Question

The solubility of an ionic compound MX (molar mass = \(346 \mathrm{~g}\) ) is \(4.63 \times 10^{-3} \mathrm{~g} / \mathrm{L}\). What is \(K_{\mathrm{sp}}\) for this compound?

Short answer

The solubility product constant, \(K_{sp}\), for the compound is \(1.80 \times 10^{-10}\).

Step-by-step solution

Convert Solubility to Moles per Liter

First, determine the molar solubility of the compound MX by converting its solubility from grams per liter to moles per liter. Use the formula: \( \text{Molar Solubility} = \frac{\text{Solubility in grams per liter}}{\text{Molar Mass}} \). Substitute the given values: \( \frac{4.63 \times 10^{-3} \text{ g/L}}{346 \text{ g/mol}} = 1.34 \times 10^{-5} \text{ mol/L} \).

Discuss Solubility Equilibrium

Understand that the dissolution of the ionic compound MX can be represented as: \( \text{MX}_{(s)} \rightleftharpoons \text{M}^{+}_{(aq)} + \text{X}^{-}_{(aq)} \). Each mole of MX that dissolves results in one mole of M⁺ and one mole of X⁻ ion in solution, as they dissociate completely.

Calculate Ion Concentrations

Given that the molar solubility of MX is \( 1.34 \times 10^{-5} \text{ mol/L} \), the equilibrium concentrations of both ions, M⁺ and X⁻, in the solution will also be \( 1.34 \times 10^{-5} \text{ mol/L} \) since the stoichiometry is 1:1.

Calculate the Solubility Product Constant \(K_{sp}\)

The formula for the solubility product constant \(K_{sp}\) is \(K_{sp} = [\text{M}^{+}][\text{X}^{-}]\). Substitute the equilibrium concentrations into the formula: \(K_{sp} = (1.34 \times 10^{-5})(1.34 \times 10^{-5}) = 1.80 \times 10^{-10}\).

Key concepts

Ionic Compound

Ionic compounds are substances made up of charged particles called ions. These ions are held together by strong electrostatic forces in a solid lattice structure.
The compound MX in our exercise is an example of an ionic compound. It consists of cations (positively charged ions) and anions (negatively charged ions).

• The cation in this exercise is M⁺.
• The anion is X⁻.

Once this compound is added to water, it can dissolve and dissociate the ions into the solution. This dissociation is key to understanding many of the following steps, as each step depends on the behavior of these ions in solution. When working with ionic compounds, always remember that their behavior in water involves the dissolution and separation of ions.

Molar Solubility

Molar solubility represents the number of moles of a compound that can dissolve to form a liter of solution. It helps us understand how much of the compound can saturate a given volume of solvent.
In the exercise, the solubility of the compound MX is given in grams per liter, and it needs to be converted into molar solubility units (moles per liter).
To do this conversion, use the formula:\[ \text{Molar Solubility} = \frac{\text{Solubility in g/L}}{\text{Molar Mass}} \]This conversion is essential because it allows us to relate the physical amount of substance that dissolves with the chemical stoichiometry in the solution.
For MX, with the given values:\[ \frac{4.63 \times 10^{-3} \text{ g/L}}{346 \text{ g/mol}} = 1.34 \times 10^{-5} \text{ mol/L} \]Thus, the molar solubility of MX is \(1.34 \times 10^{-5} \text{ mol/L}\). This value is crucial as it determines the ion concentrations at equilibrium.

Solubility Equilibrium

Solubility equilibrium occurs when a solid compound dissolves in a solvent until no more can dissolve, reaching a state of balance between dissolution and precipitation. For ionic compounds like MX, this means balancing between the solid and its ions in solution.
The dissociation reaction of MX can be written as:\[ \text{MX}_{(s)} \rightleftharpoons \text{M}^{+}_{(aq)} + \text{X}^{-}_{(aq)} \]In this scenario, the process is reversible, allowing ions to go back to solid form and vice versa, maintaining a dynamic equilibrium.
At equilibrium, the concentration of the dissolved ions remains constant because the rate of dissolution equals the rate of precipitation.
Understanding this concept is key to grasping how solubility products work, as the concentrations of the individual ions are crucial inputs in the calculation of the solubility product constant.

Ksp Calculation

The solubility product constant, or \(K_{sp}\), is a value that indicates the degree to which a compound will dissolve in water. It is determined at equilibrium concentrations of the ions in the solution.
For the ionic compound MX, whose dissociation is given as:\[ \text{MX}_{(s)} \rightleftharpoons \text{M}^{+}_{(aq)} + \text{X}^{-}_{(aq)} \]The expression for \(K_{sp}\) is:\[ K_{sp} = [ \text{M}^{+}][ \text{X}^{-}] \]In our exercise, both ion concentrations are equal due to the 1:1 stoichiometry, and equal to the molar solubility \(1.34 \times 10^{-5} \text{ mol/L}\).
Substituting these concentrations into the \(K_{sp}\) formula, we compute:\[ K_{sp} = (1.34 \times 10^{-5})(1.34 \times 10^{-5}) = 1.80 \times 10^{-10} \]This value helps predict how much of the compound can dissolve and form ions in solution under specific conditions. It’s a fundamental concept in chemistry that aids in understanding solubility and reactions in aqueous solutions.