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Question: The solubility product of \(BaS{O_4}\) at \(25 \circ C\) is \(1.0 \times {10^{ - 9}}\) .What would b...

The solubility product of BaSO4BaS{O_4} at 25∘C25 \circ C is 1.0×10−91.0 \times {10^{ - 9}} .What would be the concentration of H2SO4{H_2}S{O_4} required to precipitate BaSO4BaS{O_4} from a solution of 0.01M0.01M Ba2+B{a^{2 + }} ions.

Explanation

Solution

BaSO4BaS{O_4} is the chemical formula for Barium sulphate.It’s almost insoluble in water at room temperature and is used as oil well drilling fluid. H2SO4{H_2}S{O_4} is the chemical formula for Sulphuric acid which is a very strong mineral acid.

Complete answer:
For the above given question let us analyse the given compound Barium sulphate.We know that it is formed of two ions-Barium and sulphate.The chemical reaction for the same is given as
BaSO4→Ba2++SO42−BaS{O_4} \to B{a^{2 + }} + SO_4^{2 - } .In the above question it is given that the solubility product for this reaction at 25∘C25 \circ C is 1.0×10−91.0 \times {10^{ - 9}} ,i.e Ksp=1.0×10−9{K_{sp}} = 1.0 \times {10^{ - 9}}
But for the above reaction Ksp=[Ba2+][SO42−]{K_{sp}} = [B{a^{2 + }}][SO_4^{2 - }]
Substituting the values which we have obtained from the above question.
10−9=[Ba2+][SO42−]{10^{ - 9}} = [B{a^{2 + }}][SO_4^{2 - }]
But the value of [Ba2+][B{a^{2 + }}] is given as 0.01.Therefore substituting the value in the equation we get.
10−9=0.01×[SO42−]{10^{ - 9}} = 0.01 \times [SO_4^{2 - }]
Taking the constant on one side and others one one side we get,
10−90.01=[SO42−]\frac{{{{10}^{ - 9}}}}{{0.01}} = [SO_4^{2 - }]
Rearranging the terms and solving it we can get the concentration of sulphate ions.
[SO42−]=10−7[SO_4^{2 - }] = {10^{ - 7}}
Hence we now know that we need 10−7M{10^{ - 7}}M concentration of H2SO4{H_2}S{O_4} to precipitate BaSO4BaS{O_4} from a solution of 0.01M0.01M Ba2+B{a^{2 + }} ions.

Note:
The solubility product constant can be defined as the equilibrium constant for the dissolution of a solid substance into an aqueous solution.The symbol Ksp{K_{sp}} is used to denote the solubility product of an given chemical equation.The general formula for it is given as,
Ksp=[A+]a[B−]b{K_{sp}} = {[{A^ + }]^a}{[{B^ - }]^b} where A+{A^ + } and B−{B^ - } are the cations and anions respectively
For the above given equation,
A= Ba2+B{a^{2 + }} ,B= SO42−SO_4^{2 - } and a and b are equal to 1.