Question

The molar solubility(s) of zirconium phosphate with molecular formula \( \text{Zr}^{4+} \text{PO}_4^{3-} \) is given by relation:

• A. \(\left( \frac{K_{sp}}{9612} \right)^{\frac{1}{3}}\)
• B. \(\left( \frac{K_{sp}}{6912} \right)^{\frac{1}{7}}\)
• C. \(\left( \frac{K_{sp}}{5348} \right)^{\frac{1}{6}}\)
• D. \(\left( \frac{K_{sp}}{8435} \right)^{\frac{1}{7}}\)

Hint:

When calculating the solubility product, remember that the powers of the concentrations depend on the stoichiometry of the dissociation reaction. In this case, the molar solubility raised to the appropriate powers yields the Ksp expression.

Answer 1

The Correct Option is A

Step 1: The solubility product constant \( \text{K}_{sp} \) for zirconium phosphate is defined by the equilibrium:

\[\text{Zr}_3(PO_4)_4 \rightleftharpoons 3\text{Zr}^{4+} + 4\text{PO}_4^{3-}\]

Step 2: If the molar solubility of zirconium phosphate is \( s \), the ion concentrations are:

\[\text{[Zr}^{4+}\text{]} = 3s, \quad \text{[PO}_4^{3-}\text{]} = 4s\]

Step 3: The \( \text{K}_{sp} \) expression is:

\[\text{K}_{sp} = [\text{Zr}^{4+}]^3 [\text{PO}_4^{3-}]^4 = (3s)^3 (4s)^4\]

Step 4: Simplifying the expression yields:

\[\text{K}_{sp} = 27s^3 \cdot 256s^4 = 6912s^7\]

Step 5: Therefore, the molar solubility expression is \( \text{K}_{sp} = 6912s^7 \), corresponding to option (1).