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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}).

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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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