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$20 \mathrm{~mL}$ of $0.1 \mathrm{M}$ acetic acid is mixed with $50 \mathrm{~mL}$ of potassium acetate. $K_a$ 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:
0.04 M
Let the concentration of potassium acetate is $x$. From Henderson's equation,
$\begin{aligned}
& \mathrm{pH}=\mathrm{p} K_a+\log \frac{[\text { salt }]}{[\text { acid }]} \\
& 4.8=-\log \left(1.8 \times 10^{-5}\right)+\log \frac{x \times 50}{20 \times 0.1 \mathrm{M}} \\
& 4.8=4.74+\log 25 x
\end{aligned}$
$\begin{array}{rlrl}
\text { or } & \log 25 x=0.06 \\
\therefore 25 x & =1.148 \\
& x=0.045 \mathrm{M}
\end{array}$
$\begin{aligned}
& \mathrm{pH}=\mathrm{p} K_a+\log \frac{[\text { salt }]}{[\text { acid }]} \\
& 4.8=-\log \left(1.8 \times 10^{-5}\right)+\log \frac{x \times 50}{20 \times 0.1 \mathrm{M}} \\
& 4.8=4.74+\log 25 x
\end{aligned}$
$\begin{array}{rlrl}
\text { or } & \log 25 x=0.06 \\
\therefore 25 x & =1.148 \\
& x=0.045 \mathrm{M}
\end{array}$
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