Solubility product of BaSO4 is 1.1 × 10–10. Calculate the volume of its solution containing 15 mg salt in the
solution.
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BaSO
4
(s)⟶Ba
2+
(aq)+SO
4
2−
(aq)
∴K
sp
=[Ba
2+
][SO
4
2−
]=x
Then, 1.5×10
−9
=x×x;x
2
=15×10
−10
or3.87×10
−5
Then, solubility of BaSO
4
in pure water is 3.87×10
−5
.
Let the solubility of BaSO
4
in 0.1MBaCl
2
be
′
s
′
Allsolid
BaSO
4
(s)
⇒
0.1M
Ba
2+
(aq)
+SO
4
−
(aq)
Initial (from BaCl
2
) 0
At equilibrium (0.1M+s) s
Hence, 1.5×10
−9
=(s+0.1)×s=s×0.1(Since s<<1)
s=1.5×10
−8
Thus the solubility of BaSO
4
in presence of 0.1M BaCl
2
is 1.5×10
−8
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