_a=-\log K_a=-\log 10^{-5}=5 \\
&\log \mathrm{c}=\log 10^{-1}=-1 \\
&\mathrm{pH}=7+\frac{5}{2}-\frac{1}{2}=9.0
\end{aligned}
$$
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