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I was doing the following question yesterday:

Calculate the pH of a $\pu{0.1 M}$ solution of $\ce{NH4HS}$, given $\mathrm{p}K_\mathrm{b}$ of $\ce{NH3}$ is $4.74$, and $\mathrm{p}K_\mathrm{a1}$ and $\mathrm{p}K_\mathrm{a2}$ of $\ce{H2S}$ are $7.04$ and $11.96$, respectively.

My approach

The following equilibria exist in the solution $$\ce{NH4HS -> NH4+ + HS-} \tag{Eqn:1} $$ $$\ce{NH4+ + H2O <=> NH3 + H3O+} \tag{Eqn:2}$$ $$\ce{HS- + H2O <=> H2S + OH-} \tag{Eqn:3}$$ $$\ce{HS- + H2O <=> S^{2-} + H3O+} \tag{Eqn:4}$$ $$\ce{H2O + H2O <=> H3O+ + OH-} \tag{Eqn:5}$$

Setting up the equations

  1. Material Balancing:

$$\pu{0.1 M} = \ce{[NH_4 ^+] + [NH_3]} \tag{Eqn:M1}$$ $$\pu{0.1 M} = \ce{[H2S] + [HS-] + [S^2-]} \tag{Eqn:M2}$$

  1. Charge Balancing: $$\ce{[H3O+] + [NH4+] = [OH-] + [HS-] + 2[S^2-]} \tag{Eqn:C}$$

  2. Writing $K_\mathrm{eq}$ expressions:

For $(\text{Eqn:2})$, $$\frac{K_\mathrm{w}}{K_\mathrm{b}} = \frac{\ce{[NH_3}][\ce{H_3O^+}]}{\ce{[NH_4^+]}}$$ For $(\text{Eqn:3})$, $$\frac{K_\mathrm{w}}{K_\mathrm{a1}} = \frac{[\ce{H_2S}][\ce{OH^-}]}{[\ce{HS^-}]}$$ For $(\text{Eqn:4})$, $$K_\mathrm{a2} = \frac{[\ce{S^2-}][\ce{H_3O^+}]}{[\ce{HS^-}]}$$ For $(\text{Eqn:5})$, $$K_\mathrm{w} = [\ce{H_3O^+}][\ce{OH^-}]$$

These equations are driving me crazy and any approximation(s) that can be used to reduce number of variables here are appreciated.

Answer

$\ce{[H+] =\sqrt{K_{a1}.[\frac{K_\mathrm{w}}{K_\mathrm{b}} + K_\mathrm{a2}]}}$ $\implies$ pH =$8.14$

I have seen this question but in that case, the $\ce{NH4+}$ doesn't undergo hydrolysis.

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    $\begingroup$ Actually, in given example, $\ce{HSO4-}$ has under gone hydrolysis to give the final pH. The hydrolysis of $\ce{NH4+}$ is the one in question. $\endgroup$
    – Mathew Mahindaratne
    Jun 22, 2021 at 18:43
  • $\begingroup$ Consider simplifications based on assumed strong inequalities. E.g. the trivial formula for pH of weak acid solution pH=1/2*(pKa - log c ) assumes c >> [H+] >> [OH-] $\endgroup$
    – Poutnik
    Jun 22, 2021 at 19:04
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    $\begingroup$ Second pKa of H2S is much lower than that and even if this value was correct, it should still be ignored. $\endgroup$
    – Mithoron
    Jun 22, 2021 at 22:27
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    $\begingroup$ I did some calculations and we just have to prove $\ce{\frac{[NH_4^+]}{[NH_3]} = \frac{[HS^-]}{[H_2S] - [S^2-]}}$ $\endgroup$
    – M.L
    Jun 23, 2021 at 5:00
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    $\begingroup$ You can neglect HS- as an acid, as the Ka2 is too low. With some simplifying assumptions, you can consider the system consisting a weak acid NH4+ and a weak base HS- and the hydrolysis balance [NH3] = [H2S] >> max( [H+],[OH-]) $\endgroup$
    – Poutnik
    Jun 23, 2021 at 7:55

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From here:

$$\mathrm{pH} = \frac{1}{2}(\mathrm{p}K_\mathrm{a} + \mathrm{p}K_\mathrm{w} - \mathrm{p}K_\mathrm{b})$$

Plugging in $\mathrm{p}K_\mathrm{a} = 7.04$ and $\mathrm{p}K_\mathrm{b} = 4.74$, you get $\mathrm{pH} = 8.15$. As mentioned by Poutnik and Mithoron in the comments, $\ce{HS-}$ can be neglected as an acid due to a low $K_\mathrm{a2}$ value. So this is just the classic case of the salt of a weak acid and a weak base.

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