Question

If 'S' is the solubility of $X_{3}Y_{2}$ in pure water, the solubility product ($K_{sp}$) is ________.

• A. $27~S^{4}$
• B. $108~S^{5}$
• C. $108~S^{2}$
• D. $27~S^{6}$
• E. $27~S^{2}$

Hint:

For $A_xB_y$, $K_{sp} = x^x y^y S^{(x+y)}$.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
The solubility product, \(K_{sp}\), is an equilibrium constant that describes the extent to which a sparingly soluble ionic compound dissolves in a solvent. It is defined in terms of the equilibrium concentrations of the ions. The molar solubility, S, is the number of moles of the compound that can dissolve per liter of solution. We need to derive the relationship between \(K_{sp}\) and S for the given generic salt \(X_2Y_3\).
Step 2: Key Formula or Approach
For a generic salt \(A_mB_n\), the dissolution equilibrium is \(A_mB_n(s) \rightleftharpoons mA^{n+}(aq) + nB^{m-}(aq)\). The solubility product expression is \(K_{sp} = [A^{n+}]^m [B^{m-}]^n\). We will relate the ion concentrations to the molar solubility S based on the stoichiometry of the dissolution.
Step 3: Detailed Explanation
1. Write the dissolution equilibrium equation.
For the salt \(X_2Y_3\), it dissociates into 2 cations and 3 anions. Let the cation be \(X^{3+}\) and the anion be \(Y^{2-}\) to ensure charge neutrality. \[ X_2Y_3(s) \rightleftharpoons 2X^{3+}(aq) + 3Y^{2-}(aq) \] 2. Relate ion concentrations to molar solubility (S).
If S moles of \(X_2Y_3\) dissolve per liter, then according to the stoichiometry of the reaction, the equilibrium concentrations of the ions will be: - \([X^{3+}] = 2 \times S = 2S\) - \([Y^{2-}] = 3 \times S = 3S\) 3. Write the \(K_{sp}\) expression and substitute the concentrations.
The expression for the solubility product is: \[ K_{sp} = [X^{3+}]^2 [Y^{2-}]^3 \] Now, substitute the expressions in terms of S: \[ K_{sp} = (2S)^2 (3S)^3 \] 4. Simplify the expression.
\[ K_{sp} = (4S^2) (27S^3) \] \[ K_{sp} = (4 \times 27) (S^2 \times S^3) \] \[ K_{sp} = 108 S^{2+3} \] \[ K_{sp} = 108 S^5 \] Step 4: Final Answer
The solubility product, \(K_{sp}\), is \(108S^5\).