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Question

The solubility product of a sparingly soluble salt BaSO4​ is 1.1×1010 at 25oC . What will be the solubility of BaSO4​ in presence of 0.02 M H2SO4 ? (assume complete dissociation of H2SO4)

A
2.25×109 mol L1
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B
5.5×109 mol L1
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C
7.5×109 mol L1
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D
1.5×109 mol L1
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Solution

The correct option is B 5.5×109 mol L1
BaSO4 is a sparingly soluble salt , whereas H2SO4 is a strong acid.
Let the solubility of BaSO4 be s. Then

BaSO4Ba2++SO24
t=teq cs s s

Sulphuric acid is a strong electrolyte and is completely ionised. It shall provide SO24 ion concentration= 0.02 M.

[Ba2+]=s
[SO24]=(s+0.02) M

Ksp=[Ba2+][SO24]=s×(s+0.02)

Since H2SO4 is highly soluble .
So s+0.020.02, because of the sapringly soluble salt (BaSO4) where Ksp<<103
Given Ksp=1.1×1010
1.1×1010=0.02×s
or
s=1.1×10100.02=5.5×109 mol L1

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