Hydrolysis of $A_3B$ type weak acid-weak base salt

Question

I know the formulae for weak acid-weak base salt of AB type. A peculiar question made me ask this. Do the formula for derived for AB type also hold for A3B type sal. For example: This is the question, I got from my teacher

My attempt at solution:

Equation-1:$$\ce{3NH^+_4 +PO_4^{-3} + 3H_2O\rightleftharpoons 3NH_4OH + H_3PO_4}$$

$$K_h=\ce{\cfrac{[NH_4OH]^3[H_3PO_4]}{[NH_4^+]^3[PO_4^{-3}]}}=\cfrac{[3ch]^3×[ch]}{[3c(1-h)]^3[c(1-h)]}$$ $$\implies K_h=\frac{27c^4h^4}{27c^4(1-h)^4}\approx h^4$$ Equation-2: $$\ce{NH_4OH\rightleftharpoons NH_4^+ + OH^-}$$ $$K_b=\ce{\cfrac{[NH_4^+][OH^-]}{[NH_4OH]}}$$ Equation-3: $$\ce{H_3PO_4\rightleftharpoons 3H^+ + PO_4^{-3}}$$ $$K_a=\ce{\cfrac{[H^+]^3[PO_4^{-3}]}{[H_3PO_4]}}$$ Equation-4: $$K_w=[H^+][OH^-]$$

From the above equations, $$K_h×K_a×K_b^3=K_w^3$$$$\implies h^4=\cfrac{K^3_w}{K^3_b×K_a}$$

$$[H^+]^3=\cfrac{K_ach}{c-ch}\approx K_ah\implies [H^+]=\left(\cfrac{K_wK_a}{K_b}\right)^{(1/4)}$$ $$pH=\frac{1}{4}\left(pK_w+pK_a-pK_b\right)$$

This gives half of the correct answer.

Teacher's Solution

The formula given here is applicable to all scenarios so use $$pH=\frac{1}{2}\left(pK_w+pK_a-pK_b\right).$$ This gives $7.24$ which is the correct answer.

My question is :

Is the formula for $AB$ type weak acid-weak base also applicable to $A_3B$ type weak acid-weak base salt? What is the correct formula for $A_3B$ salts?
How to derive pH of $ A_xB_y$ type weak acid-weak base salt??

Answer 1

$\ce{NH4OH}$ assumes $\ce{N}$ is able to form 5 bonds. It is not.

The $\mathrm{p}K_\mathrm{a} = 5.23$ for phosphoric acid is nonsense; the task is made up. Additionally, the above provided formulat is not applicable on salts of multiprotic acids like $\ce{(NH4)3PO4}$

The respective $\mathrm{p}K_\mathrm{a}$ for each of $\ce{H3PO4}$ hydrogen are $\mathrm{p}K_\mathrm{a1}=2.14$, $\mathrm{p}K_\mathrm{a2}=7.20$ and $\mathrm{p}K_\mathrm{a3}=12.37$.
It is possible to have a combined $\mathrm{p}K_\text{a,tot}$ directly for $\ce{H3PO4 \to PO4^3-}$, but it would be their sum, 21.73.

The task can be simplified to ( informed guess assuming $1 \lt x \lt 2$):

$$\ce{3 NH4+ + PO4^3- -> (3-x) NH4+ + x NH3 + (2-x) HPO4^2- + (x-1)H2PO4-}$$

and

$$\mathrm{pH} \approx \mathrm{p}K_\ce{a,NH4+} + \log {\left(\frac{x}{3-x} \right)} \approx \mathrm{p}K_\ce{a2,H3PO4} + \log {\left(\frac{2-x}{x-1}\right)} $$

$$\mathrm{p}K_\ce{a,NH4+} - \mathrm{p}K_\ce{a2,H3PO4} = \log {\left(\frac{2-x}{x-1}\right)} - \log {\left(\frac{x}{3-x} \right)}=\log{\left(\frac {(2-x)(3-x)}{(x-1)x}\right)}$$

Let substitute $A = \mathrm{p}K_\ce{a,NH4+} - \mathrm{p}K_\ce{a2,H3PO4}= +2.05$

$$10^{A} \cdot x(x-1) = 6-5x+x^2$$

$$(10^{A}-1)x^2 + (5 -10^A)x - 6 = 0$$

$$(10^{2.05}-1)x^2 + (5 -10^{2.05})x - 6 = 0$$

$$111.2 \cdot x^2 - 107.2x - 6 \approx 0$$

$$x \approx 1.01708$$

$$\mathrm{pH} \approx \mathrm{p}K_\ce{a,NH4+} + \log {\left(\frac{1.01708}{3-1.01708} \right)} \approx 9.25 - 0.30 = 8.95$$

$$\mathrm{pH} \approx \mathrm{p}K_\ce{a2,H3PO4} + \log {\left(\frac{2-1.01708}{1.01708 - 1} \right)} \approx 7.20 + 1.76 = 8.96$$