Problem 31 - Chemistry

Question

Calculate the solubility of each of the following compounds in moles per liter. Ignore any acid–base properties. a. \(A g_{3} P O_{4}, K_{s p}=1.8 \times 10^{-18}\) b. \(\mathrm{CaCO}_{3}, K_{\mathrm{sp}}=8.7 \times 10^{-9}\) c. \(\mathrm{Hg}_{2} \mathrm{Cl}_{2}, K_{\mathrm{sp}}=1.1 \times 10^{-18}\left(\mathrm{Hg}_{2}^{2+} \right.\) is the cation in is the cation in solution.\()\)

Step-by-Step Solution

Answer

The solubility of the compounds is as follows: a. \(Ag_3PO_4\): \(1.32 \times 10^{-5}\, mol/L\) b. \(CaCO_3\): \(2.95 \times 10^{-5}\, mol/L\) c. \(Hg_2Cl_2\): \(3.0 \times 10^{-7}\, mol/L\)

Step 1: Write the balanced chemical equation and the Ksp expression.

The balanced chemical equation for the dissolution of \(Ag_{3}PO_{4}\) is: \[Ag_3PO_4(s) \rightleftharpoons 3Ag^+(aq) + PO^{3-}_4(aq)\] Now, we can write the Ksp expression as follows: \[K_{sp}= [Ag^+]^3 [PO_4^{3-}]\]

Step 2: Define solubility in terms of molar concentrations and express Ksp in terms of solubility.

Let S be the solubility of \(Ag_3PO_4\) in moles per liter. When 1 mole of \(Ag_3PO_4\) dissolves, it produces 3 moles of \(Ag^+\) and 1 mole of \(PO_4^{3-}\) ions. Therefore, the molar concentrations of the ions in the solution at equilibrium are: \[ [Ag^+] = 3S \] \[ [PO_4^{3-}] = S \] Now we can substitute these expressions into the Ksp expression from step 1: \[1.8 \times 10^{-18} = (3S)^3(S)\]

Step 3: Solve for solubility (S).

Now, we need to solve the equation for S: \[1.8 \times 10^{-18} = 27S^4\] \[S^4 = \frac{1.8 \times 10^{-18}}{27}\] \[S = \sqrt[4]{6.67 \times 10^{-20}}\] \[S = 1.32 \times 10^{-5}\] The solubility of \(Ag_3PO_4\) is approximately \(1.32 \times 10^{-5}\, mol/L\). b. Calculate the solubility of \(CaCO_3\), with \(K_{sp} = 8.7 \times 10^{-9}\).

Step 1: Write the balanced chemical equation and the Ksp expression.

The balanced chemical equation for the dissolution of \(CaCO_3\) is: \[CaCO_3(s) \rightleftharpoons Ca^{2+}(aq) + CO^{2-}_3(aq)\] Now, we can write the Ksp expression as follows: \[K_{sp} = [Ca^{2+}][CO_3^{2-}]\]

Step 2: Define solubility in terms of molar concentrations and express Ksp in terms of solubility.

The molar concentrations of the ions in the solution at equilibrium are: \[ [Ca^{2+}] = S \] \[ [CO_3^{2-}] = S \] Now we can substitute these expressions into the Ksp expression from step 1: \[8.7 \times 10^{-9} = S^2\]

Step 3: Solve for solubility (S).

Now, we need to solve the equation for S: \[S = \sqrt{8.7 \times 10^{-9}}\] \[S = 2.95 \times 10^{-5}\] The solubility of \(CaCO_3\) is approximately \(2.95 \times 10^{-5}\, mol/L\). c. Calculate the solubility of \(Hg_2Cl_2\), with \(K_{sp} = 1.1 \times 10^{-18}\).

Step 1: Write the balanced chemical equation and the Ksp expression.

The balanced chemical equation for the dissolution of \(Hg_2Cl_2\) is: \[Hg_2Cl_2(s) \rightleftharpoons Hg_2^{2+}(aq) + 2Cl^-(aq)\] Now, we can write the Ksp expression as follows: \[K_{sp} = [Hg_2^{2+}][Cl^-]^2\]

Step 2: Define solubility in terms of molar concentrations and express Ksp in terms of solubility.

The molar concentrations of the ions in the solution at equilibrium are: \[ [Hg_2^{2+}] = S \] \[ [Cl^-] = 2S \] Now we can substitute these expressions into the Ksp expression from step 1: \[1.1 \times 10^{-18} = S(2S)^2\]

Step 3: Solve for solubility (S).

Now, we need to solve the equation for S: \[1.1 \times 10^{-18} = 4S^3\] \[S^3 = \frac{1.1 \times 10^{-18}}{4}\] \[S = \sqrt[3]{2.75 \times 10^{-19}}\] \[S = 3.0 \times 10^{-7}\] The solubility of \(Hg_2Cl_2\) is approximately \(3.0 \times 10^{-7}\, mol/L\).

Key Concepts

Chemical Equilibrium

Chemical equilibrium is a fundamental concept in chemistry that occurs when a chemical reaction reaches a state where the concentrations of reactants and products remain unchanged over time. In the context of solubility, it applies to the dissolution of ionic compounds in a solvent, typically water.
When an ionic compound dissolves, its ions are initially released into the solution. Eventually, the rate of ions dissolving equals the rate at which they precipitate back into the solid, reaching equilibrium.
Understanding chemical equilibrium helps predict how much of a substance will dissolve in a solution. It is essential for calculating the solubility of compounds using the solubility product constant, or Ksp.

Molar Solubility

Molar solubility represents the maximum number of moles of a solute that can dissolve in one liter of a solvent to form a saturated solution. It is typically represented by the symbol S and is expressed in moles per liter (mol/L).
In the process of finding molar solubility, we first write the balanced chemical equation of a compound's dissolution. For example, for the compound silver phosphate, the equation is: \[Ag_3PO_4(s) \rightleftharpoons 3Ag^+(aq) + PO^{3-}_4(aq)\]In this equation, molar solubility considers how many moles of ionic species are produced when the compound dissolves. As shown, each mole of \(Ag_3PO_4\) dissolves into three moles of \(Ag^+\) and one mole of \(PO^{3-}_4\). Thus, it's crucial to express the equilibrium concentrations of ions in terms of S when calculating the solubility product.

Ksp Calculation

The solubility product constant, or Ksp, is an equilibrium constant that applies to the dissolution of sparingly soluble ionic compounds. It measures the product of the ions' concentrations, each raised to the power of its coefficient in the balanced equation.
To calculate Ksp, we use the reaction quotient of the ions in the saturated solution. Consider the equation for silver phosphate: \[K_{sp}= [Ag^+]^3 [PO_4^{3-}]\]Given a compound's Ksp value, we can solve for the ion concentrations, and subsequently, its solubility in the solvent.
For example, if the Ksp of \(Ag_3PO_4\) is \(1.8 \times 10^{-18}\), we can express the Ksp in terms of molar solubility: \[1.8 \times 10^{-18} = (3S)^3(S)\]Solving for S gives us the molar solubility of the compound.

Ionic Compounds Dissolution

When ionic compounds dissolve in water, they split into their respective ions. This dissolution is crucial for understanding their solubility behavior at equilibrium. Ionic compounds differ in how readily they dissolve, which depends on their lattice energy and the solvation energy provided by the solvent.
Dissolution involves the breaking of ionic bonds and the formation of new interactions between the ions and water molecules. The process reaches equilibrium when the ions in solution and the undissolved solid coexist in a balanced state.
Let's take calcium carbonate as an example:\[CaCO_3(s) \rightleftharpoons Ca^{2+}(aq) + CO_3^{2-}(aq)\]This equation illustrates that a single mole of \(CaCO_3\) produces one mole of \(Ca^{2+}\) and one mole of \(CO_3^{2-}\). The solubility product expression for this dissolution, \[K_{sp} = [Ca^{2+}][CO_3^{2-}]\], captures this equilibrium state.
By understanding these principles, we can predict the solubility behavior of various ionic compounds in different conditions.