# Question 18.9: Finding the Percent Dissociation of a Weak Acid Problem In 2......

Finding the Percent Dissociation of a Weak Acid

Problem In 2011, researchers showed that hypochlorous acid (HClO) generated by white blood cells kills bacteria. Calculate the percent dissociation of (a) 0.40 M HClO; (b) 0.035 M HClO ($K_a = 2.9×10^{−8}$).

Step-by-Step
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Plan We know the $K_a$ of HClO and need $[HClO]_{\text{dissoc}}$ to find the percent dissociation at two different initial concentrations. We write the balanced equation and the expression for $K_a$ and then set up a reaction table, with $x = [HClO]_{\text{dissoc}} = [ClO^−] = [H_3O^+]$. We assume that because HClO has a small $K_a$, it dissociates very little. Once $[HClO]_{\text{dissoc}}$ is known, we use Equation 18.5 to find the percent dissociation and check the assumption.

$\text{Percent HA dissociated }= \frac{[HA]_{\text{dissoc}}}{[HA]_{\text{init}}} × 100$        (18.5)

Solution (a) Writing the balanced equation and the expression for $K_a$:
$HClO(aq) + H_2O(l) \xrightleftharpoons[] H_3O^+(aq) + ClO^−(aq) K_a = \frac{[ClO^−][H_3O^+]}{[HClO]} = 2.9×10^{−8}$
Setting up a reaction table (Table 1) with  $x = [HClO]_{\text{dissoc}} = [ClO^−] = [H_3O^+]$:

Making the assumption: $K_a$ is small, so x is small compared with $[HClO]_{\text{init}}$; therefore, $[HClO]_{\text{init}} − x ≈ [HClO]_{\text{init}}$, or 0.40 M − x ≈ 0.40 M. Substituting into the $K_a$ expression and solving for x:

$K_a = \frac{[ClO^−][H_3O^+]}{[HClO]} = 2.9×10^{−8} ≈ \frac{(x)(x)}{ 0.40}$

Thus,
$x^2 = (0.40)(2.9×10^{−8})$
$x = \sqrt{(0.40)(2.9×10^{−8})} = 1.1×10^{−4} M = [HClO]_{\text{dissoc}}$
Finding the percent dissociation:

$\text{Percent dissociation }= \frac{[HClO]_{\text{dissoc}}}{[HClO]_{\text{init}}} × 100 = \frac{1.1×10^{−4} M}{0.40 M} × 100 = 0.028\%$
Since the percent dissociation is <5%, the assumption is justified.
(b) Performing the same calculations using $[HClO]_{\text{init}} = 0.035 M$:

$K_a = \frac{[ClO^−][H_3O^+]}{[HClO]} = 2.9×10^{−8} ≈ \frac{(x)(x)}{0.035}$
$x = \sqrt{(0.035)(2.9×10^{−8})} = 3.2×10^{−5} M = [HClO]_{\text{dissoc}}$

Finding the percent dissociation:
$\text{Percent dissociation }= \frac{[HClO]_{\text{dissoc}}}{[HClO]_{\text{init}}} × 100 = \frac{3.2×10^{−5} M}{0.035 M} × 100 = 0.091\%$
Since the percent dissociation is <5%, the assumption is justified.

Check The percent dissociation is very small, as we expect for an acid with such a low $K_a$. Note, however, that the percent dissociation is larger for the lower initial concentration, as we also expect.

Table 1

 Concentration (M) $\mathbf{HClO(aq) + H_2O(l) \xrightleftharpoons[]{} H_3O^+(aq) + ClO^−(aq)}$ Initial 0.40                               —                                   0                                      0 Change −x                                  —                                  +x                                   +x Equilibrium 0.40 − x                        —                                   x                                       x

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