Finding the pOH of a buffer from the scratch without the Henderson Hasselbalch equation
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, $$\text{pOH} = -\log{[\ce{OH^{-}}]}= -\log{\left(\frac{10^{-5}}{2}\right)}= -[-\log(2)-5]=\boxed{5.3}$$