JEE Mains · Chemistry · STD 11 - 6.2. Equilibrium - II (icon Equilibrium)
The solubility product constants of \(Ag_2CrO_4\) and \(AgBr\) are \(32x\) and \(4y\) respectively at \(298\) K. The value of \(\left(\dfrac{\text{molarity of } Ag_2CrO_4}{\text{molarity of } AgBr}\right)\) 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 & Solution
Correct Answer
(D) \(\dfrac{\sqrt[3]{x}}{\sqrt{y}}\)
Step-by-step Solution
Detailed explanation
Let the solubility of \(Ag_2CrO_4\) be \(S_1\). For \(Ag_2CrO_4 \rightleftharpoons 2Ag^+ + CrO_4^{2-}\) \(K_{sp} = (2S_1)^2(S_1) = 4S_1^3\) \(4S_1^3 = 32x \Rightarrow S_1^3 = 8x \Rightarrow S_1 = 2\sqrt[3]{x}\) Let the solubility of \(AgBr\) be \(S_2\). For…
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