Calculating pH of an unknown base, and the pKb
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.