Question

$pH$ of $0.0002M$ formic acid $[K_a=2\times 10^{-4}]$ approximately is

• A. 1.35
• B. 0.5
• C. 3.7
• D. 1.85

Answer 1

Answer: (C)

${\left[H^{+}\right]}_{HCOOH}=\sqrt{K_a.C}$
$=\sqrt{2\times 10^{-4}\times 0.0002}$
$pH=-\log \left[H^{+}\right]=-\log \left(2\times 10^{-4}\right)$
$=3.69\approx 3.7$