A sparingly soluble salt gets precipitated only when the product of the concentration of its ions in the solution (Qsp) becomes greater than its solubility product. If the solubility of BaSO4 in water is 8 × 10-4 mol dm-3, calculate its solubility in 0.01 mol dm-3 of H2SO4.
Given, the solubility of BaSO4 in water= 8 ×10-4 g/L
The equation of disassociation of BaSO4 will be-
BaSO4 ⇌ Ba2+ + SO42-
(S’ is the solubility of Ba2+ in 0.01 HCl)
S <<< 0.01, so it can be neglected
We know that Ksp = S2
Ksp = (8×10-8)2
= 64×10-8
Now, Ksp= (S’) (0.01)
S’ = 64.8×10-8/0.01 = 6.4×10-5
Hence solubility of BaSO4 in 0.01 mol dm-3 of H2SO4 is 6.4×10-5
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.