1
$\begingroup$

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.

$\endgroup$
8
  • 1
    $\begingroup$ 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. $\endgroup$ Jun 22 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 at 19:04
  • 2
    $\begingroup$ 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 at 22:27
  • 1
    $\begingroup$ 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 at 5:00
  • 3
    $\begingroup$ 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 at 7:55
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.