# Calculating pH of aqueous ammonium hydrogen sulfide solution

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.

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

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

• 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. Jun 22 at 18:43
• 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-] Jun 22 at 19:04
• Second pKa of H2S is much lower than that and even if this value was correct, it should still be ignored. Jun 22 at 22:27
• I did some calculations and we just have to prove $\ce{\frac{[NH_4^+]}{[NH_3]} = \frac{[HS^-]}{[H_2S] - [S^2-]}}$
– M.L
Jun 23 at 5:00
• 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-]) Jun 23 at 7:55

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