Question

In a 0.730 M solution, a weak acid is 12.5% dissociated. Calculate Ka of the acid.

Answer 1

Approximately
`1.30 * 10^(-2)`, assuming that this acid is monoprotic.

Step-by-step explanation:

Assume that this acid is monoprotic. Let
`\rm HA` denote this acid.

`\rm HA \rightleftharpoons H^(+) + A^(-)`.

Initial concentration of
`\rm HA` without any dissociation:

`[{\rm HA}] = 0.730\; \rm mol \cdot L^(-1)`.

After
`12.5\%` of that was dissociated, the concentration of both
`\rm H^(+)` and
`\rm A^(-)` (conjugate base of this acid) would become:

`12.5\% * 0.730\; \rm mol \cdot L^(-1) = 0.09125\; \rm mol \cdot L^(-1)`.

Concentration of
`\rm HA` in the solution after dissociation:

`(1 - 12.5\%) * 0.730\; \rm mol \cdot L^(-1) = 0.63875\; \rm mol\cdot L^(-1)`.

Let
`[{\rm HA}]`,
`[{\rm H}^(+)]`, and
`[{\rm A}^(-)]` denote the concentration (in
`\rm mol \cdot L^(-1)` or
`\rm M`) of the corresponding species at equilibrium. Calculate the acid dissociation constant
`K_(\rm a)` for
`\rm HA`, under the assumption that this acid is monoprotic:

`\begin{aligned}K_(\rm a) &= \frac{[{\rm H}^(+)] \cdot [{\rm A}^(-)]}{[{\rm HA}]} \\ &= ((0.09125\; \rm mol \cdot L^(-1)) * (0.09125\; \rm mol \cdot L^(-1)))/(0.63875\; \rm mol \cdot L^(-1))\\[0.5em]&\approx 1.30 * 10^(-2) \end{aligned}`.