Question

The solubility product of $BaS{O_4}$ is $4 \times {10^{ - 10}}$ . The solubility of $BsS{O_4}$ in presence of $0.02N$ ${H_2}S{O_4}$ will be:
A.$4 \times {10^{ - 8}}M$
B.$2 \times {10^{ - 8}}M$
C.$2 \times {10^{ - 5}}M$
D.$2 \times {10^{ - 4}}M$

Answer 1

Hint: First calculate the molarity ${H_2}S{O_4}$ by using the equation $Molarity = \dfrac{{Normality}}{2}$ . Assuming that there are S moles of $BaS{O_4}$ we will have S moles of $B{a^{2 + }}$ ions and S moles of $SO_4^{2 - }$ ions. So solubility of $BaS{O_4}$ is given by the equation ${K_{sp}} = \left[ {B{a^{2 + }}} \right]\left[ {SO_4^{2 - }} \right]$ . Solving this equation, you get ${K_{sp}} = {S^2} + 0.01S$ , you can neglect ${S^2}$ 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.
$Molarity = \dfrac{{Normality}}{{No.{\text{ }}of\;{H^ + }ion{\text{ }}in{\text{ }}{H_2}S{O_4}}}$
So, the molarity of ${H_2}S{O_4} = \dfrac{{0.02}}{2} = 0.01M$
So, when ${H_2}S{O_4}$ is added we add $0.01M$ $SO_4^{2 - }$ ions
The solubility reaction is represented below –
$BaS{O_4}\left( s \right)\overset {} \leftrightarrows B{a^{2 + }} + SO_4^{2 - }$
If we consider that there are S moles of $BaS{O_4}$ , then we will have S moles of $B{a^{2 + }}$ ions and S moles of $SO_4^{2 - }$ ions.
So, the total moles of $SO_4^{2 - }$ ions $ = \left( {0.01 + S} \right)moles$
We know by the law of solubility that the total solubility of $BaS{O_4}$ is equal to the product of solubility of $B{a^{2 + }}$ ions and the solubility of $SO_4^{2 - }$ ions.
$\therefore {K_{sp}} = \left[ {B{a^{2 + }}} \right]\left[ {SO_4^{2 - }} \right]$
${K_{sp}} = S \times \left( {0.01 + S} \right)$
${K_{sp}} = {S^2} + 0.01S$
Given in the problem, ${K_{sp}} = 4 \times {10^{ - 10}}$
${S^2} + 0.01S = 4 \times {10^{ - 10}}$
Since S mole is small, ${S^2}$ being the product of two small quantities and can be neglected
$0.01S = 4 \times {10^{ - 10}}$
$S = 4 \times {10^{ - 8}}M$
Hence, $4 \times {10^{ - 8}}M$ will be dissolved when we add $0.02N$ ${H_2}S{O_4}$

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.