Question

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

• A. 3
• B. 2
• C. 4
• D. 5

Answer 1

The correct option is A 3

Since C is low and K is high, we will use the following formula:
$K_a=\frac{C{\alpha }^2}{1-\alpha }$

$\Rightarrow 2\times {10}^{-4}=\frac{\left(0.005\right){\alpha }^2}{1-\alpha }$

$\Rightarrow \alpha =0.18$

$\Rightarrow pH=-logC\alpha$

$\Rightarrow -log(0.05\times 0.18)\approx 3$

Hence, option $A$ is correct.