The solubility product of a sparingly soluble salt BaSO4 is 1.1×10−10 at 25oC . What will be the solubility of BaSO4 in presence of 0.02MH2SO4 ? (assume complete dissociation of H2SO4)
A
2.25×10−9mol L−1
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B
5.5×10−9mol L−1
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C
7.5×10−9mol L−1
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D
1.5×10−9mol L−1
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Solution
The correct option is B5.5×10−9mol L−1 BaSO4 is a sparingly soluble salt , whereas H2SO4 is a strong acid. Let the solubility of BaSO4 be s. Then
BaSO4⇌Ba2++SO2−4 t=teqc−sss
Sulphuric acid is a strong electrolyte and is completely ionised. It shall provide SO2−4 ion concentration= 0.02M.
[Ba2+]=s [SO2−4]=(s+0.02)M
Ksp=[Ba2+][SO2−4]=s×(s+0.02)
Since H2SO4 is highly soluble . So s+0.02≈0.02, because of the sapringly soluble salt (BaSO4) where Ksp<<10−3 Given Ksp=1.1×10−10 ⇒1.1×10−10=0.02×s or s=1.1×10−100.02=5.5×10−9molL−1