Problem 99 - Chemistry

Question

Assuming that the solubility of \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)\) is \(1.6 \times 10^{-7} \mathrm{mol} / \mathrm{L}\) at \(25^{\circ} \mathrm{C},\) calculate the \(K_{\mathrm{sp}}\) for this salt. Ignore any potential reactions of the ions with water.

Step-by-Step Solution

Answer

The solubility product constant \(K_{sp}\) for \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) at \(25^{\circ} \mathrm{C}\) is approximately \(1.47 \times 10^{-26}\).

Step 1: 1. Write the dissolution equation

First, we write the equation for the dissolution of calcium phosphate in water: \[\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s) \rightleftharpoons 3\mathrm{Ca}^{2+}(aq) + 2\mathrm{PO}_{4}^{3-}(aq)\]

Step 2: 2. Define the solubility and solubility product

We are given the solubility of calcium phosphate, which is the concentration of the ions in a saturated solution at equilibrium. Let the solubility be represented by 's', which is given as \(1.6 \times 10^{-7} \mathrm{mol}/\mathrm{L}\). The solubility product \(K_{sp}\) is defined as the product of the concentrations of the ions in a saturated solution at equilibrium: \[K_{sp} = [\mathrm{Ca}^{2+}]^3 [\mathrm{PO}_{4}^{3-}]^2\]

Step 3: 3. Calculate ion concentrations

We see that for each mole of \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) that dissolves, 3 moles of \(\mathrm{Ca}^{2+}\) and 2 moles of \(\mathrm{PO}_{4}^{3-}\) are formed. From the solubility, we can find the concentration of each ion: \[[\mathrm{Ca}^{2+}] = 3s = 3(1.6 \times 10^{-7}\,\mathrm{mol}/\mathrm{L}) = 4.8 \times 10^{-7} \,\mathrm{mol}/\mathrm{L}\] \[[\mathrm{PO}_{4}^{3-}] = 2s = 2(1.6 \times 10^{-7} \,\mathrm{mol}/\mathrm{L}) = 3.2 \times 10^{-7} \,\mathrm{mol}/\mathrm{L}\]

Step 4: 4. Calculate the solubility product \(K_{sp}\)

Now we can plug these concentrations back into the equation for the solubility product: \[K_{sp} = (4.8 \times 10^{-7})^3 (3.2 \times 10^{-7})^2\] With a calculator, we determine the value of \(K_{sp}\): \[K_{sp} = 1.47 \times 10^{-26}\] The solubility product constant \(K_{sp}\) for \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) at \(25^{\circ} \mathrm{C}\) is approximately \(1.47 \times 10^{-26}\).

Key Concepts

Calcium Phosphate Solubility

Calcium phosphate is a compound that has limited solubility in water. Solubility refers to the maximum amount of solute that can dissolve in a solvent at a specific temperature to form a saturated solution. In this case, we are dealing with calcium phosphate \(\text{Ca}_3(\text{PO}_4)_2\) . This complex salt separates into its individual ions, calcium \(\text{Ca}^{2+}\) and phosphate \(\text{PO}_4^{3-}\), when dissolved in water. The process reaches a point called equilibrium, where the rate of dissolution equals the rate of precipitation.In our example, the solubility of calcium phosphate is given as \(1.6 \times 10^{-7} \text{mol/L}\). This value indicates how much of the compound can dissolve before the solution becomes saturated. Understanding the solubility is crucial because it informs us about the concentration of ions present in the solution, which further helps in calculating the solubility product constant, \(K_{sp}\).

Ion Concentration Calculation

The ion concentration in a solution formed by dissolving calcium phosphate is derived by recognizing the stoichiometry from its dissociation reaction. When \(\text{Ca}_3(\text{PO}_4)_2\) dissolves, it produces 3 moles of \(\text{Ca}^{2+}\) ions and 2 moles of \(\text{PO}_4^{3-}\) ions per mole of calcium phosphate.Given the solubility (s) is \(1.6 \times 10^{-7} \text{mol/L}\), we can calculate the concentration of each ion: - \([\text{Ca}^{2+}] = 3s = 3 \times 1.6 \times 10^{-7} = 4.8 \times 10^{-7} \text{mol/L}\)- \([\text{PO}_4^{3-}] = 2s = 2 \times 1.6 \times 10^{-7} = 3.2 \times 10^{-7} \text{mol/L}\)These concentrations are key to determining the solubility product constant, \(K_{sp}\), because they reflect the saturation levels of ions in the solution once equilibrium has been established.

Solubility Equilibrium

Solubility equilibrium is a dynamic state where the rate of dissolution of a solute matches the rate of precipitation. For calcium phosphate, this equilibrium is depicted by the reaction: \(\text{Ca}_3(\text{PO}_4)_2 \rightleftharpoons 3\text{Ca}^{2+} + 2\text{PO}_4^{3-}\).At this balance point, the concentration of ions remains steady, even if their individual particles are constantly dissolving and precipitating. The equilibrium state is crucial when calculating the solubility product constant, \(K_{sp}\).The solubility product constant is defined as: \[K_{sp} = [\text{Ca}^{2+}]^3 [\text{PO}_4^{3-}]^2\]Using the calculated concentrations of \(\text{Ca}^{2+}\) and \(\text{PO}_4^{3-}\), the \(K_{sp}\) of calcium phosphate is determined: \[K_{sp} = (4.8 \times 10^{-7})^3 \times (3.2 \times 10^{-7})^2 \approx 1.47 \times 10^{-26}\]This \(K_{sp}\) value helps us quantify the solubility equilibrium, representing the extent to which the salt can dissolve before reaching an undissolved and dissolved particle balance.