Question

The concentration of hydrogen ions in a $0.2M$ solution of formic acid is $6.4\times {10}^{-3}mol{L}^{-1}$. To this solution, sodium formate is added so as to adjust the concentration of sodium formate to $1mol{L}^{-1}$. What will be the pH of this solution?

The dissociation constant of formic acid is $2.4\times {10}^{-4}$ (write the value to the nearest integer).

Answer 1

	$HCOOH$ $\to$	$HCO{O}^{-}+$	$H^{+}$
Initial concentration (M)	$0.2$	$6.4\times {10}^{-3}$	$6.4\times {10}^{-3}$
Equilibrium concentration (M)	$0.2+x$	$1$	$6.4\times {10}^{-3}-x$

$K=\frac{\left[HCO{O}^{-}\right]\left[H^{+}\right]}{\left[HCOOH\right]}=\frac{6.4\times {10}^{-3}-x}{0.2+x}=2.4\times {10}^{-4}$

$⟹x=0.00635$

$\left[H^{+}\right]=6.4\times {10}^{-3}-0.00635$

$=4.9\times {10}^{-5}$

$pH=-log\left[H^{+}\right]$

$=-log4.9\times {10}^{-5}=4.305$

Hence, the answer to the nearest integer is $4$.