Question

\[Pb{(I{O_3})_2}\] is a sparingly soluble salt \[({K_{sp}} = 2.6 \times {10^{ - 13}})\]. To 35ml of 0.15M \[Pb{(N{O_3})_2}\] solution, 15ml of 0.8M \[KI{O_3}\] solution is added and a precipitate of \[Pb{(I{O_3})_2}\] is formed. What will be the molarity of \[I{O_3}^ - \] ions in the solution after completion of the reaction?
A) 0.152
B) 0.081
C) 0.41
D) 0.03

Answer 1

Hint: We have to know the reaction mechanism involved in the reaction.
We have to calculate the amount of reactants required to complete the reaction in equilibrium condition.

Complete step by step answer:
In the question, it is given that, \[Pb{(I{O_3})_2}\] is a sparingly soluble salt whose solubility product is \[({K_{sp}} = 2.6 \times {10^{ - 13}})\].
When lead nitrate [\[Pb{(N{O_3})_2}\]] reacts with potassium iodate [\[KI{O_3}\]], it will give lead iodate [\[Pb{(I{O_3})_2}\]]. When a reaction takes place, at first the compounds dissociate into different ions and react with each other to give new products.
Here, lead nitrate dissociates into lead ion and nitrate ion. We can write the respective equation as,
\[Pb{(N{O_3})_2} \rightleftharpoons P{b^{2 + }} + 2NO_3^ - \]
This implies that one mole of lead nitrate forms one mole of lead ion and two moles of nitrate ion. According to the question, volume of \[Pb{(N{O_3})_2}\] is \[35ml\] and \[0.15M\]of \[Pb{(N{O_3})_2}\] takes part in the reaction. So, the amount of lead ion produced from lead nitrate\[ = 35ml \times 0.15M = 5.25mM\].
Potassium iodate dissociates into potassium ion and iodate ion. We can write the respective equation as,
\[KI{O_3} \rightleftharpoons {K^ + } + IO_3^ - \]
This implies that one mole of potassium iodate forms one mole potassium ion and one mole iodate ion. According to the question, the volume of \[KI{O_3}\] is \[15ml\] and \[0.8M\] of \[KI{O_3}\] takes part in the reaction.
So, the amount of iodate ion produced from potassium iodate\[ = 15ml \times 0.8M = 12mM\].
Now, lead ion and iodate ion react with each other to give lead iodate, which can be written as,
\[P{b^{2 + }} + 2IO_3^ - \rightleftharpoons Pb{(I{O_3})_2}\]
That means to complete the reaction for one mole of lead ion, two moles of iodate ion is required.
After dissociation, we have 5.25mM lead ion and 12mM iodate ion present in the solution.
For 5.25mM lead ion we require \[5.25mM \times 2 = 10.5mM\] iodate ion to complete the reaction.
So, the excess amount of iodate ion remains unreacted= 12mM - 10.5mM = 1.5mM.
The total volume of the solution = 35ml + 15ml = 50ml.
So, the molarity of iodate ion remains unreacted\[ = \dfrac{{1.5mM}}{{50ml}} = 0.03M\].

Hence, the correct answer is (D) 0.03.

Note: In this question, the value of solubility product is of no use. It is given here only for additional information regarding the sparingly soluble salt.
When a sparingly soluble salt is dissolved in water, the associated equilibrium constant is known as the solubility product (\[{K_{sp}}\]).
Molarity is the number of moles of solute present in one litre of solution.
\[1mM = {10^{ - 3}}M\]