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 determine the molar solubility of zirconium phosphate, let's first understand the dissolution process of zirconium phosphate, \( \text{ZrPO}_4 \), in water. 

The dissolution of zirconium phosphate can be represented by the following equilibrium equation:

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

Given the complexity of zirconium often being highly hydrolysable and not forming simple ion complexes easily, let's focus on the simplified attribute: each mole of zirconium phosphate dissolves to give one mole of \(\text{Zr}^{4+}\) ions and one mole of \(\text{PO}_4^{3-}\) ions into the solution.

The equilibrium constant for this reaction, often referred to as the solubility product constant \(K_{\text{sp}}\), is expressed in terms of molar solubility \(s\) as follows:

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

At equilibrium:

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

 and 

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

Thus, 

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

To solve for the molar solubility \(s\) in terms of \(K_{\text{sp}}\), we take the square root of both sides:

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

<.p>From the given options, the correct expression for the molar solubility \((s)\) of zirconium phosphate matches the option:

 

Answer 2

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:

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

katex.render("\\text{Zr}_3(\\text{PO}_4)_4", document.body); katex.render("[\\text{Zr}^{4+}] = 3S", document.body); katex.render("[\\text{PO}_4^{3-}] = 4S", document.body); katex.render("K_{sp} = (3S)^3 (4S)^4 = 6912(S)^7", document.body); katex.render("S = \\left( \\frac{K_{sp}}{3^3 \\times 4^4} \\right)^{1/7} = \\left( \\frac{K_{sp}}{6912} \\right)^{1/7}", document.body);