Answer to Question #136744 in General Chemistry for Ann

Question #136744
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.
1
Expert's answer
2020-10-12T13:32:17-0400

The equation of solubility is


BaSO4(s)Ba2+(aq)+SO42(aq).\text{BaSO}_4(s)\rightleftarrows\text{Ba}^{2+}(aq)+\text{SO}_4^{2-}(aq).

The expression for solubility product is


k=[Ba2+][SO42]=1.081010k=[\text{Ba}^{2+}][\text{SO}_4^{2-}]=1.08\cdot10^{-10}

The equilibrium reactions and equilibrium constant equations:


H2SO4(aq)+H2O(l)HSO4(aq)+H3O+(aq) ka1=[HSO4][H3O+][H2SO4]=103, HSO4(aq)+H2O(l)SO42(aq)+H3O+(aq) ka2=[SO42][H3O+][HSO4]=0.0102.\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 ka2 in equation for ka1:

[H2SO4]=[SO42][H3O+]0.0102103[\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=[Ba2+]=[SO42]+[HSO4]+[H2SO4]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 ka2 and ka1, numerical values, and obtain expression for s:

s=(1.081010)(1+98.039[H3O]+0.98039[H3O]2).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=8.975105 Ms=8.975\cdot10^{-5}\text{ M}



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