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Question: The solubility product of \(BaS{O_4}\) is \(4 \times {10^{ - 10}}\) . The solubility of \(BsS{O_4}\)...

The solubility product of BaSO4BaS{O_4} is 4×10−104 \times {10^{ - 10}} . The solubility of BsSO4BsS{O_4} in presence of 0.02N0.02N H2SO4{H_2}S{O_4} will be:
A.4×10−8M4 \times {10^{ - 8}}M
B.2×10−8M2 \times {10^{ - 8}}M
C.2×10−5M2 \times {10^{ - 5}}M
D.2×10−4M2 \times {10^{ - 4}}M

Explanation

Solution

First calculate the molarity H2SO4{H_2}S{O_4} by using the equation Molarity=Normality2Molarity = \dfrac{{Normality}}{2} . Assuming that there are S moles of BaSO4BaS{O_4} we will have S moles of Ba2+B{a^{2 + }} ions and S moles of SO42−SO_4^{2 - } ions. So solubility of BaSO4BaS{O_4} is given by the equation Ksp=[Ba2+][SO42−]{K_{sp}} = \left[ {B{a^{2 + }}} \right]\left[ {SO_4^{2 - }} \right] . Solving this equation, you get Ksp=S2+0.01S{K_{sp}} = {S^2} + 0.01S , you can neglect S2{S^2} as it is small and calculate the result.

Complete answer:
Let’s start by discussing molarity and normality first.
Molarity - Molar concentration (also called molarity, amount concentration, or substance concentration) is a measure of the concentration of a chemical species, in particular of a solute in a solution, in terms of the amount of substance per unit volume of solution. The most commonly used unit for molarity is moles/liter.
Normality – A way to measure solvent concentration is normality (N). It is similar to molarity, but in its expression of the solute quantity in a liter (L) of the solution, it uses the gram-equivalent weight of a liquid rather than the gram molecular weight expressed in molarity. A 1N solution contains 1 gram of solute equal weight per liter of solution.
Molarity=NormalityNo. of  H+ion in H2SO4Molarity = \dfrac{{Normality}}{{No.{\text{ }}of\;{H^ + }ion{\text{ }}in{\text{ }}{H_2}S{O_4}}}
So, the molarity of H2SO4=0.022=0.01M{H_2}S{O_4} = \dfrac{{0.02}}{2} = 0.01M
So, when H2SO4{H_2}S{O_4} is added we add 0.01M0.01M SO42−SO_4^{2 - } ions
The solubility reaction is represented below –
BaSO4(s)⇆Ba2++SO42−BaS{O_4}\left( s \right)\overset {} \leftrightarrows B{a^{2 + }} + SO_4^{2 - }
If we consider that there are S moles of BaSO4BaS{O_4} , then we will have S moles of Ba2+B{a^{2 + }} ions and S moles of SO42−SO_4^{2 - } ions.
So, the total moles of SO42−SO_4^{2 - } ions =(0.01+S)moles = \left( {0.01 + S} \right)moles
We know by the law of solubility that the total solubility of BaSO4BaS{O_4} is equal to the product of solubility of Ba2+B{a^{2 + }} ions and the solubility of SO42−SO_4^{2 - } ions.
∴Ksp=[Ba2+][SO42−]\therefore {K_{sp}} = \left[ {B{a^{2 + }}} \right]\left[ {SO_4^{2 - }} \right]
Ksp=S×(0.01+S){K_{sp}} = S \times \left( {0.01 + S} \right)
Ksp=S2+0.01S{K_{sp}} = {S^2} + 0.01S
Given in the problem, Ksp=4×10−10{K_{sp}} = 4 \times {10^{ - 10}}
S2+0.01S=4×10−10{S^2} + 0.01S = 4 \times {10^{ - 10}}
Since S mole is small, S2{S^2} being the product of two small quantities and can be neglected
0.01S=4×10−100.01S = 4 \times {10^{ - 10}}
S=4×10−8MS = 4 \times {10^{ - 8}}M
Hence, 4×10−8M4 \times {10^{ - 8}}M will be dissolved when we add 0.02N0.02N H2SO4{H_2}S{O_4}

Hence, option A is the correct choice.

Note:
Solubility is defined as a property of a substance called solute to get dissolved in a solvent to form a solution. The solubility of ionic compounds (which dissociate to form cations and anions) in water varies a great deal. Some compounds are highly soluble and may even absorb moisture from the atmosphere whereas others are highly insoluble.