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 $x$.
According to Henderson's equation,
$pH = pK _{ a }+\log \frac{[\text{ salk} ]}{[\text{acid}]}$
$4.8=-\log \left(1.8 \times 10^{-5}\right)+\log \frac{x \times 50}{20 \times 0.1 M}$
$4.8=4.74+\log 25 x$
or $\log 25 x=0.06$
$25 x=1.148$
$\therefore x=0.045\, M$