Question

Calculate the solubility of solid \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\left(K_{\mathrm{sp}}=1.3 \times 10^{-32}\right)\) in a \(0.20-M \mathrm{Na}_{3} \mathrm{PO}_{4}\) solution.

Short answer

The solubility of \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) in a \(0.20\text{-M}\ \mathrm{Na}_{3} \mathrm{PO}_{4}\) solution is approximately \(6.59\times10^{-12}\text{ M}\).

Step-by-step solution

Write the balanced dissolution reaction

The balanced dissolution reaction for \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) is: \[ \mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2(s)} \rightleftarrows 3\,\mathrm{Ca^{2+}_{(aq)}}+2\,\mathrm{PO}_{4}^{3-_{(aq)}}. \]

Express the solubility product constant

The solubility product constant (\(K_{sp}\)) expression for the dissolution of \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) is: \[ K_{sp}=[\mathrm{Ca^{2+}}]^{3}[\mathrm{PO}_{4}^{3-}]^{2}, \] where \([\mathrm{Ca^{2+}}]\) is the concentration of \(\mathrm{Ca^{2+}}\) ions, and \([\mathrm{PO}_{4}^{3-}]\) is the concentration of \(\mathrm{PO}_{4}^{3-}\) ions in the solution.

Set up an ICE table

To determine the changes in concentration, set up an ICE table with the initial, change, and equilibrium concentrations of the ions involved: ``` Ca3(PO4)2 Ca^2+ PO4^3- Initial S 0.00M 0.20M Change -S +3xS -2xS Equilibrium S-9xS 3xS 0.20M-2xS ``` The value of \(S\) represents the solubility (in molarity) of \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) in the solution, and \(x\) represents the change in its concentration.

Substitute the equilibrium concentrations into the solubility product expression

Substitute the equilibrium concentrations from the ICE table into the \(K_{sp}\) expression: \[K_{sp}=(3x\text{S})^{3}(0.20\text{M}-2x\text{S})^{2}.\] Since the \(K_{sp}\) value of \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) is given as \(1.3\times10^{-32}\), we get: \[1.3\times10^{-32}=(3x\text{S})^{3}(0.20\text{M}-2x\text{S})^{2}.\]

Solve for the solubility

Solving the above equation for \(x\) involves simplifying, then iterating with approximations, and checking them: We take the following approximation: Since \(K_{sp}\) is very small, we can assume that \(S\) is also very small. In this case, we can approximate that \(2x\text{S}\) is negligible compared to \(0.20\text{M}\): \[1.3\times10^{-32}\approx(3x\text{S})^{3}(0.20\text{M})^{2}.\] Now we can solve for \(x\): \[x\approx6.59\times10^{-12}\text{M}.\] Finally, we calculate that the solubility of \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) in \(0.20\text{-M}\ \mathrm{Na}_{3} \mathrm{PO}_{4}\) solution is approximately \(6.59\times10^{-12}\text{ M}\).

Key concepts

Chemical Equilibrium

Understanding chemical equilibrium is vital when learning about solubility products. It's the state in which the rate of the forward reaction (the dissolution of a solid into ions) equals the rate of the backward reaction (the reformation of the solid from its ions). At equilibrium, the concentrations of the reactants and products remain constant. The solubility product constant (\( K_{sp} \)) is a special type of equilibrium constant that applies to the dissolution of sparingly soluble ionic compounds.

An important aspect to consider is that the presence of a common ion, like \( \text{PO}_{4}^{3-} \) in the provided exercise, will affect the equilibrium position due to Le Chatelier's principle. This principle states that if you change the conditions of a reaction in equilibrium, the system will respond to counteract that change. In this case, the addition of \( \text{Na}_{3} \text{PO}_{4} \) will shift the equilibrium to the left, hence decreasing the solubility of the \( \text{Ca}_{3}(\text{PO}_{4})_{2} \) in solution.

ICE Table

The ICE table is a tool used to keep track of the Initial concentrations, the Change in concentrations, and the Equilibrium concentrations of species in a chemical reaction. It's short for Initial, Change, Equilibrium, which are the stages of the reaction you need to consider. This tool is exceedingly useful in solving equilibrium problems because it helps you visualize the changes occurring throughout the reaction.

In our exercise, the initial concentration of \( \text{Ca}^{2+} \) is zero and of \( \text{PO}_{4}^{3-} \) is 0.20M because of the added \( \text{Na}_{3} \text{PO}_{4} \) solution. Assuming a solubility (\( S \) ) for the solid, when it starts dissolving, we indicate the decrease of solid and increase of ions with a change (\( x \) ) in concentration. By understanding this system, we realize that the expression for solubility product constant can be written using the equilibrium concentrations of the dissolved ions.

Solubility Calculations

When tackling solubility calculations, it's important to express the relationship between solubility and the solubility product constant. The steps followed in the provided solution serve to simplify the process of computing the solubility of a sparingly soluble salt in a solution already containing a common ion.

The approximation that was made in step 5 – ignoring \( 2xS \) compared to 0.20M in the solubility product expression – might seem small, but is crucial for solving equations with very small constants like \( K_{sp} \). Simplifications like these are often necessary in chemistry to make problems solvable without a computer. After making the approximation, we end up with an equation that is much more manageable and can provide us with a numerical solution for the solubility of the compound, in this case, \( \text{Ca}_{3}(\text{PO}_{4})_{2} \).

It is essential for students to understand the relationship between the solubility product and the ion concentrations at equilibrium – and to remember that the product of these concentrations at equilibrium equals the constant \( K_{sp} \), so long as the system is at equilibrium and the approximations made are valid.