Given that the solubility product constant (K_{sp}) of C_2D_3 is 9.14 × 10^{-9}, what is the molar solubility (S) of C_2D_3 in pure water?

• A. S = 1.32 × 10^{-3} mol/L
• B. S = 1.32 × 10^{-3} mol/L
• C. S = 1.32 × 10^{-2} mol/L
• D. S = 1.32 × 10^{-2} mol/L

Answer: (A)

Verified step by step guidance

1. Write the dissociation equation for the compound C_2D_3 in water. Assume it dissociates into its ions as: C_2D_3 (s) \rightarrow aA^{m+} + bB^{n-}, where a and b are the stoichiometric coefficients of the ions produced. Identify these coefficients based on the formula of C_2D_3.
2. Express the solubility product constant (K_{sp}) in terms of the molar solubility (S). If the compound dissociates into a moles of cation and b moles of anion, then the ion concentrations at equilibrium are aS and bS respectively. The expression for K_{sp} is: \(K_{sp} = [A^{m+}]^{a} \times [B^{n-}]^{b} = (aS)^{a} \times (bS)^{b}\).
3. Substitute the known value of K_{sp} = 9.14 \times 10^{-9} into the expression and simplify it to a single equation in terms of S. This will typically be of the form \(K_{sp} = k \times S^{a+b}\), where k is a constant derived from the stoichiometric coefficients.
4. Solve the equation for S by isolating it: \(S = \left( \frac{K_{sp}}{k} \right)^{\frac{1}{a+b}}\). This step involves taking the appropriate root depending on the total number of ions produced.
5. Calculate the numerical value of S using the above expression to find the molar solubility of C_2D_3 in pure water.