Question

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

• A. $ 3.5 \times 10^{-4} $
• B. $ 1.1 \times 10^{-5} $
• C. $ 1.8 \times 10^{-5} $
• D. $ 9.0 \times 10^{-6} $

Answer 1

Answer: (D)

$CH _{3} COOH$ (weak acid) and $CH _{3} COONa$ (conjugated salt) form acidic buffer and for acidic buffer
$p H=p K_{a}+\log \frac{[\text { salt }]}{[\text { acid }]} $
and $\left[H^{+}\right]=-\text {antilog } p H$
$p H=-\log K_{a}+\log \frac{[\text { salt }]}{[\text { acid }]} $
${\left[\therefore p K_{a}=-\log K_{a}\right]}$
$=-\log \left(1.8 \times 10^{-5}\right)+\log \frac{0.20}{0.10}$
$=4.74+\log 2 $
$=4.74+0.3010=5.041 $
Now, $\left[H^{+}\right]=$ antilog $(-5.045) $
$=9.0 \times 10^{-6} \text { mol } / L$