Question

If $50\text{mL}$ of $0.2MKOH$ is added to $40\text{mL}$ of $0.5MHCOOH$. The pH of the resulting solution is $(K=1.8\times {10}^{-4})$ :

• A. 3.75
• B. 5.6
• C. 7.5
• D. 3.4

Answer 1

The correct option is A 3.75

Given $(K=1.8\times {10}^{-4})$
$HCOOH+KOH$
$40\text{mL},0.5M\stackrel{\text{50 mL, 0.2 M}}{\to }HCOONa+H_{2}O$ after reaction is forms buffer solution
Millimoles of $H^{+}$ = 0.5 $\times$ 40 = 20
Millimoles of $O{H}^{-}$ = 0.2 $\times$ 50 = 10
$\text{Total millimoles of}{H}^{+}\text{in mixture = 20 - 10 = 10}$
$\left[HCOOH\right]=\frac{10}{90},\left[HCOOK\right]=\frac{10}{90}$
Using Henderson Equation
$pH=p{K}_a+\text{log}\frac{\text{[salt]}}{\text{acid}}$
$pH=p{K}_a$
$pH=4-\text{log}\left(1.8\right)pH=3.75$