Question

The solubility in water of a sparingly soluble salt \(\mathrm{AB}_{2}\) is \(1.0 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1}\). Its solubility product number will be: (a) \(4 \times 10^{-15}\) (b) \(4 \times 10^{-10}\) (c) \(1 \times 10^{-15}\) (d) \(1 \times 10^{-10}\)

Short answer

The solubility product \( K_{sp} \) is \( 4.0 \times 10^{-15} \), matching option (a).

Step-by-step solution

Identify the expression for solubility product

The sparingly soluble salt \( \mathrm{AB}_{2} \) dissociates into its ions in water as follows: \( \mathrm{AB}_{2} (s) \rightleftharpoons \mathrm{A}^{2+} (aq) + 2\mathrm{B}^{-} (aq) \). This means that for each mole of \( \mathrm{AB}_{2} \) that dissolves, 1 mole of \( \mathrm{A}^{2+} \) ions and 2 moles of \( \mathrm{B}^{-} \) ions are produced. The solubility product \( K_{sp} \) expression is given by \( K_{sp} = [\mathrm{A}^{2+}][\mathrm{B}^{-}]^{2} \).

Determine concentrations based on solubility

Given the solubility of \( \mathrm{AB}_{2} \) is \( 1.0 \times 10^{-5} \) moles per liter, this is the concentration of \( \mathrm{A}^{2+} \) ions in solution. Hence, \( [\mathrm{A}^{2+}] = 1.0 \times 10^{-5} \) mol/L. Since two \( \mathrm{B}^{-} \) ions are produced for each \( \mathrm{A}^{2+} \) ion, \( [\mathrm{B}^{-}] = 2 \times 1.0 \times 10^{-5} = 2.0 \times 10^{-5} \) mol/L.

Substitute concentrations into the solubility product expression

Substitute the concentrations into the \( K_{sp} \) expression:\[K_{sp} = (1.0 \times 10^{-5}) \times (2.0 \times 10^{-5})^2.\]

Calculate the solubility product

Calculate \((2.0 \times 10^{-5})^2 = 4.0 \times 10^{-10}\). Therefore,\[K_{sp} = (1.0 \times 10^{-5}) \times (4.0 \times 10^{-10}) = 4.0 \times 10^{-15}.\]

Select the correct answer

The calculated value for \( K_{sp} \) is \( 4.0 \times 10^{-15} \), which corresponds to option (a).

Key concepts

Sparingly Soluble Salt

In chemistry, a sparingly soluble salt is one that does not dissolve well in water. This means only a small amount of the salt can dissolve to form a solution. The solubility of such salts is measured in terms of their solubility product constant, termed as \( K_{sp} \). This constant gives insight into how much of the salt can actually dissolve before the solution becomes saturated.
For instance, if we consider the salt \( \mathrm{AB}_{2} \), its low solubility means that only a minimal concentration of \( \mathrm{AB}_{2} \) will dissolve in water before reaching equilibrium. Sparingly soluble salts are important in various chemical reactions because their dissolution limits can affect how the reaction proceeds or the products formed.

Ionic Equilibrium

Ionic equilibrium refers to the state where the rate of solids turning into ions equals the rate of ions returning to the solid form. This happens in solutions containing sparingly soluble salts. In our exercise, \( \mathrm{AB}_{2} \) reaches ionic equilibrium as it dissolves in water. This salt dissociates into its ions \( \mathrm{A}^{2+} \) and \( 2\mathrm{B}^{-} \).
At this point, while more solid can dissolve, the system reaches an equilibrium where the concentration of ions involved stays constant because the dissolution and precipitation processes balance out. Ionic equilibrium is crucial in predicting the concentration of ions and managing reactions. It is a foundational concept for understanding reactions in aqueous solutions.

Solubility Calculation

Solubility calculation involves determining how much of a substance can dissolve in a solvent at equilibrium. For salts like \( \mathrm{AB}_{2} \), calculating solubility helps to establish the point at which the solution becomes saturated.
With the sparingly soluble salt \( \mathrm{AB}_{2} \), one approaches solubility calculation by first determining the concentration of ions in the solution. Given, the solubility as \( 1.0 \times 10^{-5} \) mol/L, the concentration of ion \( \mathrm{A}^{2+} \) becomes \( 1.0 \times 10^{-5} \) mol/L, while the concentration of \( \mathrm{B}^{-} \) will be twice that due to stoichiometry.
Employing these values in the \( K_{sp} \) expression allows for the calculation of the solubility product, revealing the extent of solubility and aiding in understanding of the chemical behavior of the salt.

Chemical Equilibria

Chemical equilibria is the state in which the concentrations of reactants and products remain constant over time. In the case of \( \mathrm{AB}_{2} \), chemical equilibria is achieved when the salt dissolves to form ions at a rate that equals the rate at which ions return to solid form. This concept is vital as it dictates the dynamics of ionic solutions, helping chemists understand the composition of a solution at saturation.
The equilibrium involves multiple factors, including the concentrations of \( \mathrm{A}^{2+} \) and \( \mathrm{B}^{-} \) ions in our salt reaction. By understanding chemical equilibria, one can predict the outcome of a reaction, determine shifts in equilibrium under different conditions, and apply this knowledge in practical scenarios like titrations and buffer solutions. This equilibrium concept extends to various areas like thermodynamics and kinetics, making it a central theme in chemistry learning.