Finding the pH of a mixture of weak acid and strong base What are the steps we go about in finding the pH of a solution of strong base and weak acid?
Here is the question that I've been given

What is the pH of a solution made by mixing $\pu{50 ml}$ of $\pu{0.2 M}$ $\ce{NH4Cl}$ and $\pu{75 ml}$ of $\pu{0.1 M}$ $\ce{NaOH}$, when $\mathrm pK_\mathrm b(\ce{NH3}) = 4.74$?
A. 7.02
B. 13.0
C. 9.73
D. 6.31

What I did to solve it was to use the Henderson equation for buffers,
$$\begin{align}
\mathrm{pOH} &amp;= \mathrm pK_\mathrm b + \log{\frac{[\text{salt}]}{[\text{base}]}} \\
&amp;= 4.74 + \log{\left(\frac{0.2 \cdot 50}{0.1 \cdot 75}\right)} \\
&amp;= 4.74 + \log{\frac{4}{3}} \\
&amp;= 4.86
\end{align}$$
and thus $\mathrm{pH} = 14 - \mathrm{pOH} = 9.14.$
The correct answer is C. But on putting values I'm getting the wrong answer. Can you help me figure out why? Also, shouldn't the buffer solution have a common ion?
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.