Question

The solubility of sparingly soluble salts \(\text{MX}_1\), \(\text{MX}_2\) and \(\text{MX}_3\) is \(1\times 10^{-3}\) mol/L. Hence, their respective solubility products are

• A. \(1\times 10^{-6}\), \(4\times 10^{-9}\) and \(27\times 10^{-12}\)
• B. \(1\times 10^{-9}\), \(4\times 10^{-9}\) and \(32\times 10^{-12}\)
• C. \(1\times 10^{-9}\), \(8\times 10^{-8}\) and \(32\times 10^{-12}\)
• D. \(1\times 10^{-6}\), \(8\times 10^{-8}\) and \(27\times 10^{-12}\)

Hint:

Use Ksp = n^n s^(n+1) for each salt with s = 1e-3.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The solubility product \(K_{sp}\) is the product of ion concentrations raised to their coefficients at saturation. If the solubility is \(s\) mol/L, each ion concentration is its coefficient times \(s\).

Step 2: Key Formula:
For \(MX_n \rightleftharpoons M^{n+} + nX^-\), we get \([M^{n+}] = s\) and \([X^-] = ns\). So \(K_{sp} = s\,(ns)^n = n^n s^{n+1}\).

Step 3: MX (n = 1):
\[ K_{sp} = s^2 = (10^{-3})^2 = 1\times 10^{-6} \]

Step 4: MX2 (n = 2):
\[ K_{sp} = 4s^3 = 4\times (10^{-3})^3 = 4\times 10^{-9} \]

Step 5: MX3 (n = 3):
\[ K_{sp} = 27s^4 = 27\times (10^{-3})^4 = 27\times 10^{-12} \]

Step 6: Compare with options:
The sequence \(1\times 10^{-6}\), \(4\times 10^{-9}\), \(27\times 10^{-12}\) matches option (A). Options (B) and (C) use \(10^{-9}\) for the first salt, which is wrong because \(s^2\) gives \(10^{-6}\). Option (D) has \(8\times 10^{-8}\), which would come from squaring instead of cubing.

Final Answer:
The three values are 1e-6, 4e-9 and 27e-12, option (A). \[ \boxed{\text{(A)}\ 1\times 10^{-6},\ 4\times 10^{-9},\ 27\times 10^{-12}} \]