Question

A solution of HCOOH has 0.22M HCOOH at equilibrium. The Ka for HCOOH is 1.8×10−4. What is the pH of this solution at equilibrium?

Answer 1

Answer : The pH of the solution is, 2.21

Solution :  Given,

Concentration (c) = 0.22 M

Acid dissociation constant = $k_a=1.8\times 10^{-4}$

The equilibrium reaction for dissociation of $HCOOH$ (weak acid) is,

                           $HCOOH\rightleftharpoons HCOO^-+H^+$

initially conc.         c                       0         0

At eqm.              $c(1-\alpha)$                $c\alpha$        $c\alpha$

First we have to calculate the concentration of value of dissociation constant $(\alpha)$.

Formula used :

$k_a=\frac{(c\alpha)(c\alpha)}{c(1-\alpha)}$

Now put all the given values in this formula ,we get the value of dissociation constant $(\alpha)$.

$1.8\times 10^{-4}=\frac{(0.22\alpha)(0.22\alpha)}{0.22(1-\alpha)}$

By solving the terms, we get

$\alpha=0.0282$

Now we have to calculate the concentration of hydrogen ion.

$[H^+]=c\alpha=0.22\times 0.0282=6.204\times 10^{-3}M$

Now we have to calculate the pH.

$pH=-\log [H^+]$

$pH=-\log (6.204\times 10^{-3})$

$pH=2.21$

Therefore, the pH of the solution is, 2.21