Question

The solubility product constants of \(Ag_2CrO_4\) and \(AgBr\) are \(32x\) and \(4y\) respectively at 298 K. The value of} \[ \left(\frac{\text{molarity of } Ag_2CrO_4}{\text{molarity of } AgBr}\right) \] can be expressed as:

• A. \(\dfrac{2\sqrt[3]{x}}{y}\)
• B. \(2\sqrt{\dfrac{x}{y}}\)
• C. \(\sqrt{\dfrac{x}{y}}\)
• D. \(\dfrac{\sqrt[3]{x}}{\sqrt{y}}\)

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The molarity of a salt in its saturated solution is equivalent to its solubility (\(s\)).
Solubility depends on the solubility product constant (\(K_{sp}\)) and the stoichiometry of the salt.
Step 2: Key Formula or Approach:
For a salt of the type \(A_2B\), \(K_{sp} = 4s^3\).
For a salt of the type \(AB\), \(K_{sp} = s^2\).
Step 3: Detailed Explanation:
1. For Silver Chromate (\(Ag_2CrO_4\)):
The dissociation is: \(Ag_2CrO_4(s) \rightleftharpoons 2Ag^+(aq) + CrO_4^{2-}(aq)\).
Let its solubility be \(s_1\).
\[ K_{sp} = [Ag^+]^2 [CrO_4^{2-}] = (2s_1)^2 (s_1) = 4s_1^3 \]
Given \(K_{sp} = 32x\):
\[ 4s_1^3 = 32x \implies s_1^3 = 8x \implies s_1 = 2x^{1/3} = 2\sqrt[3]{x} \]

2. For Silver Bromide (\(AgBr\)):
The dissociation is: \(AgBr(s) \rightleftharpoons Ag^+(aq) + Br^-(aq)\).
Let its solubility be \(s_2\).
\[ K_{sp} = [Ag^+][Br^-] = (s_2)(s_2) = s_2^2 \]
Given \(K_{sp} = 4y\):
\[ s_2^2 = 4y \implies s_2 = 2y^{1/2} = 2\sqrt{y} \]

3. Calculation of the Ratio:
\[ \text{Ratio} = \frac{\text{molarity of } Ag_2CrO_4}{\text{molarity of } AgBr} = \frac{s_1}{s_2} \]
\[ \text{Ratio} = \frac{2\sqrt[3]{x}}{2\sqrt{y}} = \frac{\sqrt[3]{x}}{\sqrt{y}} \]
Step 4: Final Answer:
The ratio of the molarities is \(\frac{\sqrt[3]{x}}{\sqrt{y}}\).