Question

If $50mL$ of $0.2MKOH$ is added to $40mL$ of $0.5MHCOOH$, the $pH$ of the resulting solution is:
$(K_a=1.8\times {10}^{-4},\text{log}18=1.26)$

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

Answer 1

The correct option is A 3.74

We know molarity $=\frac{\text{moles}}{\text{volume}}$
moles = molarity $\times$ volume
Given Molarity $KOH=0.2M$
Molarity of $HCOOH=0.5M$
volume of $KOH=50\text{ml}$
volume of $HCOOH=40\text{ml}$
$n_{KOH}=0.2\times \frac{50}{1000}=0.01\text{mol}$
$n_{HCOOH}=\frac{40\times 0.5}{1000}=0.02\text{mol}$
Moles of salt from $HCOOK=\frac{50\times 0.2}{1000}$
$=0.01\text{mol}$
Moles of left acid $=0.02-0.01$
$=0.01\text{mol}$
$pH=pKa+\text{log}\frac{\text{salt}}{\text{acid}}$
$pH=-\text{log}Ka+\text{log}\frac{0.01}{0.01}$
$pH=-\text{log}\left(18\right)-\text{log}\left({10}^{-5}\right)+0$
$pH=3.74$