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.

Short Answer

Expert verified
(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 by step solution

01

(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
02

(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
03

(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.
04

(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
05

(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.

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Most popular questions from this chapter

The solubility of two slightly soluble salts of \(\mathrm{M}^{2+}, \mathrm{MA}\) and \(\mathrm{MZ}_{2}\), is the same, $4 \times 10^{-4} \mathrm{~mol} / \mathrm{L} .(\mathbf{a})$ Which has the larger numerical value for the solubility product constant? (b) In a saturated solution of each salt in water, which has the higher concentration of \(\mathrm{M}^{2+} ?(\mathbf{c})\) If you added an equal volume of a solution saturated in MA to one saturated in \(\mathrm{MZ}_{2}\), what would be the equilibrium concentration of the cation, \(\mathrm{M}^{2+}\) ?

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(a) True or false: "solubility" and "solubility-product constant" are the same number for a given compound. (b) Write the expression for the solubility- product constant for each of the following ionic compounds: \(\mathrm{MnCO}_{3}, \mathrm{Hg}(\mathrm{OH})_{2},\) and \(\mathrm{Cu}_{3}\left(\mathrm{PO}_{4}\right)_{2}\).

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