Question

Use the following data to calculate the \(K_{\mathrm{sp}}\) value for each solid. a. The solubility of \(\mathrm{Pb}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) is \(6.2 \times 10^{-12} \mathrm{mol} / \mathrm{L}\). b. The solubility of \(\mathrm{Li}_{2} \mathrm{CO}_{3}\) is \(7.4 \times 10^{-2} \mathrm{mol} / \mathrm{L}\).

Short answer

The \(K_{\mathrm{sp}}\) value for: a) \(\mathrm{Pb}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) is \(2.47 \times 10^{-60}\). b) \(\mathrm{Li}_{2} \mathrm{CO}_{3}\) is \(6.44 \times 10^{-5}\).

Step-by-step solution

a. Calculating \(K_{\mathrm{sp}}\) for \(\mathrm{Pb}_{3}\left(\mathrm{PO4}\right)_{2}\)

Step 1: Write the balanced dissolution equation. \[ \mathrm{Pb}_{3}\left(\mathrm{PO}_{4}\right)_{2(s)} \rightarrow 3\mathrm{Pb}^{2+}_{(aq)} + 2\mathrm{PO}_{4}^{3-}_{(aq)} \] Step 2: Write the \(K_{\mathrm{sp}}\) expression. \[K_{\mathrm{sp}} = [\mathrm{Pb}^{2+}]^{3}[\mathrm{PO}_{4}^{3-}]^{2}\] Step 3: Using the given solubility value, plug into the \(K_{\mathrm{sp}}\) expression Let the molar solubility of Pb3(PO4)2 be x. \[\begin{aligned} [\mathrm{Pb}^{2+}] &= 3x \\ [\mathrm{PO}_{4}^{3-}] &= 2x \end{aligned}\] Given that the solubility of \(\mathrm{Pb}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) is \(6.2 \times 10^{-12} \mathrm{mol} / \mathrm{L}\), we have: \[x = 6.2 \times 10^{-12}\] Step 4: Calculate the \(K_{\mathrm{sp}}\) value \[K_{\mathrm{sp}} = (3x)^{3}(2x)^{2} = (3(6.2 \times 10^{-12}))^{3}(2(6.2 \times 10^{-12}))^{2}\] \[K_{\mathrm{sp}} = 2.47 \times 10^{-60}\]

b. Calculating \(K_{\mathrm{sp}}\) for \(\mathrm{Li}_{2} \mathrm{CO}_{3}\)

Step 1: Write the balanced dissolution equation. \[\mathrm{Li}_{2}\mathrm{CO_{3(s)}} \rightarrow 2\mathrm{Li}^{+}_{(aq)} + \mathrm{CO}_{3}^{2-}_{(aq)}\] Step 2: Write the \(K_{\mathrm{sp}}\) expression. \[K_{\mathrm{sp}} = [\mathrm{Li}^{+}]^{2}[\mathrm{CO}_{3}^{2-}]\] Step 3: Using the given solubility value, plug into the \(K_{\mathrm{sp}}\) expression Let the molar solubility of Li2CO3 be y. \[\begin{aligned} [\mathrm{Li}^{+}] &= 2y \\ [\mathrm{CO}_{3}^{2-}] &= y \end{aligned}\] Given that the solubility of \(\mathrm{Li}_{2} \mathrm{CO}_{3}\) is \(7.4 \times 10^{-2} \mathrm{mol} / \mathrm{L}\), we have: \[y = 7.4 \times 10^{-2}\] Step 4: Calculate the \(K_{\mathrm{sp}}\) value \[K_{\mathrm{sp}} = (2y)^{2}(y) = (2(7.4 \times 10^{-2}))^{2}(7.4 \times 10^{-2})\] \[K_{\mathrm{sp}} = 6.44 \times 10^{-5}\] The \(K_{\mathrm{sp}}\) value for: a) \(\mathrm{Pb}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) is \(2.47 \times 10^{-60}\). b) \(\mathrm{Li}_{2} \mathrm{CO}_{3}\) is \(6.44 \times 10^{-5}\).

Key concepts

Understanding Chemical Solubility

Chemical solubility refers to the ability of a substance to dissolve in a solvent, forming a homogeneous mixture at a specific temperature and pressure. When it comes to ionic solids such as salts, solubility involves dissociation into ions. For example, when lead phosphate, \(\mathrm{Pb}_{3}\left(\mathrm{PO}_{4}\right)_{2}\), dissolves, it separates into lead ions, \(\mathrm{Pb}^{2+}\), and phosphate ions, \(\mathrm{PO}_{4}^{3-}\). The quantitative measure of solubility is often given as the maximum amount of solute that can dissolve in a unit volume of solvent, commonly expressed in moles per liter (mol/L).

In the exercise provided, we see the solubility of \(\mathrm{Pb}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) and \(\mathrm{Li}_{2}\mathrm{CO}_{3}\) in terms of how many moles can dissolve to form a saturated solution. A saturated solution is one in which no more solute can dissolve at the given temperature and pressure, and it's at this point that equilibrium is established between the solid and its respective ions in solution.

Ionic Product and its Role in Solubility

The ionic product is the product of the concentrations of the ions of a solute in a saturated solution. It's crucially important because it determines whether a solution is unsaturated, saturated, or supersaturated with respect to the solute. An unsaturated solution has an ionic product less than the solubility product constant (\(K_{\mathrm{sp}}\)), while a supersaturated solution has an ionic product greater than the \(K_{\mathrm{sp}}\).

When we calculate \(K_{\mathrm{sp}}\), we're finding the point at which the solution is just saturated. For instance, in the given exercise for \(\mathrm{Pb}_{3}\left(\mathrm{PO}_{4}\right)_{2}\), the ionic product at solubility is \[\left(3x\right)^{3}\left(2x\right)^{2}\], where x is the molar solubility. This equation represents the point of equilibrium where the rate at which \(\mathrm{Pb}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) dissolves is equal to the rate at which the ions come together and form the solid. The solubility, and therefore the ionic product, will differ for different substances and under different conditions.

Equilibrium Constant and Solubility Product Constant

The equilibrium constant (\(K_{\mathrm{eq}}\)) in a chemical reaction quantifies the ratio of the concentration of the products to the concentrations of the reactants once the reaction has reached equilibrium. In the context of solubility, the specific type of equilibrium constant we use is the solubility product constant (\(K_{\mathrm{sp}}\)), which applies to the dissolution of sparingly soluble salts. It is unique for every solid substance and dependent on temperature.

The \(K_{\mathrm{sp}}\) expression is derived from the balanced equation for the dissolution of the ionic solid. For example, in the dissolution of \(\mathrm{Li}_{2}\mathrm{CO}_{3}\) into \(\mathrm{Li}^{+}\) and \(\mathrm{CO}_{3}^{2-}\), the \(K_{\mathrm{sp}}\) expression is \[\left(\mathrm{Li}^{+}\right)^{2}\left(\mathrm{CO}_{3}^{2-}\right)\], reflecting the stoichiometry of the solid's dissociation. Knowing the \(K_{\mathrm{sp}}\) value enables chemists to predict whether a precipitate will form under certain conditions, which has practical applications in fields such as chemistry and environmental science.