Answer to Question #136744 in General Chemistry for Ann

Question

Follow the systematic procedure (9 steps, if applicable) to calculate for the
molar solubility of BaSO4 in a solution in which [H3O+] is 0.75 M.

Answer 1

The equation of solubility is

$$\text{BaSO}_4(s)\rightleftarrows\text{Ba}^{2+}(aq)+\text{SO}_4^{2-}(aq).$$

The expression for solubility product is

$$k=[\text{Ba}^{2+}][\text{SO}_4^{2-}]=1.08\cdot10^{-10}$$

The equilibrium reactions and equilibrium constant equations:

$$\text{H}_2\text{SO}_4(aq)+\text{H}_2\text{O}(l)\rightleftarrows\text{HSO}_4^-(aq)+\text{H}_3\text{O}^+(aq)\\\space\\ k_{a1}=\frac{[\text{HSO}_4^-][\text{H}_3\text{O}^+]}{[\text{H}_2\text{SO}_4]}=10^3,\\\space\\ \text{HSO}_4^-(aq)+\text{H}_2\text{O}(l)\rightleftarrows\text{SO}_4^{2-}(aq)+\text{H}_3\text{O}^+(aq)\\\space\\ k_{a2}=\frac{[\text{SO}_4^{2-}][\text{H}_3\text{O}^+]}{[\text{H}\text{SO}_4^-]}=0.0102.$$

Now we can substitute the last equation for k_a2 in equation for k_a1:

$$[\text{H}_2\text{SO}_4]=\frac{[\text{SO}_4^{2-}][\text{H}_3\text{O}^+]}{0.0102\cdot10^3}$$

Assume that the solubility is

$$s=[\text{Ba}^{2+}]=[\text{SO}_4^{2-}]+[\text{H}\text{SO}_4^-]+[\text{H}_2\text{SO}_4]$$

Substitute the previous equation and equations for k_a2 and k_a1, numerical values, and obtain expression for s:

$$s=\sqrt{(1.08\cdot10^{-10})(1+98.039[\text{H}_3\text{O}]+0.98039[\text{H}_3\text{O}]^2)}.$$

For [H₃O⁺] = 0.75 M the calculations give us

$$s=8.975\cdot10^{-5}\text{ M}$$