Question

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

• A.
• B.
• C.
• D.

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

Solution for Solubility of Zirconium Phosphate

Consider the salt zirconium phosphate with the molecular formula:

\(\text{Zr}_3(\text{PO}_4)_4\)

This salt dissociates into 3 zirconium cations (\(\text{Zr}^{4+}\)) with a charge of +4 and 4 phosphate anions (\(\text{PO}_4^{3-}\)) with a charge of -3.

The concentration of the zirconium cation is:

\([\text{Zr}^{4+}] = 3S \end{p}

The concentration of the phosphate anion is:

\([\text{PO}_4^{3-}] = 4S\)

The solubility product constant \(K_{sp}\) is given by:

\( K_{sp} = (3S)^3 (4S)^4 = 6912(S)^7 \)

Solving for \(S\), the solubility:

\( S = \left( \frac{K_{sp}}{3^3 \times 4^4} \right)^{1/7} = \left( \frac{K_{sp}}{6912} \right)^{1/7} \)