Calculating pH of an unknown base, and the pKb The percent dissociation of a base $\ce{X}$ in a $0.002\ \mathrm{M}$ solution is $4.7 \cdot 10^{-3}\ \%$. What is the pH of the solution and what is the $\mathrm{p}K_\text{b}$ of base $\ce{X}$? The autoionization of water may not be neglected.


Attempt at solution: We have the following reactions: $$ \ce{X + H2O(l) &lt;=&gt; XH+ + OH-} \\ \ce{H2O(l) &lt;=&gt; H+ + OH-}$$

Since the percent dissociation is $4.7 \times 10^{-3}\ \%$, this enables us to find the concentration dissociated: $$ \frac{x}{0.002\ \mathrm{M}} \cdot 100 = 4.7 \times 10^{-3},$$ so $x = 9.4 \times 10^{-8}$. Now I did: $$ K_\text{a} = \frac{x^2}{0.002 - x}$$ and I let $(0.002 - x) \approx 0.002$ since $x$ is so small. But this gives me the wrong answer, and it seems I haven't taken the autoionization of water into account yet. How should I solve this problem then?
In such an solution, since the autoionization of water may not be neglected, you should expect $x=9.4 \times 10^{-8}$ do not mean $[\ce{OH-}]=9.4 \times 10^{-8}$. So you can only leave it unknown.

Then, $K_\text{b} = \frac {[\ce{XH+}][\ce{OH-}]}{[\ce{X}]} \approx \frac {9.4 \times 10^{-8} [\ce{OH-}]}{0.002} = 4.7 \times 10^{-5}[\ce{OH-}]$.

However, we still do not know what $[\ce{OH-}]$ is.

If only to consider the autoionization of water balanced, you will get

$[\ce{OH-}][\ce{H+}] = K_\text{w} = 1 \times 10^{-14}$

and this is still not enough.

The answer can only consider by the fact that the solution will be electrically neutral,

$[\ce{OH-}]=[\ce{H+}] + [\ce{XH+}] = \frac {1 \times 10^{-14}}{[\ce{OH-}]} + 9.4 \times 10^{-8}$

$[\ce{OH-}]^2 - 9.4 \times 10^{-8} [\ce{OH-}] - 1 \times 10^{-14}=0$

Since $\Delta = (9.4 \times 10^{-8})^2 + 4 \times 10^{-14} \approx 4 \times 10^{-14}$,

$[\ce{OH-}] \approx \frac {9.4 \times 10^{-8} + 2 \times 10^{-7}}{2} = 2.47 \times 10^{-7}$.

Thus, $\mathrm{pH} = -\log \left(\frac {1 \times 10^{-14}}{2.47 \times 10^{-7}}\right) = 7.39$

$\mathrm{p}K_\text{b} = -\log (4.7 \times 10^{-5} \times 2.47 \times 10^{-7}) = 10.94$.

Sometimes the “electrically neutral” trait will be very useful, this is a good example.