Question

Calculate the percent ionization of nitrous acid in a solution that is 0.222 M in nitrous acid (HNO3) and 0.278 M in potassium nitrite (KNO2). The acid dissociation constant of nitrous acid is 4.50 x 10-4 A) 55.6 B) 0.162 C) 15.5 D) 2.78 * 10-3

Answer 1

Explanation:

Concentration of $KNO_{2}$ is 0.278 M and it is completely ionized into $K^{+}$ and $NO^{-}_{2}$.

This means that $[KNO_{2}]$ = $[NO_{2}]$ = 0.278 M

It is given that concentration of $HNO_{2}$ is 0.222 M.

As $HNO_{2}$ is a weak acid. Therefore, its dissociation will be as follows.

              $HNO_{2}(aq) \rightarrow H^{+}(aq) + NO^{-}_{2}(aq)$

Initially :    0.222 M            0        0.278 M

Change :    - x                    +x           +x

Equilibrium : 0.222 M - x    x         0.278 M + x

Therefore, dissociation constant for this reaction will be as follows.

        $K_{a} = \frac{[H^{+}][NO^{-}_{2}]}{[HNO_{2}]}$

Hence, putting the given values into the above formula as follows.

        $K_{a} = \frac{[H^{+}][NO^{-}_{2}]}{[HNO_{2}]}$

      $4.50 \times 10^{-4} = \frac{x \times (0.278 + x)}{(0.222 - x)}$

                          x = 0.00036

As, $[H^{+}]$ is 0.00036 M. Therefore, percentage ionization of $HNO_{2}$ will be calculated as follows.

              % ionization of $HNO_{2}$ = $\frac{[H^{+}]}{[HNO_{2}]} \times 100$

                               = $\frac{0.00036 M}{0.222 M} \times 100$

                               = 0.16 %

Thus, we can conclude that % ionization of $HNO_{2}$ is 0.16%.