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

To ascertain the molar solubility of zirconium phosphate, \( \text{ZrPO}_4 \), we first examine its dissolution in water.

The dissolution equilibrium of zirconium phosphate is expressed as:

\[\text{ZrPO}_4 \left(s\right) \rightleftharpoons \text{Zr}^{4+} \left(aq\right) + \text{PO}_4^{3-} \left(aq\right)\]

For simplification, assuming each mole of zirconium phosphate yields one mole of \(\text{Zr}^{4+}\) and one mole of \(\text{PO}_4^{3-}\) ions upon dissolution, we proceed.

The solubility product constant (\(K_{\text{sp}}\)) for this reaction is defined using molar solubility \(s\) as:

\[\text{K}_{\text{sp}} = \left[\text{Zr}^{4+}\right]\left[\text{PO}_4^{3-}\right]\]

At equilibrium, the concentrations are:

\[\left[\text{Zr}^{4+}\right] = s\]

and 

\[\left[\text{PO}_4^{3-}\right] = s\]

Consequently,

\[K_{\text{sp}} = s^{2}\]

Solving for molar solubility \(s\) in terms of \(K_{\text{sp}}\) yields:

\[s = \sqrt{K_{\text{sp}}}\]

The correct expression for the molar solubility \((s)\) of zirconium phosphate among the provided options is: