Question

Teeth and bones are composed, to a first approximation, of calcium phosphate, \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s) .\) The \(K_{\mathrm{sp}}\) for this salt is \(1.3 \times 10^{-32}\) at \(25 \mathrm{C}\). Calculate the concentration of calcium ion in a saturated solution of \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\).

Short answer

The concentration of calcium ions in a saturated solution of calcium phosphate is approximately \(2.20 \times 10^{-7} \, M\).

Step-by-step solution

Write the balanced dissolution equation.

First, write the balanced equation for the dissolution of calcium phosphate: \[Ca_3(PO_4)_2 (s) \rightleftharpoons 3Ca^{2+} (aq) + 2PO_4^{3-} (aq)\]

Create the solubility product constant expression.

Next, write the expression for the solubility product constant (Ksp) in terms of the concentrations of the ions in the solution: \[K_{sp} = [Ca^{2+}]^3[PO_4^{3-}]^2\] We are given that \(K_{sp} = 1.3 \times 10^{-32}\).

Define concentration variables.

Let's denote the molar solubility of \(Ca_3(PO_4)_2\) as \(s\). Since there are three moles of \(Ca^{2+}\) ions produced for every mole of \(Ca_3(PO_4)_2\) that dissolves, we will have a concentration of \(3s\) for \(Ca^{2+}\). Similarly, the concentration of \(PO_4^{3-}\) ions will be \(2s\).

Substitute the concentration variables into the Ksp expression.

Replace the concentrations in the Ksp expression with the variables defined in Step 3: \[(1.3 \times 10^{-32}) = (3s)^3(2s)^2\]

Solve for the molar solubility, s.

Multiply and simplify the equation before solving for \(s\): \[(1.3 \times 10^{-32}) = 108s^5\] \[s = \sqrt[5]{\frac{1.3 \times 10^{-32}}{108}} \approx 7.34 \times 10^{-8} \, M\]

Calculate the concentration of calcium ions.

We have found the molar solubility (\(s\)) of \(Ca_3(PO_4)_2\), and the concentration of \(Ca^{2+}\) ions is \(3s\). Multiply the value of \(s\) by 3 to obtain the calcium ion concentration: \[3s = 3(7.34 \times 10^{-8} \, M) \approx 2.20 \times 10^{-7} \, M\] The concentration of calcium ions in a saturated solution of calcium phosphate is approximately \(2.20 \times 10^{-7} \, M\).

Key concepts

Ksp Calculation

Understanding the solubility product constant, or Ksp, is fundamental for predicting how much of a substance will dissolve in water to create a saturated solution. The Ksp value indicates just how much product formation is energetically favorable at equilibrium.

For a salt such as calcium phosphate, \(Ca_{3}(PO_{4})_{2}\), the Ksp calculation begins with the establishment of a dissolution equation. Take the Ksp value given, \(1.3 \times 10^{-32}\), and relate it to the concentrations of the ions at equilibrium, considering their stoichiometric ratios from the dissolution equation. Multiply the concentrations of the products raised to their coefficients in the balanced chemical equation. The smaller the Ksp, the less soluble the compound is; hence, calcium phosphate has very low solubility in water.

Molar Solubility

Molar solubility, represented by \(s\), refers to the number of moles of a compound that can dissolve in one liter of solution at equilibrium. It helps us determine the concentration of the ions in a saturated solution.

From the Ksp expression, you can calculate this value by introducing the stoichiometry of the dissociation process into the equation, as seen where calcium ions have a coefficient of 3 and phosphate ions have a coefficient of 2. The algebraic manipulation involves solving for \(s\) to find how many moles per liter dissolve before the system reaches equilibrium and no more solid can dissolve. With a compound like calcium phosphate, the molar solubility is very low due to its low Ksp value, reflecting its limited solubility in water.

Dissolution Equations

Dissolution equations represent the breakdown of solid compounds into their aqueous ionic constituents. They are essential for understanding how salts dissociate in water, which is crucial in the study of solubility. For calcium phosphate, the dissolution equation is presented as:
\[Ca_3(PO_4)_2 (s) \rightleftharpoons 3Ca^{2+} (aq) + 2PO_4^{3-} (aq)\]

Every mole of calcium phosphate produces three moles of calcium ions \(Ca^{2+}\) and two moles of phosphate ions \(PO_4^{3-}\). Establishing and balancing this equation enables you to grasp the stoichiometry, which directly contributes to the calculation of Ksp and to interpreting the molar solubility of the substance. Through these equations, you can also visualize the dynamic of ions being released into solution and the equilibrium which establishes the limit of this process.