Finding the pOH of a buffer from the scratch without the Henderson Hasselbalch equation

Question

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 "common ion" 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

$$\ce{NH4OH <<=> NH4+ + OH-}$$

\begin{array}{l|c|c|c} \hline n_\mathrm{initial} & 0.1& 0 &0\\ n_\mathrm{equilibrium} &0.1(1-\alpha) &0.1\alpha& 0.1\alpha \\ \hline \end{array}

$$\ce{(NH4)2SO4 -> 2 NH4+ + SO4^2-}$$

\begin{array}{l|c|c|c} \hline n_\mathrm{initial} & 0.1& 0 &0\\ n_\mathrm{dissociation} &0 &0.2 & 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?

Answer 1

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.

$$\ce{NH4OH <<=> NH4+ + OH-}$$

\begin{array}{l|c|c|c} \hline n_\mathrm{initial} & 0.1& 0 &0\\ n_\mathrm{equilibrium} &0.1(1-\alpha) &0.1\alpha& 0.1\alpha \\ \hline \end{array}

$$\ce{(NH_4)_2SO_4 -> 2 NH_4^{+} + SO4^{2-}}$$

\begin{array}{l|c|c|c} \hline n_\mathrm{initial} & 0.1& 0 &0\\ n_\mathrm{dissociation} &0 &0.2 & 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, $$\text{pOH} = -\log{[\ce{OH^{-}}]}= -\log{\left(\frac{10^{-5}}{2}\right)}= -[-\log(2)-5]=\boxed{5.3}$$

Answer 2

Find the $\mathrm{pOH}$ of the solution obtained by mixing $\pu{0.1 mol}$ of $\ce{NH4OH}$ $(K_\mathrm{b} = 10^{-5})$ and $\pu{0.1 mol}$ of $\ce{(NH4)2SO4}$ in a $\pu{500 mL}$ solution.

OP wants to find $\mathrm{pOH}$ without applying Henderson-Hasselbalch equation, and to do it from the scratch. Here is how I would did it:

Aqueous ammonia solution, still erroneously called it as $\ce{NH4OH}$, is a weak base, and $\ce{(NH4)2SO4}$ is a strong electrolyte of conjugate acid of aqueous ammonia. In aqueous solution, $\ce{(NH4)2SO4}$ is completely dissociate:

$$\ce{(NH4)2SO4(aq) -> 2 NH4+(aq) + SO4^2-(aq)}$$

The molarity of each of $\ce{(NH4)2SO4}$ and aqueous ammonia is $\dfrac{0.1}{0.5} = \pu{0.2 M}$. Accordingly, initial concentration of $\ce{NH4+}$ due to $\ce{(NH4)2SO4}$ dissociation is $\pu{0.4 M}$. Hence:

$$\begin{array}{l|c|c|c} & [\ce{(NH4)2SO4}] & [\ce{NH4+}] & [\ce{SO4^2-}]\\ \hline n_\mathrm{ini.} & 0.2 & 0 & 0\\ n_\mathrm{diss.} & 0 & 0.4 & 0.2\\ \hline \end{array}$$

The ionization of ammonia in water can be written as:

$$\ce{NH3 (aq) + H2O (l) <=> NH4+ (aq) + OH- (aq)}$$

If $\alpha$ is the amount of ammonia ionized to give $\ce{NH4+}$ ion in $\pu{M}$:

$$\begin{array}{l|c|c|c} & [\ce{NH3}] & [\ce{NH4+}] & [\ce{OH-}]\\ \hline n_\mathrm{ini.} & 0.2 & 0.4 & 0\\ n_\mathrm{equi.} & (0.2 - \alpha) & (0.4 + \alpha) & \alpha \\ \hline \end{array} $$

By definition,

$$K_\mathrm b = \frac{[\ce{NH4+}] [\ce{OH-}]}{[\ce{NH3}]}$$

$$10^{-5} = \frac{(0.4 + \alpha) \times \alpha}{0.2 - \alpha}$$

Since $\alpha \lt \lt 0.2$ for a weak base, $(0.2 - \alpha) \approx 0.2$ and $(0.4 + \alpha) \approx 0.4$.

$$\therefore \ 10^{-5} = \frac{0.4 \times \alpha}{0.2} \ \Rightarrow \ \alpha = 10^{-5} \times \frac{0.2}{0.4} \tag1$$

Since $\alpha = [\ce{OH-}]$, we can rewrite the equation $(1)$ as:

$$ [\ce{OH-}] = 10^{-5} \times \frac{0.2}{0.4} \tag2$$

Take $-\log$ on both sides of the equation $(2)$:

$$ -\log [\ce{OH-}] = -\log (10^{-5}) - \log \left( \frac{0.2}{0.4}\right) = -\log (10^{-5}) + \log \left( \frac{0.4}{0.2}\right) \tag3$$

The equation $(3)$ is technically the Henderson-Hasselbalch equation for the basic buffer since $ -\log [\ce{OH-}] = \mathrm{pOH}$, $ -\log (10^{-5}) = \mathrm{p}K_\mathrm{b}$, $0.2 = [\ce{NH3}]$, and $0.4 = [\ce{NH4+}]$. If rewrite the equation $(3)$ with these notation, you got:

$$ \mathrm{pOH} = \mathrm{p}K_\mathrm{b} + \log \left( \frac{[\ce{NH4+}]}{[\ce{NH3}]}\right) \tag4$$

So, you have end up with The Henderson-Hasselbalch equation after all. That's because, it is the way the Henderson-Hasselbalch equation derived. If it's not given, you can derived it and use it. Remember the major assumption, $\alpha \lt \lt [\ce{NH3}]$ and $\alpha \lt \lt [\ce{NH4+}]$ in this case, has to be valid.