Question

A\(_3\)B\(_2\) is a sparingly soluble salt of molar mass M (g mol\(^{-1}\)) and solubility \(x\text{ g L}^{-1}\). The solubility product satisfies \(K_{sp} = a \left(\frac{x}{M}\right)^5\). The value of \(a\) is \(\dots\dots\dots\). (Integer answer)

Hint:

For a general salt \(A_xB_y\), the solubility product is \(K_{sp} = x^x y^y S^{(x+y)}\). For \(A_3B_2\), \(K_{sp} = 3^3 2^2 S^5 = 108 S^5\).

Answer 1

Correct Answer: 108

Step 1: Understanding the Concept:
The solubility product constant (\(K_{sp}\)) is the equilibrium constant for the dissolution of a sparingly soluble ionic compound. We must first convert the mass solubility (\(x\text{ g/L}\)) to molar solubility (\(S\text{ mol/L}\)).
Step 2: Key Formula or Approach:
\[ \text{Molar solubility } S = \frac{\text{Solubility in g/L}}{\text{Molar mass in g/mol}} = \frac{x}{M} \]
Step 3: Detailed Explanation:
The dissociation of salt \(\text{A}_3\text{B}_2\) is:
\[ \text{A}_3\text{B}_2(s) \rightleftharpoons 3\text{A}^{2+}(aq) + 2\text{B}^{3-}(aq) \]
If the molar solubility is \(S\), the concentrations at equilibrium are:
\[ [\text{A}^{2+}] = 3S \]
\[ [\text{B}^{3-}] = 2S \]
Expression for \(K_{sp}\):
\[ K_{sp} = [\text{A}^{2+}]^3 [\text{B}^{3-}]^2 = (3S)^3 (2S)^2 \]
\[ K_{sp} = (27S^3) (4S^2) = 108 S^5 \]
Substituting \(S = \frac{x}{M}\):
\[ K_{sp} = 108 \left(\frac{x}{M}\right)^5 \]
Comparing with the given form \(K_{sp} = a \left(\frac{x}{M}\right)^5\), we find \(a = 108\).
Step 4: Final Answer:
The value of \(a\) is 108.