Question

0.25 mol of formic acid $\left(H C O_{2} H\right)$ 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 $=2.19=-log \left[\right.H^{+}\left]\right.$

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

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

$K_{a}\left(H C O O H\right)=\frac{\left[H^{+}\right] \left[H C O O^{-}\right]}{\left[H C O O H\right]}$

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

$\left[H C O O H\right]=0.25$ mol/1L

$=0.25\text{ M}$ and $K_{a}\left(H C O O H\right)=?$

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