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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