Question

Relating ldeas Let s equal the solubility, in mol/L, of \(\mathrm{AB}_{2}\) . In terms of \(s\) , what is the molar concentration of \(\mathrm{A} ?\) of \(\mathrm{B} ?\) What is the \(K_{\mathrm{sp}}\) of \(\mathrm{A} \mathrm{B}_{2}\) ?

Short answer

The molar concentration of \( \mathrm{A}^{2+} \) is \( s \). The molar concentration of \( \mathrm{B}^{-} \) is \( 2s \). \( K_{\mathrm{sp}} = 4s^3 \).

Step-by-step solution

Identify the Dissociation Equation

The compound \( \mathrm{AB}_{2} \) dissociates in water according to the equation: \[ \mathrm{AB}_{2} \rightleftharpoons \mathrm{A}^{2+} + 2 \mathrm{B}^{-} \]

Determine the Solubility

Let the solubility of \( \mathrm{AB}_{2} \) be \( s \) mol/L. This means at equilibrium, the concentration of \( \mathrm{A}^{2+} \) ions will be \( s \) mol/L.

Calculate the Concentration of \( \mathrm{B}^{-} \) Ions

Since each \( \mathrm{AB}_{2} \) produces 2 \( \mathrm{B}^{-} \) ions, the concentration of \( \mathrm{B}^{-} \) will be \( 2s \) mol/L.

Formulate the Expression for \( K_{\mathrm{sp}} \)

The solubility product constant \( K_{\mathrm{sp}} \) is given by: \[ K_{\mathrm{sp}} = [\mathrm{A}^{2+}][\mathrm{B}^{-}]^{2} \]

Substitute the Concentrations into the Expression

Substitute \( [\mathrm{A}^{2+}] = s \) and \[ [\mathrm{B}^{-}] = 2s \] into the \( K_{\mathrm{sp}} \) expression: \[ K_{\mathrm{sp}} = s (2s)^{2} = 4s^{3} \]

Key concepts

Dissociation Equation

When a compound dissolves in water, it breaks down into its ions. This process is represented by a dissociation equation. It shows how the compound splits into positively and negatively charged ions. For example, for the compound \(\text{AB}_{2}\), the dissociation equation is:
\[ \text{AB}_{2} \rightleftharpoons \text{A}^{2+} + 2 \text{B}^{-} \]
This equation tells us that one formula unit of \(\text{AB}_{2}\) breaks down into one \(\text{A}^{2+}\) ion and two \(\text{B}^{-}\) ions. Understanding this equation is important because it helps us predict how much of each ion will be present at equilibrium. It also serves as the basis for calculating the solubility and the solubility product constant.

Molar Concentration

Molar concentration, commonly referred to as molarity, is a way to express the concentration of a solute in a solution. It is defined as the number of moles of solute per liter of solution (mol/L).
If we let the solubility of \(\text{AB}_{2}\) be \(\text{s}\) mol/L, we can determine the concentrations of the ions at equilibrium. From the dissociation equation, for every 1 mole of \(\text{AB}_{2}\) that dissolves, 1 mole of \(\text{A}^{2+}\) is produced. Therefore, the concentration of \(\text{A}^{2+}\) will be \(\text{s}\) mol/L.
Similarly, since each \(\text{AB}_{2}\) compound produces 2 moles of \(\text{B}^{-}\), the concentration of \(\text{B}^{-}\) will be \(\text{2s}\) mol/L. Molar concentration is crucial for calculating the solubility product constant and understanding the extent of dissociation.

Equilibrium Expressions

Equilibrium expressions are used to describe the concentrations of reactants and products in a chemical equilibrium. For the dissociation of \(\text{AB}_{2}\), the equilibrium constant related to solubility is called the solubility product constant (\text{K_sp}).
The expression for \text{K_sp} is: \(\text{K_sp} = [\text{A}^{2+}][\text{B}^{-}]^{2} \)
Here, \[ \text{K_sp} \] represents how much \( \text{AB}_{2} \) will dissolve in water. Substitute \( [\text{A}^{2+}] = \text{s} \) and \( [\text{B}^{-}] = 2\text{s}\) into the \text{K_sp} expression: \[ \text{K_sp} = \text{s} (2\text{s})^{2} = 4\text{s}^{3} \]
This equation helps in predicting and understanding the solubility of compounds in various conditions. The lower the \[ \text{K_sp},\] the less soluble the compound is in water. Equilibrium expressions are essential tools for chemists as they analyze reaction behaviors.