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$20 \mathrm{ml}$ of $0.1 \mathrm{M}$ acetic acid is mixed with $50 \mathrm{~mL}$ of potassium acetate. $\mathrm{K}_{\mathrm{a}}$ of acetic acid $=1.8 \times 10^{-5}$ at $27^{\circ} \mathrm{C} .$ Calculate concentration of potassium acetate if $\mathrm{pH}$ of the mixture is $4.8$.
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Verified Answer
The correct answer is:
$0.1 \mathrm{M}$
Let the concentration of potassium acetate is $x$. According to Henderson's equation,
$$
\begin{array}{l}
\mathrm{pH}=\mathrm{pK}_{\mathrm{a}}+\log \frac{[\mathrm{salt}]}{[\mathrm{acid}]} \\
4.8=-\log \left(1.8 \times 10^{-5}\right)+\log \frac{x \times 50}{20 \times 0.1 \mathrm{M}} \\
4.8=4.74+\log 25 x \\
\text { or } \log 25 x=0.06
\end{array}
$$
$$
\begin{array}{l}
25 x=1.148 \\
\therefore x=0.045 \mathrm{M}
\end{array}
$$
$$
\begin{array}{l}
\mathrm{pH}=\mathrm{pK}_{\mathrm{a}}+\log \frac{[\mathrm{salt}]}{[\mathrm{acid}]} \\
4.8=-\log \left(1.8 \times 10^{-5}\right)+\log \frac{x \times 50}{20 \times 0.1 \mathrm{M}} \\
4.8=4.74+\log 25 x \\
\text { or } \log 25 x=0.06
\end{array}
$$
$$
\begin{array}{l}
25 x=1.148 \\
\therefore x=0.045 \mathrm{M}
\end{array}
$$
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