Question

What concentration of HCOO is present in a solution of 0.01 M HCOOH(Ka = 1.8 x 10^-4) and 0.01 M HCI?
(a) 1.8 x 10^-3
(b) 10^-2
(C) 1.8 x 10^-4
(d)10^-4​

Answer 1

Answer: The concentration of $HCOO^-$ is $1.8\times 10^{-4}M$

Explanation:

Hydrochloric acid is a strong acid and will completely dissociate into its ions when dissolved in water.

Formic acid is a weak acid and will partially dissociate into its ions when dissolved in water.

Concentration of $H^+$ ion initially = 0.01 M  (from HCl)

The chemical equation for the dissociation of HCOOH follows:

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

Initial:                0.01            -           0.01

At eqllm:         0.01-x            x         0.01+x

The expression of $K_a$ for above equation follows:

$K_a=\frac{[HCOO^-][H^+]}{[HCOOH]}$

We are given:

$K_a=1.8\times 10^{-4}$

Putting values in above equation, we get:

$1.8\times 10^{-4}=\frac{x\times (0.01+x)}{0.01-x}\\\\x^2+0.01018x-1.8\times 10^{-6}=0\\\\x=1.8\times 10^{-4},-0.0104$

Neglecting the negative value of 'x', we get:

$[HCOO^-]=x=1.8\times 10^{-4}M$

Hence, the concentration of $HCOO^-$ is $1.8\times 10^{-4}M$