Question

Calculate $pH$ change when $0.01 \,mol \,CH _{3} KCOO\, Na$ solution is added to $1\, L$ of $0.01 \,M \,CH _{3} COOH$ solution.
$K_{a}\left( CH _{3} COOH \right)=1.8 \times 10^{-5}, pK _{ a }=4.74$

• A. 3.37
• B. 1.37
• C. 4.74
• D. 8.01

Answer 1

Answer: (B)

$pH$ of $0.01 \,MCH _{3} COOH$
$pH =\frac{1}{2}\left[ p K_{ a }-\log C\right]$
$=\frac{1}{2}(4.74-\log 0.01)$
$=\frac{1}{2}(4.74+2)$
$=3.37$
When $0.01 \,mol \,CH _{3} COONa$ is added to it,
it is now a buffer and $\left[\left[ CH _{3} COONa \right]=0.01\right] M$.
Now from
$pH = p K_{ a }+\log \frac{\left[ CH _{3} COO \right]}{\left[ CH _{3} COOH \right]}$
$=4.74+\log \frac{0.01}{0.01}=4.74$
$\therefore $ Change in $pH =4.74-3.37=1.37$