Question

0.25 mol of formic acid $\left(HC{O}_{2}H\right)$ is dissolved in enough water to make one litre of solution. The pH of that solution is 2.19. The $K_a$ of formic acid is

• A. $6.5\times 10^{-3}$
• B. $4.3\times 10^{-4}$
• C. $1.7\times 10^{-4}$
• D. $5.3\times 10^{-2}$

Answer 1

Answer: (C)

pH $=2.19=-log[H^{+}]$

$\left[H^{+}\right]=10^{-2.19}=$ Antilog $\left(-2.19\right)$

$=6.46\times 10^{-3}M$

$K_a\left(HCOOH\right)=\frac{\left[H^{+}\right]\left[HCO{O}^{-}\right]}{\left[HCOOH\right]}$

Here $\left[H^{+}\right]=\left[HCO{O}^{-}\right]=6.46\times 10^{-3}M$

$\left[HCOOH\right]=0.25$ mol/1L

$=0.25\text{ M}$ and $K_a\left(HCOOH\right)=?$

So ${\mathrm{K}}_{\mathrm{a}}(\mathrm{HCOOH})=\frac{{\left(6.46\times 10^{-3}\right)}^2}{0.25}=1.7\times 10^{-4}$.