Problem 59 - Chemistry The Molecular Nature of Matter

Question

Calculate the molar solubility of \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\) at \(25^{\circ} \mathrm{C}\) in (a) \(0.200 \mathrm{M} \mathrm{AgNO}_{3}\) and (b) \(0.200 \mathrm{M} \mathrm{Na}_{2} \mathrm{CrO}_{4}\). For \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\) at \(25^{\circ} \mathrm{C}, K_{\mathrm{cn}}=1.1 \times 10^{-12}\)

Step-by-Step Solution

Answer

For (a) the approximated molar solubility of \(\mathrm{Ag}_2\mathrm{CrO}_4\) is \(s = 2.78 \times 10^{-12} \mathrm{M}\), for (b) the approximated molar solubility is \(s' = 3.74 \times 10^{-6} \mathrm{M}\).

Step 1 - Write down the dissolution equation

First, write down the balanced chemical equation for the dissolution of \(\mathrm{Ag}_2\mathrm{CrO}_4\) in water: \[\mathrm{Ag}_2\mathrm{CrO}_4(s) \rightleftharpoons 2\mathrm{Ag}^{+}(aq) + \mathrm{CrO}_4^{2-}(aq)\].

Step 2 - Write the expression for the solubility product constant (Ksp)

The solubility product constant expression for the dissolution of \(\mathrm{Ag}_2\mathrm{CrO}_4\) is \[K_{sp} = [\mathrm{Ag}^{+}]^2[\mathrm{CrO}_4^{2-}]\], where \(K_{sp} = 1.1 \times 10^{-12}\) at \(25^{\text{\textdegree}}C\).

Step 3 - Solve for molar solubility in \(0.200 \mathrm{M} \mathrm{AgNO}_3\) solution

In the presence of a common ion \(\mathrm{Ag}^{+}\), the solubility of \(\mathrm{Ag}_2\mathrm{CrO}_4\) will be reduced. Assume the molar solubility of \(\mathrm{Ag}_2\mathrm{CrO}_4\) in \(0.200 \mathrm{M} \mathrm{AgNO}_3\) is \(s\), then \[ [\mathrm{Ag}^{+}] = 0.200 + 2s \] and \[ [\mathrm{CrO}_4^{2-}] = s \]. Substitute these into the \(K_{sp}\) expression and solve for \(s\). Because \(s\) is very small compared to \(0.200\mathrm{M}\), it can often be neglected.

Step 4 - Calculate the molar solubility in \(0.200 \mathrm{M} \mathrm{Na}_2\mathrm{CrO}_4\) solution

Similarly, the presence of \(\mathrm{CrO}_4^{2-}\) ions from \(\mathrm{Na}_2\mathrm{CrO}_4\) will reduce the solubility of \(\mathrm{Ag}_2\mathrm{CrO}_4\). Assume the molar solubility of \(\mathrm{Ag}_2\mathrm{CrO}_4\) in \(0.200 \mathrm{M} \mathrm{Na}_2\mathrm{CrO}_4\) is \(s'\), with \[ [\mathrm{CrO}_4^{2-}] = 0.200 + s' \] and \[ [\mathrm{Ag}^{+}] = 2s' \].

Step 5 - Neglect \(s\) and \(s'\) if small and solve for Ksp

Due to the low solubility of \(\mathrm{Ag}_2\mathrm{CrO}_4\), \(s\) and \(s'\) can be regarded as negligible in comparison with \(0.200 \mathrm{M}\). Therefore, for part (a), \( [\mathrm{Ag}^{+}] \approx 0.200 \mathrm{M}\), plug this into the \(K_{sp}\) equation and solve for \([\mathrm{CrO}_4^{2-}]\) or \(s\). For part (b), \( [\mathrm{CrO}_4^{2-}] \approx 0.200 \mathrm{M}\), solve for \( [\mathrm{Ag}^{+}]\) or \(2s'\).

Key Concepts

Solubility Product Constant

The solubility product constant, represented as Ksp, is a pivotal term in understanding the principles of solubility at an equilibrium state. Highly relevant in the context of sparingly soluble salts, Ksp provides a quantitative measure of a compound's solubility in a solution. It is defined as the product of the molar concentrations of the ions when a solid is in equilibrium with its ions in a saturated solution, each concentration raised to the power of its stoichiometric coefficient.
For example, consider the compound \(Ag_2CrO_4\), which dissociates into two silver ions (\(Ag^+\)) and one chromate ion (\(CrO_4^{2-}\)). The equilibrium expression for its Ksp would be \(K_{sp} = [Ag^+]^2[CrO_4^{2-}]\), where the brackets indicate the concentration of ions in mol/L. The Ksp value is unique for every salt and depends on the temperature, illustrated by the given Ksp of \(1.1 \times 10^{-12}\) for \(Ag_2CrO_4\). This information enables us to calculate the extent to which the salt can dissolve in a particular solvent at a given temperature.

Common Ion Effect

The common ion effect is a phenomenon where the solubility of a salt is reduced in a solution that already contains one of the ions present in that salt. This occurs because the addition of a common ion shifts the equilibrium position as per Le Chatelier's principle, favoring the formation of the undissolved solid.
Taking our earlier example of \(Ag_2CrO_4\), if we introduce it into a solution containing \(Ag^+\) ions from another source like \(AgNO_3\), the increased concentration of \(Ag^+\) decreases the solubility of \(Ag_2CrO_4\). In the exercise, we saw this effect in action when calculating the molar solubility in a \(0.200 M AgNO_3\) solution. As the silver ion concentration is already significant, less \(Ag_2CrO_4\) can dissolve to reach equilibrium, consequently decreasing its molar solubility. This concept is critical when predicting the solubility of ionic compounds in solutions with shared ions and has broad applications in fields such as pharmacology, environmental science, and chemical manufacturing.

Equilibrium Concentration

Equilibrium concentration refers to the concentration of reactants and products when a chemical reaction has reached a state of balance, and no further change in concentration is observed. At this point, the rate of the forward reaction equals the rate of the reverse reaction, resulting in a constant concentration of the reacting species.
In our dissolution equation for \(Ag_2CrO_4\), the molar solubility is directly linked to the equilibrium concentrations of the resultant ions. This concept not only helps in calculating the extent of solubility but also in understanding how various external factors, such as the common ion effect, temperature, and pressure, can shift the equilibrium. Particularly, it is important for carrying out precise calculations in the lab, for example, when a given concentration of \(Na_2CrO_4\) in the solution influences the dissociation of \(Ag_2CrO_4\), as seen in part (b) of the exercise. Understanding equilibrium concentration is vital in predicting the behavior of systems in dynamic equilibrium, allowing chemists to manipulate conditions to favor desired outcomes in synthesis and to explore reaction mechanisms.