Question

The ionic substance \(\mathrm{T}_{3} \mathrm{U}_{2}\) ionizes to form \(\mathrm{T}^{2+}\) and \(\mathrm{U}^{3-}\) ions. The solubility of \(\mathrm{T}_{3} \mathrm{U}_{2}\) is \(3.8 \times 10^{-10} \mathrm{mol} / \mathrm{L}\) . What is the value of the solubility product constant?

Short answer

The solubility product constant is \(1.48 \times 10^{-27} \).

Step-by-step solution

- Write the dissociation equation

The ionic compound \(\text{T}_3 \text{U}_2\) dissociates into ions. The dissociation can be represented as: \(\text{T}_3 \text{U}_2 \rightarrow 3 \text{T}^{2+} + 2 \text{U}^{3-}\).

- Define solubility

The solubility of \(\text{T}_3 \text{U}_2\) is given as \(3.8 \times 10^{-10} \text{ mol/L}\). This is the concentration of \(\text{T}_3 \text{U}_2\) that dissolves in water.

- Calculate the concentration of ions

When \(-10^{-10}\) mol/L of \(\text{T}_3 \text{U}_2\) dissolves, it produces concentrations of \(3 \times (3.8 \times 10^{-10} \text{ mol/L})\) of \(\text{T}^{2+}\) and \(2 \times (3.8 \times 10^{-10} \text{ mol/L})\) of \(\text{U}^{3-}\). Therefore: \( [ \text{T}^{2+} ] = 1.14 \times 10^{-9} \text{ mol/L} \) \( [ \text{U}^{3-} ] = 7.6 \times 10^{-10}\) \(\text{ mol/L} \).

- Write the expression for the solubility product constant

The solubility product constant, \(K_{sp}\) is given by: \( K_{sp} = [ \text{T}^{2+}]^3 [ \text{U}^{3-}]^2 \).

- Substitute the ion concentrations into the \(K_{sp}\) expression

Substitute the calculated ion concentrations into the \(K_{sp}\) expression: \(\begin{aligned} K_{sp} &= (1.14 \times 10^{-9})^3 \times (7.6 \times 10^{-10})^2 \ &= 1.48 \times 10^{-27} \end{aligned}\).

Key concepts

ionic compounds

Ionic compounds are substances composed of ions held together by electrostatic forces known as ionic bonds.
These ions are atoms or molecules that have gained or lost electrons and thus bear a charge.
In our example, the ionic substance is \(\text{T}_3 \text{U}_2\).

When discussing ionic compounds, it is essential to understand a few key points:

• Ionic compounds generally consist of cations (positive ions) and anions (negative ions).
• They usually form crystalline structures and have high melting and boiling points.
• Ionic compounds are often soluble in water and other polar solvents.

Recognizing that \(\text{T}_3 \text{U}_2\) dissociates into \(\text{T}^{2+}\) and \(\text{U}^{3-}\) ions is the first step in solving solubility-related problems.

dissociation equation

The dissociation equation shows how an ionic compound breaks apart into its constituent ions in a solution.
For \(\text{T}_3 \text{U}_2\), the dissociation can be written as:
\(\text{T}_3 \text{U}_2 \rightarrow 3 \text{T}^{2+} + 2 \text{U}^{3-} \).

To understand a dissociation equation:

• Identify the ions that form when the compound dissociates.
• Balance the equation to ensure the number of atoms and the charges are equal on both sides.

In our example, the dissociation shows that one unit of \(\text{T}_3 \text{U}_2\) produces three \(\text{T}^{2+}\) ions and two \(\text{U}^{3-}\) ions.
This step is crucial for calculating ion concentrations and further deriving the solubility product constant.

chemical equilibrium

A solution reaches chemical equilibrium when the rate of the forward reaction (dissolution of the compound) equals the rate of the reverse reaction (re-formation of the solid compound).
For the dissolution of \(\text{T}_3 \text{U}_2\) in water, this happens when the ions produced in the solution are in a steady state.

Some points to understand chemical equilibrium:

• At equilibrium, the concentration of reactants and products remains constant but not necessarily equal.
• The equilibrium condition is described by the equilibrium constant (\text{K}
• For solubility, the relevant equilibrium constant is the solubility product constant (\text{K}_{sp}).

Knowing this helps when using the concentrations of \(\text{T}^{2+}\) and \(\text{U}^{3-}\) ions to find the \text{K}_{sp}.
It provides the basis for further calculations.

solubility calculations

Solubility calculations involve determining how much of a substance can dissolve in a solvent, leading to a saturated solution.
With \(\text{T}_3 \text{U}_2\), the solubility is given as \(\text{3.8} \times \text{10}^{-10} \text{mol/L}\).
Using this, we calculate the concentration of the ions:

• \([ \text{T}^{2+} ] = 3 \times (3.8 \times 10^{-10} \text{mol/L}) = 1.14 \times 10^{-9} \text{mol/L} \).
• \([ \text{U}^{3-} ] = 2 \times (3.8 \times 10^{-10} \text{mol/L}) = 7.6 \times 10^{-10} \text{mol/L}\text{mol/L} \).

Finally, substituting these values into the expression for the solubility product constant \(K_{sp}\) yields:
\( K_{sp} = [ \text{T}^{2+}]^3 [ \text{U}^{3-}]^2 = (1.14 \times 10^{-9})^3 \times (7.6 \times 10^{-10})^2 = 1.48 \times 10^{-27} \).
This value of \(\text{K}_{sp}\) gives us an understanding of the solubility of the compound in water.