To find the solubility product constant ([tex]\(K_{sp}\)[/tex]) of the salt [tex]\(AB_3\)[/tex], we need to follow these steps:
1. Determine the molar solubility of [tex]\(AB_3\)[/tex]:
- The molar mass of [tex]\(AB_3\)[/tex] is given as 289 g/mol.
- The solubility of [tex]\(AB_3\)[/tex] is given as 3.20 g/L.
We can convert the solubility from grams per liter to moles per liter (molar solubility) by using the molar mass:
[tex]\[
\text{Molar Solubility} = \frac{\text{Solubility (g/L)}}{\text{Molar Mass (g/mol)}} = \frac{3.20\ \text{g/L}}{289\ \text{g/mol}} \approx 0.0110727\ \text{mol/L}
\][/tex]
2. Write the dissociation equation and express solubility in terms of its ions:
The dissociation of [tex]\(AB_3\)[/tex] can be represented as:
[tex]\[
AB_3(s) \rightleftharpoons A^{3+}(aq) + 3B^-(aq)
\][/tex]
From the stoichiometry of the dissociation:
- The concentration of [tex]\(A^{3+}\)[/tex] ions will be equal to the molar solubility of [tex]\(AB_3\)[/tex], which is [tex]\(0.0110727\ \text{mol/L}\)[/tex].
- The concentration of [tex]\(B^-\)[/tex] ions will be three times the molar solubility, which is [tex]\(3 \times 0.0110727\ \text{mol/L} \approx 0.033218\ \text{mol/L}\)[/tex].
3. Calculate the solubility product constant [tex]\(K_{sp}\)[/tex]:
The solubility product constant [tex]\(K_{sp}\)[/tex] is given by the product of the equilibrium concentrations of the ions, each raised to the power of their stoichiometric coefficients:
[tex]\[
K_{sp} = [A^{3+}] \times [B^-]^3
\][/tex]
Substituting the ion concentrations:
[tex]\[
K_{sp} = (0.0110727) \times (0.033218)^3
\][/tex]
Evaluating this expression:
[tex]\[
K_{sp} \approx 4.0586 \times 10^{-7}
\][/tex]
Therefore, the solubility product constant [tex]\(K_{sp}\)[/tex] of [tex]\(AB_3\)[/tex] at [tex]\(25^{\circ}C\)[/tex] is:
[tex]\[
K_{sp} \approx 4.06 \times 10^{-7}
\][/tex]
