Question

If 50 ml of 0.2 M $KOH$ is added to 40 ml of 0.5 M $HCOOH$. The pH of the resulting solution is:
Use : $[K_a=1.8\times {10}^{-4},log18=1.26]$

• A. 3.74
• B. 5.64
• C. 7.57
• D. 3.42

Answer 1

The correct option is A 3.74

$n_{KOH}=\frac{50\times 0.2}{1000}=0.01mol$

$n_{HCOOH}=\frac{40\times 0.5}{1000}=0.02mol$

Moles of salt form $HCOOK=\frac{50\times 0.2}{1000}=0.01mol$

Moles of left acid $=0.02-0.01=0.01mol$

$pH=p{K}_a+log\frac{Salt}{Acid}$

$pH=-log{K}_a+log\frac{0.01}{0.01}$

$pH=3.744$