The equilibrium is:
(i) BaSO4⇌Ba2++SO2−4
Let S be the solubility in mol Litre−1; then
Ks=[Ba2+][SO2−4]=S2
or 1.5×10−9=S2
so, S=3.87×10−5mol L−1
(ii) Let S′ be the solubility of BaSO4 in 0.1M BaCl2 solution.
Total Ba2+ ions concentration =(S′+c)mol L−1
and SO2−4 ions concentration =S′mol L−1
So, Ks=(S′+c)S′=(S′+0.1)S′
or 1.5×10−9=(S′+0.1)S′
or S′2+0.1S′=1.5×10−9
Neglecting (S′)2
0.1S′=1.5×10−9
or S′=1.5×10−8mol L−1