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Question: In a saturated solution of the sparingly soluble strong electrolyte \[AgI{O_3}\](molecular mass =283...

In a saturated solution of the sparingly soluble strong electrolyte AgIO3AgI{O_3}(molecular mass =283) the equilibrium which sets in isAgIO3⇌Ag+(aq)+IO3−(aq)AgI{O_3} \rightleftharpoons A{g^ + }(aq) + I{O_3}^ - (aq). If the solubility product constant Ksp{K_{sp}} of AgIO3AgI{O_3} at a given temperature is 1.0×10−81.0 \times {10^{ - 8}}, what is the mass of AgIO3AgI{O_3} contained in 100 ml of its saturated solution?
A. 28.3×10−2g28.3 \times {10^{ - 2}}g
B. 2.83×10−3g2.83 \times {10^{ - 3}}g
C. 1.0×10−7g1.0 \times {10^{ - 7}}g
D. 1.0×10−4g1.0 \times {10^{ - 4}}g

Explanation

Solution

Try to recall that solubility product is defined as the product of molar concentrations of ions in a saturated solution raised to their stoichiometric coefficients. Now, by using this you can easily find the correct option from the given ones.

Complete step by step answer:
It is known to you that AgIO3AgI{O_3} is a sparingly soluble salt and AgIO3AgI{O_3} ionizes completely in the solution as: AgIO3→Ag++IO3−AgI{O_3} \to A{g^ + } + I{O_3}^ - .

Calculation:
Given, Ksp{K_{sp}}= 1.0×10−81.0 \times {10^{ - 8}}--------1
On complete ionization: AgIO3→Ag++IO3−AgI{O_3} \to A{g^ + } + I{O_3}^ - .

So, let the solubility of [Ag+]=[IO3−]=s\left[ {A{g^ + }} \right] = \left[ {I{O_3}^ - } \right] = s.

AgI{O_3} \to A{g^ + } + I{O_3}^ - \\\ {\text{s 0 0}} \\\ {\text{0 s s}} \\\ \end{gathered} $$. Therefore, $${K_{sp}} = \left( s \right)\left( s \right) = {s^2}$$---------2 Equating eq. 1 and 2 we get, $$\begin{gathered} {s^2} = 1.0 \times {10^{ - 8}} \\\ or,s = {10^{ - 4}}mol/L \\\ \end{gathered} $$. So, $$\left[ {AgI{O_3}} \right] = s = {10^{ - 4}}mol/L$$ Given, volume of solution =100 ml=0.1L Let the number of moles of $$AgI{O_3}$$ be n. $$\begin{gathered} \dfrac{n}{{0.1}} = {10^{ - 4}} \\\ or,n = {10^{ - 5}} \\\ \end{gathered} $$ Also, given molar mass of $$AgI{O_3}$$= 283 Let the mass of $$AgI{O_3}$$ be x. $$\begin{gathered} \dfrac{x}{{283}} = {10^{ - 5}} \\\ or,x = 283 \times {10^{ - 5}} \\\ or,x = 2.83 \times {10^{ - 3}}g \\\ \end{gathered} $$ _Hence, from the above calculation we can easily conclude that option B is the correct option to the given question._ **Note:** It should be remembered to you that if to the solution of a weak electrolyte which ionizes to a small extent, a strong electrolyte having a common ion is added which ionizes almost completely, the ionization of weak electrolyte is further suppressed.Similarly, if the solution of a sparingly soluble salt if a soluble salt having a common ion is added, the solubility of the sparingly soluble salt further decreases.