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

For Ag$_2$CrO$_4$ (1), \( K_{sp} = [Ag^+]^2[CrO_4^{2-}] = (2s)^2(s) = 4s^3 \). Given \( 4s^3 = 32x \), so \( s = 2\sqrt[3]{x} \).

For AgBr (2), \( K_{sp} = [Ag^+][Br^-] = (t)(t) = t^2 \). Given \( t^2 = 4y \), so \( t = 2\sqrt{y} \).

Ratio \( s/t = 2\sqrt[3]{x} / 2\sqrt{y} = \sqrt[3]{x} / \sqrt{y} \). According to given options, we identify the match.