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

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:

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.