The equation of solubility is
BaSO4(s)⇄Ba2+(aq)+SO42−(aq).The expression for solubility product is
k=[Ba2+][SO42−]=1.08⋅10−10The equilibrium reactions and equilibrium constant equations:
H2SO4(aq)+H2O(l)⇄HSO4−(aq)+H3O+(aq) ka1=[H2SO4][HSO4−][H3O+]=103, HSO4−(aq)+H2O(l)⇄SO42−(aq)+H3O+(aq) ka2=[HSO4−][SO42−][H3O+]=0.0102.Now we can substitute the last equation for ka2 in equation for ka1:
[H2SO4]=0.0102⋅103[SO42−][H3O+]
Assume that the solubility is
s=[Ba2+]=[SO42−]+[HSO4−]+[H2SO4] Substitute the previous equation and equations for ka2 and ka1, numerical values, and obtain expression for s:
s=(1.08⋅10−10)(1+98.039[H3O]+0.98039[H3O]2). For [H3O+] = 0.75 M the calculations give us
s=8.975⋅10−5 M
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