Finding the pOH of a buffer from the scratch without the Henderson Hasselbalch equation I found this question on the internet that said:

Find the pOH of the solution obtained by mixing $\pu{0.1 mol}$ of $\mathrm{NH_4 OH}$ $(K_\mathrm b = 10^{-5})$ and $\pu{0.1 mol}$ of $\ce{(NH4)2SO4}$ in a $\pu{500 mL}$ solution.

What my approach was to do it without applying the Henderson Hasselbalch equation and doing it from the scratch using the &quot;common ion&quot; effect. Here is how I did it:
$\ce{NH4OH}$ is a weak acid and $\ce{(NH4)2SO4}$ is a strong electrolyte.
$\alpha$= The dissociation constant per mole
$$\ce{NH4OH &lt;&lt;=&gt; NH4+ + OH-}$$
\begin{array}{l|c|c|c} \hline
n_\mathrm{initial} &amp; 0.1&amp; 0 &amp;0\\
n_\mathrm{equilibrium} &amp;0.1(1-\alpha) &amp;0.1\alpha&amp; 0.1\alpha \\
\hline
\end{array}
$$\ce{(NH4)2SO4 -&gt; 2 NH4+ + SO4^2-}$$
\begin{array}{l|c|c|c} \hline
n_\mathrm{initial} &amp; 0.1&amp; 0 &amp;0\\
n_\mathrm{dissociation} &amp;0 &amp;0.2 &amp; 0.1\\
\hline
\end{array}
Now, we can say that
$$K_\mathrm b = \frac{[\ce{NH4+}] [\ce{OH-}]}{[\ce{NH_4OH}]}$$
$$10^{-5}= \frac{\dfrac{0.2+0.1 \alpha}{0.5} \times \dfrac{0.1 \alpha}{0.5}}{\dfrac{0.1}{0.5}}$$
As, $\alpha$ is very small for a weak acid, $0.1 \alpha \approx 0$.
Doing the further calculations, we get:
$$\boxed{\alpha = \frac{5}{2} \times 10^{-5}}$$
$$\implies \boxed{\ce{[OH-]} = 0.1 \alpha = \frac{5}{2} \times 10^{-6} \frac{\pu{mol}}{\pu{500 mL}}}$$
Now, $$\mathrm p\ce{OH}=\log[\ce{OH-}] = -[\log(5)-\log(2)-6] = \boxed{5.6}$$
But if we directly apply the Henderson Hasselbalch equation for the buffer, we get:
$$\mathrm p\ce{OH} = \mathrm pK_\mathrm b + \log\left(\frac{[\mathrm{salt}]}{[\mathrm{base}]}\right)$$
$$\mathrm p\ce{OH} = 5 + \log\left(\frac{\dfrac{0.2}{0.5}}{\dfrac{0.1}{0.5}} \right) = 5+ \log(2)=\boxed{5.3}$$
[.] = molarity
Where did I go wrong in the calculation or what mistake did I do while solving this question?
I found out where the mistake was in my method while solving the question. Here it goes.
• $\ce{NH4OH}$ is a weak acid and $\ce{(NH4)2SO4}$ is a strong electrolyte.
• $\alpha$= The dissociation constant per mole.
$$\ce{NH4OH &lt;&lt;=&gt; NH4+ + OH-}$$
\begin{array}{l|c|c|c} \hline
n_\mathrm{initial} &amp; 0.1&amp; 0 &amp;0\\
n_\mathrm{equilibrium} &amp;0.1(1-\alpha) &amp;0.1\alpha&amp; 0.1\alpha \\
\hline
\end{array}
$$\ce{(NH_4)_2SO_4 -&gt; 2 NH_4^{+} + SO4^{2-}}$$
\begin{array}{l|c|c|c} \hline
n_\mathrm{initial} &amp; 0.1&amp; 0 &amp;0\\
n_\mathrm{dissociation} &amp;0 &amp;0.2 &amp; 0.1\\
\hline
\end{array}
Now, we can say that:
$$K_\mathrm b = \frac{[\ce{NH4+}] [\ce{OH-}]}{[\ce{NH_4OH}]}$$
Plugging in the values mentioned and derived,
$$10^{-5}= \left(\frac{\dfrac{0.2+0.1 \alpha}{0.5} \times \dfrac{0.1 \alpha}{0.5}}{\dfrac{0.1(1-\alpha)}{0.5}}\right)$$
As $\alpha$ is very small for a weak acid, $0.1 \alpha \approx 0$ and $(1- \alpha) \approx 1$. Doing the calculations above, we get that:
$$\boxed{\alpha = \frac{5}{2} \times 10^{-5}}$$
Which further implies that, there are $0.1 \alpha$ moles of $\ce{OH^{-}}$ ions in the 0.5L solution.
$$ \therefore \ \boxed{\ce{[OH-]} = \frac{0.1 \alpha \ \text{mol}}{0.5 \ \text{L}} = \frac{0.1}{0.5} \cdot \frac{5}{2} 10^{-5} = \frac{10^{-5}}{2} \frac{\text{mol}}{\text{L}}}$$
Now, $$\text{pOH} = -\log{[\ce{OH^{-}}]}= -\log{\left(\frac{10^{-5}}{2}\right)}= -[-\log(2)-5]=\boxed{5.3}$$