Finding the pH of a mixture of weak acid and strong base
Buffer equation

The Henderson equation for buffers is:

$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log{\frac{[\ce{A-}]}{[\ce{AH}]}}$$

$\mathrm{p}K_\mathrm{a}$ and $\mathrm{p}K_\mathrm{b}$ add up to 14, as do $\mathrm{pH}$ and $\mathrm{pOH}$. So the expression for $\mathrm{pOH}$ is:

$$\mathrm{14 - pOH = 14} - \mathrm{p}K_\mathrm{b}  + \log{\frac{[\ce{A-}]}{[\ce{AH}]}}$$

or 

$$\mathrm{pOH} = \mathrm{p}K_\mathrm{b}  - \log{\frac{[\ce{A-}]}{[\ce{AH}]}}$$

Amounts of weak acid and weak base


  Also, shouldn't the buffer solution have a common ion? 


You start with $\pu{10 mmol}$ of ammonium salt (weak acid), to which you add $\pu{7.5 mmol}$ of $\ce{NaOH}$. The result is $\pu{7.5 mmol}$ of ammonia (weak base) with $\pu{2.5 mmol}$ ammonium salt remaining. So the ratio of weak base to weak acid is 1:3.

If you plug this into one or the other buffer equation, you get the answer.