Problem 57 - Chemistry The Molecular Science

Question

Calculate the \(K_{\mathrm{sp}}\) of \(\mathrm{HgI}_{2}\) given that its solubility in water is \(2.0 \times 10^{-10} \mathrm{M}\). Assume that there are no reactions other than the \(K_{\mathrm{sp}}\) reaction.

Step-by-Step Solution

Answer

The \( K_{\mathrm{sp}} \) of \( \mathrm{HgI}_2 \) is \( 3.2 \times 10^{-29} \).

Step 1: Understand the Dissolution Reaction

First, write the dissolution equation for \( \mathrm{HgI}_2 \). When \( \mathrm{HgI}_2 \) dissolves, it dissociates into its ions as follows:\[ \mathrm{HgI}_2 (s) \rightleftharpoons \mathrm{Hg}^{2+} (aq) + 2 \mathrm{I}^- (aq) \]

Step 2: Define the Solubility Product Constant (\( K_{\mathrm{sp}} \))

The solubility product constant, \( K_{\mathrm{sp}} \), for a salt is given by the product of the concentrations of the ions, each raised to the power of their respective coefficients in the balanced equation. Thus,\[ K_{\mathrm{sp}} = [\mathrm{Hg}^{2+}][\mathrm{I}^-]^2 \]

Step 3: Substitute the Solubility Values

Since the solubility of \( \mathrm{HgI}_2 \) is given as \( 2.0 \times 10^{-10} \ \mathrm{M} \), this means that the concentration of \( \mathrm{Hg}^{2+} \) ions in a saturated solution is \( 2.0 \times 10^{-10} \ \mathrm{M} \). Since each molecule of \( \mathrm{HgI}_2 \) produces two \( \mathrm{I}^- \) ions, the concentration of \( \mathrm{I}^- \) ions is \( 2 \times 2.0 \times 10^{-10} \ \mathrm{M} = 4.0 \times 10^{-10} \ \mathrm{M} \).

Step 4: Calculate the \( K_{\mathrm{sp}} \) Value

Substitute the ion concentrations into the expression for \( K_{\mathrm{sp}} \):\[K_{\mathrm{sp}} = (2.0 \times 10^{-10})(4.0 \times 10^{-10})^2\]Calculate:\[K_{\mathrm{sp}} = (2.0 \times 10^{-10})(16.0 \times 10^{-20}) = 32.0 \times 10^{-30}\]Thus, \( K_{\mathrm{sp}} = 3.2 \times 10^{-29} \).

Key Concepts

Dissolution Reactions Explained

When a solid compound like mercuric iodide (HgI_2) dissolves in water, it undergoes a process known as a dissolution reaction. In this reaction, the solid breaks apart into its constituent ions. For HgI_2, which is the chemical formula for mercuric iodide, the balanced dissolution equation is:\[ \mathrm{HgI}_2(s) \rightleftharpoons \mathrm{Hg}^{2+}(aq) + 2\mathrm{I}^-(aq) \]This equation shows that when mercuric iodide dissolves, it creates one Hg^{2+} ion and two I^- ions in the solution.The process of dissolution is essential because it determines the concentration of each ion in the solution. The dissolution reaction reaches a state of balance, known as equilibrium, where the rate of dissolution equals the rate of precipitation back to the solid form. This equilibrium is described by a special constant called the solubility product constant, K_{sp}, which helps predict how much of the salt can dissolve in water.

Understanding Ion Concentration

Ion concentration refers to the amount of ions present in a solution at equilibrium. It is usually expressed in molarity (M), which is moles of solute per liter of solution. When HgI_2 dissolves, each formula unit releases one Hg^{2+} ion and two I^- ions. If the solubility of HgI_2 is 2.0 \times 10^{-10} \ \mathrm{M}, this tells us the molar concentration of Hg^{2+} ions in the solution is the same, 2.0 \times 10^{-10} \ \mathrm{M}.For the I^- ions, since each HgI_2 creates two I^- ions, the concentration of I^- ions will be doubled. Hence, the concentration of I^- is calculated as:\[ 2 \times 2.0 \times 10^{-10} \ \mathrm{M} = 4.0 \times 10^{-10} \ \mathrm{M} \]By knowing these concentrations, we can calculate the K_{sp} for the dissolution reaction. High ion concentration in a solution tends to increase its solubility, but only up to the point where equilibrium is reached.

Navigating Chemical Equilibrium

At equilibrium, a chemical reaction, such as the dissolution of HgI_2 into Hg^{2+} and I^- ions, happens at the same rate in both directions. This means the rate at which the solid dissolves into ions is equal to the rate at which ions precipitate back into the solid form. The point of equilibrium is characterized by the constant K_{sp}, known as the solubility product constant.The K_{sp} value is unique for different compounds and provides insight into the solubility of the compound in water. It is the mathematical product of the ion concentrations at equilibrium, each raised to the power of its coefficient in the dissolution equation.For HgI_2, we have:\[ K_{sp} = [\mathrm{Hg}^{2+}][\mathrm{I^-}]^2 \]Substituting the known concentrations:\[ K_{sp} = (2.0 \times 10^{-10})(4.0 \times 10^{-10})^2 = 3.2 \times 10^{-29} \]Understanding chemical equilibrium and K_{sp} helps in predicting whether a precipitate will form under specific conditions, thus guiding us in many practical applications such as lab procedures and industrial processes.