Question

$1.0\, M$ acetic acid is mixed with $0.01\, mol\, L^{-1}\, HCl$. The $pH$ of the resultant solution is
$[K_a(CH_3COOH) = 1.8 × 10^{-5}]$

• A. $1.846$
• B. $1.04$
• C. $2.36$
• D. $2.79$

Answer 1

Answer: (A)

For weak acid, $CH_{3}COOH, \left[H^{+}\right]=\sqrt{K_{a}C}$
$\sqrt{1.8\times10^{-5}\times1}$
$\left[H^{+}\right]=4.242\times10^{-3}$
Total $\left[H^{+}\right]=0.01 + 4.242 × 10^{-3} = 1.424 × 10^{-2}$
$pH=-log\,\left[H^{+}\right]$
$pH = -log\left(1.424 × 10^{-2}\right)$
$\therefore pH=1.846$