Question

The pH of $0.005MHCOOH$ $[K_a=2\times {10}^{-4}]$ is equal to:

• A. 3
• B. 2
• C. 4
• D. 5

Answer 1

The correct option is A 3

Let $x$ be the degree of dissociation of $HCOOH$.

$HCOOH\to HCO{O}^{-}+H^{+}$

$c$ $-$ $-$

$c(1-x)$ $x$ $x$

$K=c{x}^2$, for weak acid if $x<<1$

So, $\left[H^{+}\right]=C\times x=\sqrt{K_a\times C}=\sqrt{2\times {10}^{-4}\times 0.005}={10}^{-3}$

So, $pH=-log\left[H^{+}\right]=3$