Question

$20 \,mL$ of $0.1 M$ acetic acid is mixed with $50 \,mL$ of potassium acetate. $K_{a}$ of acetic acid $=1.8 \times 10^{-5}$ at $27^{\circ} C .$ Calculate concentration of potassium acetate if $pH$ of the mixture is $4.8$.

• A. $0.1\, M$
• B. $0.04 \, M$
• C. $0.4\, M$
• D. $0.02\, M$

Answer 1

Answer: (B)

Let the concentration of potassium acetate is $x$.
From Henderson's equation,
$pH = p K_{a}+\log \frac{[ salt ]}{[\text { acid }]}$
$4.8=-\log \left(1.8 \times 10^{-5}\right)+\log \frac{x \times 50}{20 \times 0.1 M } $
$4.8=4.74+\log 25 x $
or $\log 25 x=0.06$
$25\,x = 1.148$
$\therefore x =0.045 \,M $