Question
In a saturated solution of the sparingly soluble strong electrolyte $$AgI{O_3}$$ (molecular mass = 283) the equilibrium which sets in is $$AgI{O_{3\left( s \right)}} \rightleftharpoons A{g^ + }_{\left( {aq} \right)} + IO_{3\left( {aq} \right)}^ - .$$ If the solubility
product constant $${K_{sp}}$$ of $$AgI{O_3}$$ at a given temperature is $$1.0 \times {10^{ - 8}},$$ what is the mass of $$AgI{O_3}$$ contained in $$100 ml$$ of
its saturated solution?
A.
$$1.0 \times {10^{ - 4}}g$$
B.
$$28.3 \times {10^{ - 2}}g$$
C.
$$2.83 \times {10^{ - 3}}g$$
D.
$$1.0 \times {10^{ - 7}}g$$
Answer :
$$2.83 \times {10^{ - 3}}g$$
Solution :
$${\text{Let}}\,s = {\text{solubility}}$$
\[AgI{{O}_{3}}\rightleftharpoons {{\underset{s}{\mathop{Ag}}\,}^{+}}+\underset{s}{\mathop{I{{O}_{3}}^{-}}}\,\]
$$\eqalign{
& {K_{sp}} = \left[ {A{g^ + }} \right]\left[ {I{O_3}^ - } \right] = s \times s = {s^2} \cr
& {\text{Given}}\,\,{K_{sp}} = 1 \times {10^{ - 8}} \cr
& \therefore \,\,s = \sqrt {{K_{sp}}} = \sqrt {1 \times {{10}^{ - 8}}} \cr
& = 1.0 \times {10^{ - 4}}\,mol/lit = 1.0 \times {10^{ - 4}} \times 283\,g/lit \cr
& \left( {\because \,\,{\text{Molecular mass of}}\,Ag\,I{O_3} = 283} \right) \cr
& = \frac{{1.0 \times {{10}^{ - 4}} \times 283 \times 100}}{{1000}}gm/100\,ml \cr
& = 2.83 \times {10^{ - 3}}gm/100\,ml \cr} $$