Question

K$_{sp}$ of Ag$_2$CrO$_4$ = 32x

K$_{sp}$ of AgBr = 4y Then, the ratio of molarity (solubility) of (1) to (2) is:

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

Hint:

Relate solubility 's' to Ksp using stoichiometry: \( K_{sp} = 4s^3 \) for \( A_2B \) and \( K_{sp} = s^2 \) for \( AB \).

Answer 1

The Correct Option is A

To find the ratio of molarity (solubility) of Ag₂CrO₄ to AgBr, we start by calculating the molar solubility of each compound from their given \(K_{sp}\) values.

1. For silver chromate, Ag₂CrO₄:
   • The dissociation is given by: Ag₂CrO₄ ⇌ 2Ag⁺ + CrO₄^2-
   • If 's' is the molar solubility of Ag₂CrO₄, then at equilibrium: [Ag⁺] = 2s and [CrO₄^2-] = s.
   • \(K_{sp} = [Ag^+]^2[CrO_4^{2-}] = (2s)^2 \times s = 4s^3\)
   • Given \(K_{sp}\) for Ag₂CrO₄ = 32x, so \(4s^3 = 32x \Rightarrow s^3 = 8x \Rightarrow s = \sqrt[3]{8x} = 2\sqrt[3]{x}\).
2. For silver bromide, AgBr:
   • The dissociation is given by: AgBr ⇌ Ag⁺ + Br^-
   • If 's' is the molar solubility of AgBr, then at equilibrium: [Ag⁺] = s and [Br^-] = s.
   • \(K_{sp} = [Ag^+][Br^-] = s^2\)
   • Given \(K_{sp}\) for AgBr = 4y, so \(s^2 = 4y \Rightarrow s = \sqrt{4y} = 2\sqrt{y}\).
3. The ratio of the molar solubility of Ag₂CrO₄ to AgBr is given by:

\(\frac{s_{\text{Ag}_2\text{CrO}_4}}{s_{\text{AgBr}}} = \frac{2\sqrt[3]{x}}{2\sqrt{y}} = \frac{\sqrt[3]{x}}{\sqrt{y}} \times \frac{2}{1} = \frac{2\sqrt[3]{x}}{\sqrt{y}} \times \frac{1}{1} = \frac{2\sqrt[3]{x}}{y}\)

Therefore, the correct answer is \(\frac{2 \sqrt[3]{x}}{y}\).