Question

The molar solubility of \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\) in \(0.10 \mathrm{M} \mathrm{Na}_{2} \mathrm{CrO}_{4}\) is \(1.7 \times 10^{-6} \mathrm{M}\). What is the value of \(K_{\mathrm{sp}}\) for \(\mathrm{Ag}_{2} \mathrm{CrO}_{4} ?\)

Short answer

\( K_{\mathrm{sp}} = 1.156 \times 10^{-12} \)

Step-by-step solution

Write down the dissolution equation

Start by writing down the dissolution equation of \( \mathrm{Ag}_{2} \mathrm{CrO}_{4} \) in water: \( \mathrm{Ag}_{2} \mathrm{CrO}_{4} (s) \rightleftharpoons 2\mathrm{Ag}^+ (aq) + \mathrm{CrO}_4^{2-} (aq) \). This equation shows that one mole of \( \mathrm{Ag}_{2} \mathrm{CrO}_{4} \) will produce two moles of \( \mathrm{Ag}^+ \) and one mole of \( \mathrm{CrO}_4^{2-} \) in solution.

Set up the expression for the solubility product constant (Ksp)

The expression for the solubility product constant (\( K_{\mathrm{sp}} \)) for \( \mathrm{Ag}_{2} \mathrm{CrO}_{4} \) is the product of the concentrations of the ions raised to the power of their coefficients in the balanced equation: \( K_{\mathrm{sp}} = [\mathrm{Ag}^+]^2 \times [\mathrm{CrO}_4^{2-}] \).

Substitute molar solubility and common ion concentration into Ksp expression

Since the molar solubility of \( \mathrm{Ag}_{2} \mathrm{CrO}_{4} \) is given as \(1.7 \times 10^{-6} \) M, we have \( [\mathrm{Ag}^+] = 2 \times 1.7 \times 10^{-6} \) M because for every 1 mole of \( \mathrm{Ag}_{2} \mathrm{CrO}_{4} \) that dissolves, 2 moles of \( \mathrm{Ag}^+ \) ions are formed. However, \( [\mathrm{CrO}_4^{2-}] \) will not be affected by this solubility due to the presence of common ions from \( \mathrm{Na}_{2} \mathrm{CrO}_{4} \), so \( [\mathrm{CrO}_4^{2-}] = 0.10 \) M. The revised Ksp expression with these values is \( K_{\mathrm{sp}} = (2 \times 1.7 \times 10^{-6})^2 \times 0.10 \).

Calculate the value of Ksp

Calculate the value of \( K_{\mathrm{sp}} \) using the substituted concentrations: \( K_{\mathrm{sp}} = (2 \times 1.7 \times 10^{-6})^2 \times 0.10 = (3.4 \times 10^{-6})^2 \times 0.10 = 1.156 \times 10^{-11} \times 0.10 = 1.156 \times 10^{-12} \).

Key concepts

Molar Solubility

Molar solubility is the number of moles of a solute that can dissolve in a liter of solution before the solution becomes saturated. It's a crucial concept in chemistry because it helps us understand how much of a substance can be dissolved in a given amount of solvent at equilibrium. To put it in practical terms, it's like figuring out how much sugar you can dissolve in your tea before it can’t take any more and the sugar starts to accumulate at the bottom of your cup.

When calculating molar solubility from the solubility product constant (\(K_{sp}\)), it's essential to consider the stoichiometry of the dissolution process. For instance, the solubility of \(\mathrm{Ag}_2\mathrm{CrO}_4\) leads to the formation of two \(\mathrm{Ag}^+\) ions for every \(\mathrm{CrO}_4^{2-}\) ion. Therefore, if the molar solubility is \(s\), the concentration of \(\mathrm{Ag}^+\) will be \(2s\), because each formula unit produces two silver ions.

Common Ion Effect

The common ion effect refers to the decrease in the solubility of an ionic compound when a solution already contains one of the ions present in the compound. This is due to Le Chatelier's principle, which predicts that the addition of an ion that is part of an equilibrium system will shift the equilibrium to offset the change, often resulting in less of the solid dissolving.

For instance, if you add common table salt (NaCl) to a saturated solution of silver chloride (AgCl), you'll find that less AgCl dissolves because the chloride ions from NaCl shift the equilibrium toward the solid AgCl. In the exercise, the presence of the common ion \(\mathrm{CrO}_4^{2-}\) from \(\mathrm{Na}_2\mathrm{CrO}_4\) in the solution affects the solubility of \(\mathrm{Ag}_2\mathrm{CrO}_4\), hence the molar solubility of \(\mathrm{Ag}_2\mathrm{CrO}_4\) is utilized considering the unchanged concentration of \(\mathrm{CrO}_4^{2-}\) due to the common ion effect.

Dissolution Equation

The dissolution equation gives us a snapshot of the chemical process when a solid solute dissolves in a solvent to form a solution. It represents the stoichiometry and the phase of each substance involved in the dissolution. For a solid like \(\mathrm{Ag}_2\mathrm{CrO}_4\), the dissolution equation \(\mathrm{Ag}_2\mathrm{CrO}_4(s) \rightleftharpoons 2\mathrm{Ag}^+(aq) + \mathrm{CrO}_4^{2-}(aq)\) describes the solid starting material (left) and the resulting ions in the solution (right).

In this reaction, for every mole of \(\mathrm{Ag}_2\mathrm{CrO}_4\) that dissolves, two moles of \(\mathrm{Ag}^+\) and one mole of \(\mathrm{CrO}_4^{2-}\) are formed in the solution. This stoichiometric relationship is pivotal when determining the concentrations of ions in the solution for the solubility product expression.

Equilibrium Concentration

Equilibrium concentration refers to the final concentrations of all reactants and products in a reversible reaction when the rates of the forward and reverse reactions are equal. At this point, the reaction has reached equilibrium and the concentration of the substances involved remain constant as long as the system is not disturbed.

Using the dissolution of \(\mathrm{Ag}_2\mathrm{CrO}_4\) as an example, once equilibrium is reached, the concentrations of \(\mathrm{Ag}^+\) and \(\mathrm{CrO}_4^{2-}\) in the solution will not change over time. These concentrations are crucial in determining the solubility product constant, \(K_{sp}\), which is a measure of the extent to which \(\mathrm{Ag}_2\mathrm{CrO}_4\) can dissolve. In the provided solution, the equilibrium concentration of \(\mathrm{CrO}_4^{2-}\) remains constant at 0.10 M due to the common ion effect, while the equilibrium concentration of \(\mathrm{Ag}^+\) is twice the molar solubility because of the stoichiometry explained in the dissolution equation.