Equilibrium Question 826
Question: In a saturated solution of the sparingly soluble strong electrolyte $ AgIO_3 $ (molecular mass = 283) the equilibrium which sets is $ AgIO_3(s)\rightarrow A{g^{+}}(aq)+IO_3^{-}(aq). $ If the solubility product constant $ K _{sp} $ of $ AgIO_3 $ at a given temperature is $ 1.0\times {10^{-8}} $ , what is the mass of $ AgIO_3 $ contained in 100 mL of its saturated solution'
Options:
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. $
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Answer:
Correct Answer: C
Solution:
Let s = solubility $ AgIO_3\rightarrow \underset{s}{\mathop{A{g^{+}}}},+\underset{s}{\mathop{IO_3^{-}}}, $
$ K _{sp}=[A{g^{+}}][IO_3^{-}]=s\times s=s^{2} $ Given $ K _{sp}=1\times {10^{-8}} $
$ \therefore s=\sqrt{K _{sp}}=\sqrt{1\times {10^{-8}}} $
$ =1.0\times {10^{-4}}mol/lit=1.0\times {10^{-4}}\times 283g/lit $ ( $ \therefore $ Molecular mass of $ AgIO_3=283 $ ) $ =\frac{1.0\times {10^{-4}}\times 283\times 100}{1000}g/100mL $
$ =2.83\times {10^{-3}}g/100mL $
