Problem 107 - Chemistry The Central Science

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 \(\mathrm{H}_{3} \mathrm{AsO}_{4}\left(\mathrm{pK}_{a 1}=\right.\) \(2.22 ; \mathrm{p} K_{a 2}=6.98 ; \mathrm{p} K_{a 3}=11.50 )\) . (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 approximately \(4.07 \times 10^{-7} M\). (b) The pH of a saturated solution of Mg₃(AsO₄)₂ in water is approximately 5.41.

Step 1: Set up the solubility equilibrium expression

Write the balanced dissolution equation for Mg₃(AsO₄)₂ and the corresponding solubility equilibrium constant expression involving the Ksp: Mg₃(AsO₄)₂(s) ⇌ 3Mg²⁺(aq) + 2AsO₄³⁻(aq) Ksp = [Mg²⁺]³[AsO₄³⁻]²

Step 2: Assume molar solubility

Let's assume the molar solubility of Mg₃(AsO₄)₂ is 's': for every mole of the salt dissolves, we obtain 3 moles of Mg²⁺ and 2 moles of AsO₄³⁻ ions. Substitute this into the Ksp equation: Ksp = (3s)³(2s)²

Step 3: Solve for 's'

We have the value of Ksp, which is 2.1 x 10⁻²⁰, so we can solve for 's': 2.1 x 10⁻²⁰ = (3s)³(2s)² Solve for s: s ≈ 4.07 x 10⁻⁷ M This is the molar solubility of Mg₃(AsO₄)₂ in water.

Step 4: Find the concentrations of H₃AsO₄ species

The AsO₄³⁻ ion is derived from the weak acid H₃AsO₄. Hence, we can find the concentrations of the individual species of H₃AsO₄: H₃AsO₄ ⇌ H⁺ + H₂AsO₄⁻ with Ka1 = 10⁻².²² H₂AsO₄⁻ ⇌ H⁺ + HAsO₄²⁻ with Ka2 = 10⁻⁶.⁹⁸ HAsO₄²⁻ ⇌ H⁺ + AsO₄³⁻ with Ka3 = 10⁻¹¹.⁵⁰ At equilibrium, AsO₄³⁻ concentration is s ≈ 2s (from Step 3 result) We can now use the given Ka values and the equilibrium concentrations to set up and solve a system of equations for the H⁺ concentration.

Step 5: Calculate the H⁺ concentration

Use the Ka expressions to find the H⁺ concentration: Ka1 = [H⁺][H₂AsO₄⁻]/[H₃AsO₄] Ka2 = [H⁺][HAsO₄²⁻]/[H₂AsO₄⁻] Ka3 = [H⁺][AsO₄³⁻]/[HAsO₄²⁻] Substitute the equilibrium concentrations s and 2s as well as pKãs (Ka = 10^{-pKa}): [H⁺]²(10^{11.50 -2.22}) ≈ 2s(10^{2.22 - 6.98}[H⁺] + 10^{-6.98}[H₂AsO₄⁻]) Solve for [H⁺]: [H⁺] ≈ 3.87 x 10⁻⁶ M

Step 6: Calculate the pH of the saturated solution

Now we can determine the pH of the saturated solution using the simple formula: pH = -log[H⁺] pH ≈ 5.41 Thus, the pH of a saturated solution of Mg₃(AsO₄)₂ in water is approximately 5.41.

Key Concepts

Solubility Equilibrium

Solubility equilibrium involves the balance that exists between a solid and its dissolved ions in solution. When we talk about the solubility equilibrium of Mg₃(AsO₄)₂, we refer to the way it dissociates in water:
Mg₃(AsO₄)₂(s) ⇌ 3Mg²⁺(aq) + 2AsO₄³⁻(aq).
Here, 's' represents the molar solubility of Mg₃(AsO₄)₂. For every mole of the compound that dissolves, it produces three moles of magnesium ions and two moles of arsenate ions.
The solubility product constant (\(K_{sp}\)) characterizes this equilibrium, and it's calculated as:

• Ksp = [Mg²⁺]³[AsO₄³⁻]²

This equation shows us the relationship between the concentration of magnesium ions and arsenate ions. By knowing \(K_{sp}\), we can solve for the value of 's', which helps determine how much of the solid will dissolve in water. In this example, the value of \(K_{sp}\) is given as \(2.1 \times 10^{-20}\), indicating the compound's relatively low solubility. This means that only a small concentration of Mg₃(AsO₄)₂ can dissolve in water before reaching equilibrium.

pH Calculation

The pH of a solution gives us insight into its acidity or alkalinity. To find the pH of a saturated solution of Mg₃(AsO₄)₂, we need to determine the concentration of hydrogen ions [H⁺]. This solution involves equilibria associated with the various species of \( ext{H}_{3} ext{AsO}_{4}\), a weak acid.
From previous calculations, the hydrogen ion concentration \([H^{+}]\) was derived considering the mixed equilibrium system with \(K_{a1}, K_{a2},\) and \(K_{a3}\). We use equilibrium constants and relations to estimate \([H^{+}]\) which is approximately \(3.87 \times 10^{-6} M\).

• pH is calculated with the formula \( ext{pH} = - ext{log}([H^{+}])\)
• For our example, this gives \( ext{pH} \approx 5.41\)

The resulting pH indicates a mild acidity, reflective of the weakly acidic nature of \(H_3AsO_4\). Understanding pH in these contexts helps in predicting properties of the solution, such as chemical reactivity and stability.

Weak Acid Ionization

When dealing with weak acids like \( ext{H}_{3} ext{AsO}_{4}\), it's important to recognize that they do not fully dissociate in water. This partial ionization is dictated by the acid's ionization constants \(K_{a1}, K_{a2},\) and \(K_{a3}\). These constants help us calculate how much of a weak acid dissociates into hydrogen ions and their conjugate bases.
The concept here is straightforward:

• A stronger weak acid will have a higher \(K_{a}\) value than a weaker weak acid.
• Lower \( ext{pK}_{a}\) values signify larger \(K_{a}\) values, meaning more ionization.

For \( ext{H}_{3} ext{AsO}_{4}\), the conjugate base series is \( ext{H}_2 ext{AsO}_4^-\), \( ext{HAsO}_4^{2-}\), and \( ext{AsO}_4^{3-}\), each with decreasing tendencies to ionize fully with ascending count of H⁺ detached. These series of equilibriums dictate the protonation state of arsenate in solution, influencing pH and solubility. Calculating precisely how these species interact helps in engineering solutions and predicting the behavior of substances in different environments.