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 given by the relation: \[ \text{Zr}_3(PO_4)_4 \rightleftharpoons 3\text{Zr}^{4+} + 4\text{PO}_4^{3-} \] Step 2: The molar solubility \( s \) of zirconium phosphate will lead to the following concentration expressions: \[ \text{[Zr}^{4+}\text{]} = 3s, \quad \text{[PO}_4^{3-}\text{]} = 4s \] Step 3: The expression for \( \text{K}_{sp} \) is: \[ \text{K}_{sp} = [\text{Zr}^{4+}]^3 [\text{PO}_4^{3-}]^4 = (3s)^3 (4s)^4 \] Step 4: Simplifying this expression: \[ \text{K}_{sp} = 27s^3 \cdot 256s^4 = 6912s^7 \] Step 5: Thus, the molar solubility expression is \( \text{K}_{sp} = 6912s^7 \), which corresponds to option (1).