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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?

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    $\begingroup$ Using photos/screenshots of text ( especially if handwritten ) instead of typing text itself is highly discouraged. The image text content cannot be indexed nor searched for, cannot be reused nor referred in answers. Additionally, it can be challenge to decipher. Consider copy/pasting or rewriting of the essential parts and using of MathJax for eventual formatting of mathematical/chemical formulas or equations. $\endgroup$
    – Poutnik
    May 12 at 14:23
  • $\begingroup$ Thank you for your reply @Poutnik. So shall I delete the question or edit it and try to write it down in the form of LaTeX? $\endgroup$
    – Floatoss
    May 12 at 14:25
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    $\begingroup$ Rather edit it, using MathJax with mhchem extension. Notation basics and How can I format math/chemistry expressions on Chemistry SE. See also upright vs italic and Math SE MathJax tutorial. $\endgroup$
    – Poutnik
    May 12 at 14:27
  • $\begingroup$ Alright thanks, will do. $\endgroup$
    – Floatoss
    May 12 at 14:29
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    $\begingroup$ Note that $K_\mathrm{b, \ce{NH3}}=\pu{e-5}$ is wrong and $\ce{NH4OH}$ does not exist. $\mathrm{p}K_\mathrm{b}=4.75$, so $K_\mathrm{b}=\pu{10^{-4.75}}$ $\endgroup$
    – Poutnik
    May 12 at 14:41
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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, $$\text{pOH} = -\log{[\ce{OH^{-}}]}= -\log{\left(\frac{10^{-5}}{2}\right)}= -[-\log(2)-5]=\boxed{5.3}$$

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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.

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  • $\begingroup$ Why did you do the calculations considering everything happening for 1 litre of the stuff and not considering 500 mL itself? $\endgroup$
    – Floatoss
    May 13 at 4:55
  • $\begingroup$ Didn't you see I have calculated the initial concentration in first hand so I don't have to deal with volumes? $\endgroup$ May 13 at 5:33
  • $\begingroup$ Thank you for your reply. But why did you bring the "0.2 M" is what I am not able to understand. As in you wrote 0.2 mol of $\ce{NH_4OH}$ where as it was given that it was 0.1 mol in the question. I thought what you might have done would be "if 0.1 mol in 500mL then, 0.2 mol in 1L". Am I missing something? What I wanted to discuss was the "extra factor of 0.5" I had when I solved it the way I explained. $\endgroup$
    – Floatoss
    May 13 at 5:45
  • $\begingroup$ $\pu{1 M} = \pu{1 mol L-1}$. Your solutions are $\pu{0.1 mol}$ in $\pu{0.5 L}$ for both compounds. So, $\pu{M} = \frac{\pu{mol}}{\pu{L}}$, and do your calculations. $\endgroup$ May 13 at 5:58
  • $\begingroup$ That makes sense to me now, thank you. But I still don't know the "extra factor" of 0.5 I had in the way where I did not convert into standard concentration. Is there an explanation to that? $\endgroup$
    – Floatoss
    May 13 at 6:01

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