Problem 3

a.

Note that the K values I use may differ slightly from those in your book

Dissolution of the sparingly soluble salt: $$Ag_3PO_{4(s)} \rightleftharpoons 3 Ag_{(aq)}^+ + PO^{3-}_{4(aq)}$$ $$K_{sp} = \left[Ag^+\right]^3\left[PO_4^{3-}\right] = 8.89 \times 10^{-17}$$

Autoprotolysis of water: $$2H_2O_{(l)} \rightleftharpoons H_3O_{(aq)}^+ + OH_{(aq)}^-$$ $$K_w = [OH^-][H_3O^+] = 10^{-14}$$

First basic reaction: $$PO^{3-}_{4(aq)} + H_2O_{(l)} \rightleftharpoons HPO^{2-}_{4(aq)} + OH_{(aq)}^-$$ $$K_{b1} = \frac{[OH^-][HPO_4^{2-}]}{[PO_4^{3-}]}=0.0467$$

Second basic reaction: $$HPO^{2-}_{4(aq)} + H_2O_{(l)} \rightleftharpoons H_2PO^{-}_{4(aq)} + OH_{(aq)}^-$$ $$K_{b2} = \frac{[OH^-][H_2PO_4^{-}]}{[HPO_4^{2-}]}=1.61\times 10^{-7}$$

Third basic reaction: $$H_2PO^{-}_{4(aq)} + H_2O_{(l)} \rightleftharpoons H_3PO_{4(aq)} + OH_{(aq)}^-$$ $$K_{b3} = \frac{[OH^-][H_3PO_4]}{[H_2PO_4^{-}]}=1.33\times 10^{-3}$$

b.

$$[Ag^+]+[H_3O^+] = [OH^-]+3[PO_4^{3-}] + 2[HPO_4^{2-}] + [H_2PO_4^-]$$

c.

Every 3 $Ag^+$ ions come with a $PO_4^{3-}$ ion, which then partitions into four forms: $$\frac{1}{3}[Ag^+] = [PO_4^{3-}] + [HPO_4^{2-}] + [H_2PO_4^-] + [H_3PO_4]$$

d.

$$\alpha_{PO_4^{3-}} = \frac{K_{a1}K_{a2}K_{a3}}{[H_3O^+]^3 + [H_3O^+]^2K_{a1} + [H_3O^+]K_{a1}K_{a2}+K_{a1}K_{a2}K_{a3}}$$$$~$$

$$~$$ $$K_{sp} = \left[Ag^+\right]^3\left[PO_4^{3-}\right]$$ $$~$$ $$~$$ $$K_{sp} = \left[Ag^+\right]^3\frac{K_{a1}K_{a2}K_{a3}}{[H_3O^+]^3 + [H_3O^+]^2K_{a1} + [H_3O^+]K_{a1}K_{a2}+K_{a1}K_{a2}K_{a3}}$$ $$~$$ $$~$$ $$\left[Ag^+\right]^3 = \frac{1}{K_{sp}}\frac{K_{a1}K_{a2}K_{a3}}{[H_3O^+]^3 + [H_3O^+]^2K_{a1} + [H_3O^+]K_{a1}K_{a2}+K_{a1}K_{a2}K_{a3}}$$ $$~$$ $$~$$ $$\left[Ag^+\right] = \left[\frac{1}{K_{sp}}\frac{K_{a1}K_{a2}K_{a3}}{[H_3O^+]^3 + [H_3O^+]^2K_{a1} + [H_3O^+]K_{a1}K_{a2}+K_{a1}K_{a2}K_{a3}}\right]^{\frac{1}{3}}$$ $$~$$ $$~$$ $$\left[Ag^+\right] = \left[\frac{1}{8.89 \times 10^{-17}}\frac{K_{a1}K_{a2}K_{a3}}{[H^+]^3 + [H^+]^2K_{a1} + [H^+]K_{a1}K_{a2}+K_{a1}K_{a2}K_{a3}}\right]^{\frac{1}{3}}$$ $$~$$ $$~$$ $$\left[Ag^+\right] = \left[1.125\times 10^{16}\frac{K_{a1}K_{a2}K_{a3}}{[H^+]^3 + [H^+]^2K_{a1} + [H^+]K_{a1}K_{a2}+K_{a1}K_{a2}K_{a3}}\right]^{\frac{1}{3}}$$

This equation is relatively simple! That is why adding a buffer actually makes this problem simpler

e.

First, although we used $K_b$ values in part a., we really need the $K_a$ values here. $$K_{a1} = 7.5 \times 10^{-3}$$ $$K_{a2} = 6.2 \times 10^{-8}$$ $$K_{a3} = 2.14 \times 10^{-13}$$

At $pH = 5.00$, $[H_3O^+] = 10^{-5}$ and at $pH = 9.00$, $[H_3O^+] = 10^{-9}$

It will require some care, but just plug in the numbers and go.