JEE Advanced · Chemistry · 8. Ionic Equilibrium
Solubility product constant \(\left(K_{\mathrm{sp}}\right)\) of salts of types \(M X, M X_2\) and \(M_3 X\) at temperature ' \(T\) ' are \(4.0 \times 10^{-8}, 3.2 \times 10^{-14}\) and \(2.7 \times 10^{-15}\), respectively. Solubilities \(\left(\mathrm{mol} \mathrm{dm}^{-3}\right)\) of the salts at temperature ' \(T\) ' are in the order
- A \(M X>M X_2>M_3 X\)
- B \(M_3 X>M X_2>M X\)
- C \(M X_2>M_3 X < M X\)
- D \(M X>M_3 X>M X_2\)
Answer & Solution
Correct Answer
(D) \(M X>M_3 X>M X_2\)
Step-by-step Solution
Detailed explanation
\(\begin{array}{lcl}\text { Salt } & \text { Solubility product } & \text { Solubility } \\ M X & S_1^2=4.0 \times 10^{-8} & S_1=2 \times 10^{-4} \\ M X_2 & 4 S_2^3=3.2 \times 10^{-14} & S_2=2 \times 10^{-5} \\ M_3 X & 27 S_3^4=2.7 \times 10^{-15} & S_3=1 \times 10^{-4}\end{array}\)
Thus, solubility order \(=M X>M_3 X>M X_2\).
Thus, solubility order \(=M X>M_3 X>M X_2\).
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