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Question

The solubility product of BaSO4 is 1.5×109. Find out the solubility in (i) pure water and (ii) 0.1M BaCl2 solution.

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Solution

The equilibrium is:
(i) BaSO4Ba2++SO24
Let S be the solubility in mol Litre1; then
Ks=[Ba2+][SO24]=S2
or 1.5×109=S2
so, S=3.87×105mol L1
(ii) Let S be the solubility of BaSO4 in 0.1M BaCl2 solution.
Total Ba2+ ions concentration =(S+c)mol L1
and SO24 ions concentration =Smol L1
So, Ks=(S+c)S=(S+0.1)S
or 1.5×109=(S+0.1)S
or S2+0.1S=1.5×109
Neglecting (S)2
0.1S=1.5×109
or S=1.5×108mol L1

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