Question

What is $\left[H^{+}\right]$in mol/L of a solution that is $0.20M$ in $C{H}_{3}COONa$ and $0.10M$ in $C{H}_{3}COOH$ ?
$\left(K_a\text{ for }C{H}_{3}COOH=1.8×10^{−5}\right)$

• A. $3.5×10^{−4}$
• B. $1.1×10^{−5}$
• C. $1.8×10^{−5}$
• D. $9.0×10^{−6}$

Answer 1

Answer: (D)

$C{H}_{3}COOH$ (weak acid) and $C{H}_{3}COONa$ (conjugated salt) form acidic buffer and for acidic buffer
$pH=p{K}_a+\log \frac{[\text{ salt }]}{[\text{ acid }]}$
and $\left[H^{+}\right]=−\text{antilog }pH$
$pH=−\log K_a+\log \frac{[\text{ salt }]}{[\text{ acid }]}$
$\left[∴p{K}_a=−\log K_a\right]$
$=−\log \left(1.8×10^{−5}\right)+\log \frac{0.20}{0.10}$
$=4.74+\log 2$
$=4.74+0.3010=5.041$
Now, $\left[H^{+}\right]=$ antilog $(−5.045)$
$=9.0×10^{−6}\text{ mol }/L$