Problem 100 - 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 \(\mathrm{AsO}_{4}{\underline{\phantom{xx}}}^{3-}\) ion is derived from the weak acid \(\mathrm{H}_{3} \mathrm{AsO}_{4}\) \(\left(\mathrm{pK}_{a 1}=2.22 ; \mathrm{pK}_{a 2}=6.98 ; \mathrm{pK}_{a 3}=11.50\right) .\) When asked to calculate the molar solubility of \(\mathrm{Mg}_{3}\left(\mathrm{AsO}_{4}\right)_{2}\) in water, a student used the \(K_{s p}\) expression and assumed that \(\left[\mathrm{Mg}^{2+}\right]=1.5\left[\mathrm{AsO}_{4}^{3-}\right]\). Why was this a mistake?

Step-by-Step Solution

Answer

The student's mistake was assuming that \(\left[Mg^{2+}\right]=1.5\left[AsO_{4}^{3-}\right]\), when in fact, based on the dissolution equation, the correct relationship is \(\left[Mg^{2+}\right]=3\left[AsO_{4}^{3-}\right]\). This incorrect assumption would lead to an inaccurate calculation of the molar solubility of Mg3(AsO4)2 in water.

Step 1: Write the dissolution equation for Mg3(AsO4)2

We begin by writing the equation for the dissolution of Mg3(AsO4)2 in water: \[Mg_{3}(AsO_{4})_{2}(s) \rightleftharpoons 3Mg^{2+}(aq) + 2AsO_{4}^{3-}(aq)\]

Step 2: Write the expression for Ksp

The solubility product constant (Ksp) can be expressed using the equilibrium concentrations of Mg²⁺ and AsO₄³⁻ ions. Based on the dissolution equation, we can write the Ksp expression as: \[K_{sp} = [Mg^{2+}]^3[AsO_{4}^{3-}]^2\]

Step 3: Set up the relationship between the concentrations of Mg²⁺ and AsO₄³⁻ ions

Let x be the molar solubility of Mg3(AsO4)2 in water. Then, for every one mole of Mg3(AsO4)2 that dissolves, we get 3 moles of Mg²⁺ ions and 2 moles of AsO₄³⁻ ions. We can therefore set up a relationship between their equilibrium concentrations: \[[Mg^{2+}] = 3x\] \[[AsO_{4}^{3-}] = 2x\]

Step 4: Analyze the student's mistake

The student assumed that the concentration of Mg²⁺ ions is 1.5 times the concentration of AsO₄³⁻ ions. However, looking at the dissolution equation, we see that the concentration of Mg²⁺ ions is actually 3 times the concentration of AsO₄³⁻ ions. This means the student's assumption of \(\left[Mg^{2+}\right]=1.5\left[AsO_{4}^{3-}\right]\) is incorrect, and the actual relationship is \(\left[Mg^{2+}\right]=3\left[AsO_{4}^{3-}\right]\). The student's mistake would lead to an incorrect calculation of the molar solubility of Mg3(AsO4)2 in water.

Key Concepts

Understanding the Solubility Product Constant (Ksp)

When we discuss the solubility of a compound, it is often in the context of how much of it can dissolve in water to form a homogenous solution. Solubility is a key concept in chemistry, especially when studying precipitation reactions and the chemistry of aqueous solutions. A particular value of interest is the solubility product constant, denoted by Ksp. This constant is a special type of equilibrium constant that measures the extent to which a compound can dissociate into its component ions in water.

Ksp is unique to each slightly soluble ionic compound and is temperature-dependent. The equation representing the solubility product constant involves the concentrations of the ions at equilibrium raised to the power of their respective stoichiometric coefficients from the balanced dissolution equation.

For a generic equation:
\(A_aB_b(s) \rightleftharpoons aA^{n+}(aq) + bB^{m−}(aq)\),
the Ksp expression would be:
\(K_{sp} = [A^{n+}]^a[B^{m−}]^b\).

Therefore, Ksp is directly linked to the molar solubility of the compound in question. Higher Ksp values indicate higher solubility. Understanding Ksp is vital for predicting whether a precipitate will form under certain conditions and for calculating molar solubility, which is the amount of substance that can be dissolved in a given volume of solvent to reach saturation.

Equilibrium Concentrations in Solubility Equilibria

In a solubility equilibrium scenario, the equilibrium concentrations refer to the amounts of solutes and products that are present in a solution when the dissolution process reaches a state where the rate of dissolution equals the rate of precipitation. Molar solubility, the maximum amount of solute that can dissolve in a solvent to form a saturated solution at a given temperature and pressure, plays a crucial role in determining these concentrations.

For the dissolution of an ionic compound, the equilibrium concentrations relate directly to the stoichiometry of the dissolution equation. For every mole of a solid that dissolves, a definite number of moles of ions are released into the solution. If a compound has a dissolution equation:
\(AB_{2}(s) \rightleftharpoons A^{2+}(aq) + 2B^{-}(aq)\),
and 'x' represents the molar solubility of AB2, then at equilibrium we would have:
\([A^{2+}] = x\) and \([B^{-}] = 2x\).

These equilibrium concentrations are then used to calculate the Ksp value for the compound, which can further be applied to predict whether a precipitate will form in a given solution. Incorrect assumptions about these relations, like the mistake of a student misidentifying the stoichiometric ratio, can lead to inaccurate determinations of solubility and potential issues with laboratory and industrial applications.

The Dissolution Equation and Its Role in Solubility

The dissolution equation provides vital information when it comes to understanding the solubility of an ionic compound in a solvent. It expresses the process by which a solid compound breaks down into its constituent ions in a solution. This breakdown is reversible and is represented by a double arrow in the equation, indicating the existence of a dynamic equilibrium between the undissolved solid and the dissolved ions.

For example, let's consider the dissolution of a salt AB in water:
\(AB(s) \rightleftharpoons A^{+}(aq) + B^{-}(aq)\).
According to this equation, one mole of solid AB dissociates into one mole of A+ ions and one mole of B- ions. The stoichiometry of this equation is 1:1, reflecting a direct relationship in their concentrations at equilibrium. It is this stoichiometry that must be correctly understood to avoid mistakes like the one made by the student described in the exercise, who incorrectly assumed the relationship between concentrations of the resulting ions.

Moreover, the dissolution equation forms the foundation for calculating Ksp expressions and, by extension, molar solubility. Failing to write the correct dissolution equation or misinterpreting the stoichiometry can compromise this entire process and yield erroneous results concerning solubility and concentration calculations.