Ionic EquilibriumMedium
Question
The solubility product constants of ${Ag}_{2}{CrO}_{4}$ and AgBr are 32 x and 4 y respectively at 298 K .
The value of $\left( \frac{\text{~}\text{molarity of}\text{~}{Ag}_{2}{CrO}_{4}}{\text{~}\text{molarity of}\text{~}AgBr} \right)$ can be expressed as :
Options
A.$\frac{2\sqrt[3]{x}}{y}$
B.$2\sqrt{\frac{x}{y}}$
C.$\sqrt{\frac{x}{y}}$
D.$\frac{\sqrt[3]{x}}{\sqrt{y}}$
Solution
Sol. [Key Concept: Write $K_{sp}$ in terms of solubility for each salt, then take the molarity ratio.]
For ${Ag}_{2}C_{2}O_{4}$
$${4{\text{ }S}_{1}\ ^{3} = 32x }{\Rightarrow S_{1} = 2x^{1/3} }$$For AgBr
$${S_{2}\ ^{2} = 4y }{\Rightarrow S_{2} = 2y^{1/2} }$$Ratio $\left( \frac{S_{1}}{{\text{ }S}_{2}} \right) = \frac{\sqrt[3]{x}}{\sqrt{y}}$
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