Problem 107 - Chemistry: The Central Science in SI Units, Global Edition

Question

The value of \(K_{s p}\) for \(\mathrm{Mg}_{3}\left(\mathrm{AsO}_{4}\right)_{2}\) is \(2.1 \times 10^{-20} .\) The \(\mathrm{AsO}_{4}^{3-}\) ion is derived from the weak acid \(\mathrm{H}_{3} \mathrm{AsO}_{4}\left(\mathrm{pK}_{a 1}=\right.\) \(\left.2.22 ; \mathrm{p} K_{a 2}=6.98 ; \mathrm{pK}_{a 3}=11.50\right)\) (a) Calculate the molar solubility of \(\mathrm{Mg}_{3}\left(\mathrm{AsO}_{4}\right)_{2}\) in water. (b) Calculate the pH of a saturated solution of \(\mathrm{Mg}_{3}\left(\mathrm{AsO}_{4}\right)_{2}\) in water.

Step-by-Step Solution

Answer

(a) The molar solubility of Mg₃(AsO₄)₂ in water is \(7.34 \times 10^{-3}\) M. (b) The pH of a saturated solution of Mg₃(AsO₄)₂ in water is approximately 2.22.

Step 1: (Establish the equilibrium)

Write the dissociation equilibrium of Mg3(AsO4)2 and establish an ICE table. Mg₃(AsO₄)₂ (s) ⇌ 3Mg²⁺ (aq) + 2AsO₄³⁻ (aq) Initial concentrations: [Mg²⁺] = 0 [AsO₄³⁻] = 0 Changes: [Mg²⁺] = +3s [AsO₄³⁻] = +2s Final concentrations after dissociation: [Mg²⁺] = 3s [AsO₄³⁻] = 2s

Step 2: (Use the Ksp to find solubility)

Use the given Ksp to establish the solubility equation and solve for the solubility 's' of Mg3(AsO4)2. Ksp = [Mg²⁺]³[AsO₄³⁻]² 2.1 × 10⁻²⁰ = (3s)³(2s)²

Step 3: (Find the molar solubility of Mg3(AsO4)2)

Solve for the solubility 's' of Mg3(AsO4)2. s^9 = (2.1 × 10⁻²⁰) / (3³ × 2²) s^9 = 2.59 × 10⁻²² s = 0.00734 (as ninth root of s^9) Molar solubility of Mg₃(AsO₄)₂ = 7.34 × 10⁻³ M.

Step 4: (Find the concentration of ions in the saturated solution)

With the molar solubility found, calculate the concentration of ions in the saturated solution. [Mg²⁺] = 3s = 3(7.34 × 10⁻³) = 2.20 × 10⁻² M [AsO₄³⁻] = 2s = 2(7.34 × 10⁻³) = 1.47 × 10⁻² M

Step 5: (Calculate the pH of the saturated solution)

Use the pKa values given and ion concentrations to determine the pH of the saturated solution. Considering the first proton dissociation in H₃AsO₄: H₃AsO₄ ⇌ H⁺ + H₂AsO₄⁻ pKa1 = 2.22 Ka1 = [H⁺][H₂AsO₄⁻] / [H₃AsO₄] [H⁺] = 10^(-pKa1) = 10^(-2.22) = 6.027 × 10⁻³ M pH = -log[H⁺] = -log(6.027 × 10⁻³) = 2.22 (approximately) (a) The molar solubility of Mg₃(AsO₄)₂ in water is 7.34 × 10⁻³ M. (b) The pH of a saturated solution of Mg₃(AsO₄)₂ in water is approximately 2.22.

Key Concepts

Ksp (Solubility Product Constant)

The solubility product constant, often abbreviated as \(K_{sp}\), is a crucial concept in understanding the solubility of sparingly soluble ionic compounds in water. It represents the equilibrium constant for the dissolution reaction of a solid substance. For our case, the dissolution of \(\text{Mg}_3(\text{AsO}_4)_2\) in water establishes itself in the following equilibrium:\[\text{Mg}_3(\text{AsO}_4)_2 (s) \rightleftharpoons 3\text{Mg}^{2+} (aq) + 2\text{AsO}_4^{3-} (aq)\]The value of \(K_{sp}\) is expressed as:\[K_{sp} = [\text{Mg}^{2+}]^3[\text{AsO}_4^{3-}]^2\]In the equation above, \([\text{Mg}^{2+}]\) and \([\text{AsO}_4^{3-}]\) are the molar concentrations of the ions in the solution at equilibrium. For this particular compound, \(K_{sp}\) is given as \(2.1 \times 10^{-20}\), which is quite small, indicating very low solubility in water. To calculate molar solubility, set up an equation based on how each ion's concentration changes proportionately with the dissolution of \(\text{Mg}_3(\text{AsO}_4)_2\). By determining the solubility \(s\), you find how many moles of the solid will dissolve per liter of water. Here, \(s\) was found to be \(7.34 \times 10^{-3}\) M.

Ionic Equilibrium

Ionic equilibrium deals with the state of balance between the ions in a solution when a particular compound has dissolved partially. At equilibrium, the rate of dissociation (dissolving into ions) matches the rate of precipitation (ions combining back into a solid).**Establishing Equilibrium**:For \(\text{Mg}_3(\text{AsO}_4)_2\), as it dissolves, there will be initial concentrations of ions as zero:- \([\text{Mg}^{2+}] = 0\)- \([\text{AsO}_4^{3-}] = 0\)As the equilibrium establishes, these concentrations increase:- \([\text{Mg}^{2+}] = 3s\)- \([\text{AsO}_4^{3-}] = 2s\)The equilibrium concentrations directly depend on the molar solubility \(s\), indicating how much of the salt dissolves in the solution. Use these relationships to plug into the \(K_{sp}\) equation, which emphasizes its role in buffering between dissociation and reformation of the solid under non-ideal, low solubility conditions.

Acid Dissociation Constant (pKa)

The acid dissociation constant \((pK_a)\) is instrumental in understanding how weak acids behave in solution, affecting both solubility and pH of the solution. In the given problem, we analyzed arsenic acid \(\text{H}_3\text{AsO}_4\), which can lose protons in three steps, each characterized by its own \(pK_a\):- \(pK_{a1} = 2.22\)- \(pK_{a2} = 6.98\)- \(pK_{a3} = 11.50\)These values indicate the strength and the degree to which \(\text{H}_3\text{AsO}_4\) loses its protons in solution. The first dissociation has the lowest \(pK_a\), which makes \(\text{H}_3\text{AsO}_4\) a stronger acid compared to its other forms. **pH of the Solution**:In a saturated solution of \(\text{Mg}_3(\text{AsO}_4)_2\), where \(\text{AsO}_4^{3-}\) ions are present, using the \(pK_{a1}\), we can calculate the concentration of \([H^+]\). This initial concentration helps in determining the pH of the solution using the formula:\[pH = -\log[\text{H}^+]\]Here, the calculated \([H^+]\) of \(6.027 \times 10^{-3}\) M provides a pH of approximately 2.22, showing the acidic nature of the solution. Understanding \(pK_a\) values can also inform us about the polyprotic nature of acids and their sequential ionization in different pH environments, influencing the solubility equilibrium.