Question

Calculate the molar solubility of \(\mathrm{MX}\left(K_{\mathrm{sp}}=1.27 \times 10^{-36}\right) \mathrm{in}\) each liquid or solution. a. pure water b. \(0.25 \mathrm{M} \mathrm{MCl}_{2}\) c. \(0.20 \mathrm{M} \mathrm{Na}_{2} \mathrm{X}\)

Short answer

Pure water: \(s = \sqrt{1.27 \times 10^{-36}}\). In \(0.25 M MCl_{2}\), assume \(s\) is negligible, \(s \approx \frac{1.27 \times 10^{-36}}{0.50}\). In \(0.20 M Na_{2}X\), assume \(s\) is negligible, \(s \approx \frac{1.27 \times 10^{-36}}{0.40}\).

Step-by-step solution

Establish the Dissociation Reaction in Pure Water

Write the balanced chemical equation for the dissociation of MX in water: \( MX(s) \leftrightarrows M^{+}(aq) + X^{-}(aq) \).

Express the Solubility Product Constant (Ksp)

Write the expression for the solubility product constant \(K_{sp}\) of MX: \( K_{sp} = [M^{+}][X^{-}] \).

Define Molar Solubility in Pure Water

Assume the molar solubility of MX in water is \(s\), so \([M^{+}] = [X^{-}] = s\). Rewrite the Ksp expression as \( K_{sp} = s \cdot s = s^2 \).

Calculate the Molar Solubility in Pure Water

Solve for \(s\) using the given \(K_{sp}\) value: \( s^2 = 1.27 \times 10^{-36} \), so \( s = \sqrt{1.27 \times 10^{-36}} \).

Establish the Common Ion Effect in \(0.25 M MCl_{2}\)

In a solution of \(MCl_{2}\), there is already a common ion \(M^{+}\) present at concentration \(0.50 M\) (since 1 mole of \(MCl_{2}\) releases 2 moles of \(M^{+}\)). This will affect the solubility of MX.

Adjust the Ksp Expression for \(0.25 M MCl_{2}\)

The Ksp equation for MX in the presence of \(MCl_{2}\) is now \( K_{sp} = (0.50 + s)[X^{-}] = 0.50s + s^2 \), as \(M^{+}\) concentration will be increased by the solubility \(s\) of MX.

Calculate the Molar Solubility in \(0.25 M MCl_{2}\)

Since \(s\) is very small compared to \(0.50 M\), it can be ignored in the expression making it \( K_{sp} \approx 0.50s \). Solve for \(s\) by dividing \(K_{sp}\) by \(0.50\).

Establish the Common Ion Effect in \(0.20 M Na_{2}X\)

In a solution of \(Na_{2}X\), there is already a common ion \(X^{-}\) present at concentration \(0.40 M\). This will affect the solubility of MX.

Adjust the Ksp Expression for \(0.20 M Na_{2}X\)

The Ksp equation for MX in the presence of \(Na_{2}X\) is now \( K_{sp} = [M^{+}](0.40 + s) = 0.40s + s^2 \), as \(X^{-}\) concentration will be increased by the solubility \(s\) of MX.

Calculate the Molar Solubility in \(0.20 M Na_{2}X\)

Since \(s\) is very small compared to \(0.40 M\), it can be ignored in the expression making it \( K_{sp} \approx 0.40s \). Solve for \(s\) by dividing \(K_{sp}\) by \(0.40\).

Key concepts

Solubility Product Constant

Understanding the solubility product constant, often abbreviated as Ksp, is crucial when studying the solubility of ionic compounds in solution. The Ksp is an important equilibrium constant that relates to the maximal amount of an ionic compound that can dissolve in water. It is defined for a saturated solution at a certain temperature and applies to sparingly soluble salts.

The general expression for the solubility product constant for a salt AB is given by the equation:
\[ K_{sp} = [A^+][B^-] \]
where \([A^+]\) and \([B^-]\) represent the concentrations of the ions in moles per liter (M). If the salt dissociates into multiple ions, the concentration terms are raised to the power of their coefficients in the balanced equation. In essence, Ksp reflects the degree to which a compound can dissociate into its constituent ions in a solution.

For the compound MX from the exercise, with a given Ksp value, one can deduce its molar solubility in pure water. By setting the molar solubility as 's', we establish the relationship:
\[ K_{sp} = s^2 \]
This equation is then solved to find the molar solubility 's', which gives us insight into how much of MX will dissolve before the solution becomes saturated.

Common Ion Effect

The presence of a common ion in a solution can significantly influence the solubility of a salt; this phenomenon is known as the common ion effect. It's a perfect example of Le Chatelier's principle in action, showing how a system at equilibrium reacts to a stress—in this case, the addition of an ion already present in the dissociation equilibrium.

When a salt, such as MX, is added to a solution already containing an ion it would produce upon dissociation, the equilibrium shifts to favor the formation of more undissociated MX. Consequently, the solubility of MX decreases compared to its solubility in pure water.

For example, when MX is dissolved in a solution containing MCl₂, the added M⁺ ions from MCl₂ will reduce the solubility of MX. This is calculated by adjusting the Ksp expression to account for the increased concentration of the common ion. Assuming the contribution of MX's solubility to the common ion's concentration is negligible due to its low solubility, the Ksp expression simplifies and allows for the computation of the molar solubility in such a mixed solution.

Dissociation Reaction

Dissociation reactions are at the heart of understanding solubility and the behavior of ionic compounds in water. A dissociation reaction involves the separation of a compound into ions when it dissolves in a solvent, such as water.

For a salt like MX, the general dissociation reaction can be represented as:
\[ MX(s) \leftrightarrows M^{+}(aq) + X^{-}(aq) \]
The double arrow signifies a dynamic equilibrium between the solid ionic compound and its ions in solution. At this equilibrium, the rate at which MX dissociates into M⁺ and X^– ions is equal to the rate at which these ions combine back to form the solid compound.

The dissociation reaction sets the stage for determining the solubility product constant (Ksp) as well as evaluating the impact of common ions on solubility. The molar solubility 's' used in calculating Ksp is directly obtained from the stoichiometry of this reaction. By understanding the dissociation process, students gain a better grasp on how soluble a compound is in various environments and can better predict the outcome of solubility calculations.