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: Write each dissolution:
$MX_1$ gives $M^+ + X^-$. $MX_2$ gives $M^{2+} + 2X^-$. $MX_3$ gives $M^{3+} + 3X^-$.

Step 2: Plug in $s = 10^{-3}$:
First salt: $[M][X] = s \cdot s$.
Second salt: $[M][X]^2 = s\cdot(2s)^2$.
Third salt: $[M][X]^3 = s\cdot(3s)^3$.

Step 3: Evaluate:
$s^2 = 10^{-6}$. Next, $4s^3 = 4\times 10^{-9}$. Last, $27s^4 = 27\times 10^{-12}$.

Step 4: Match:
Only option (A) lists these three numbers in this order. The others change either the power or the numerical factor.

Final Answer:
Matching the three computed values picks option (A). \[ \boxed{A} \]