Express Ratio of Molarities of Silver Chromate and Silver Bromide

To find the ratio of the molarities (molar solubilities) of $\text{Ag}_2\text{CrO}_4$ and $\text{AgBr}$ in their respective saturated solutions, we analyze the solubility product expressions for both salts separately.

1. For Silver Chromate ($\text{Ag}_2\text{CrO}_4$):

The dissolution equilibrium of $\text{Ag}_2\text{CrO}_4$ in water is represented as: $\text{Ag}_2\text{CrO}_4 (s) \rightleftharpoons 2\text{Ag}^+ (aq) + \text{CrO}_4^{2-} (aq)$

Let the molarity (solubility) of $\text{Ag}_2\text{CrO}_4$ be $S_1$. Then, at equilibrium: $[\text{Ag}^+] = 2S_1$ $[\text{CrO}_4^{2-}] = S_1$

The solubility product constant ($K_{sp1}$) is given by: $K_{sp1} = [\text{Ag}^+]^2 [\text{CrO}_4^{2-}] = (2S_1)^2 (S_1) = 4S_1^3$

Given $K_{sp1} = 32x$: $4S_1^3 = 32x$ $S_1^3 = 8x$ $S_1 = \sqrt[3]{8x} = 2\sqrt[3]{x}$

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2. For Silver Bromide ($\text{AgBr}$):

The dissolution equilibrium of $\text{AgBr}$ in water is represented as: $\text{AgBr} (s) \rightleftharpoons \text{Ag}^+ (aq) + \text{Br}^- (aq)$

Let the molarity (solubility) of $\text{AgBr}$ be $S_2$. Then, at equilibrium: $[\text{Ag}^+] = S_2$ $[\text{Br}^-] = S_2$

The solubility product constant ($K_{sp2}$) is given by: $K_{sp2} = [\text{Ag}^+] [\text{Br}^-] = S_2 \cdot S_2 = S_2^2$

Given $K_{sp2} = 4y$: $S_2^2 = 4y$ $S_2 = \sqrt{4y} = 2\sqrt{y}$

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3. Ratio of Molarities:

Now, taking the ratio of the molarity of $\text{Ag}_2\text{CrO}_4$ to the molarity of $\text{AgBr}$: $\frac{\text{molarity of Ag}_2\text{CrO}_4}{\text{molarity of AgBr}} = \frac{S_1}{S_2} = \frac{2\sqrt[3]{x}}{2\sqrt{y}} = \frac{\sqrt[3]{x}}{\sqrt{y}}$

Thus, the correct option is D.