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

Concept: Solubility product relates to solubility \(s\). For \(AgBr\): \[ AgBr \rightleftharpoons Ag^+ + Br^- \] \[ K_{sp} = s^2 \] \[ s = \sqrt{K_{sp}} \]
Step 1:Solubility of \(AgBr\)} \[ K_{sp} = 4y \] \[ s_{AgBr} = \sqrt{4y} \] \[ = 2\sqrt{y} \]
Step 2:Solubility of \(Ag_2CrO_4\)} \[ Ag_2CrO_4 \rightleftharpoons 2Ag^+ + CrO_4^{2-} \] If solubility \(=s\) \[ K_{sp} = (2s)^2(s) \] \[ =4s^3 \] Given \[ 4s^3 = 32x \] \[ s^3 = 8x \] \[ s = 2\sqrt[3]{x} \]
Step 3:Take ratio} \[ \frac{s_{Ag_2CrO_4}}{s_{AgBr}} = \frac{2\sqrt[3]{x}}{2\sqrt{y}} \] \[ = \sqrt{\frac{x}{y}} \] Thus \[ \boxed{2\sqrt{\frac{x}{y}}} \]