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Question: The solubility product of \[BaS{O_4}\] at \[{25^ \circ }C\] is \[1.0 \times {10^{ - 9}}\]. What woul...

The solubility product of BaSO4BaS{O_4} at 25∘C{25^ \circ }C is 1.0×10−91.0 \times {10^{ - 9}}. What would be the minimum concentration of H2SO4{H_2}S{O_4} necessary to precipitate BaSO4BaS{O_4} from a solution of 0.01M0.01M Ba2+B{a^{2 + }} ions?
A. 10−9{10^{ - 9}}
B. 10−8{10^{ - 8}}
C. 10−7{10^{ - 7}}
D. 10−6{10^{ - 6}}

Explanation

Solution

The precipitation of an ionic salt is observed only when the individual concentration of the constituent ions become so high that the product of concentration becomes high then the solubility product at that particular temperature.

Complete answer:
Barium sulphate is an ionic compound that is sparingly soluble in water. Physical conditions like the temperature of the solution affects its solubility. But the key factor that impacts the solubility of the compound is its concentration or the concentration of its constituent ions.
The dissociation of barium sulphate in water into its constituent ions can be written as follows:
BaSO4⇌Ba2++SO42−BaS{O_4} \rightleftharpoons B{a^{2 + }} + S{O_4}^{2 - }
The solution of barium sulphate already contains 0.01M0.01M Ba2+B{a^{2 + }} ions.
Thus the allowed concentration of sulphate ions is the concentration that does not exceed the value of the solubility product. The minimum concentration of sulphuric acid that can be added must be equal to the allowed sulphate ion concentration. Any concentration of sulphate ions above this value would result in the precipitation of barium sulphate salt as the product of concentration would exceed the solubility product.
[Ba2+][SO42−]=Ksp(BaSO4)[B{a^{2 + }}][S{O_4}^{2 - }] = {K_{sp}}(BaS{O_4})
Putting in the values of barium ion concentration and solubility product at given temperature,
[SO42−]=Ksp(BaSO4)[Ba2+]=1.0×10−90.01=10−7[S{O_4}^{2 - }] = \dfrac{{{K_{sp}}(BaS{O_4})}}{{[B{a^{2 + }}]}} = \dfrac{{1.0 \times {{10}^{ - 9}}}}{{0.01}} = {10^{ - 7}}

Hence, the correct option is (c)

Note:
The addition of sulphuric acid to an aqueous solution of barium sulphate salt affects its solubility due to the common ion effect. The excess of sulphate ions released by sulphuric acid increases the total sulphate ion concentration in the solution and the equilibrium shifts backwards i.e. the ions begin to recombine to give the parent salt which results in the precipitation.