Question

Can the solubility of \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\) be lowered to \(5.0 \times 10^{-8} \mathrm{mol} \mathrm{Ag}_{2} \mathrm{CrO}_{4} / \mathrm{L}\) by using \(\mathrm{CrO}_{4}^{2-}\) as the common ion? by using Ag+? Explain.

Short answer

Yes, the solubility of \(Ag_2CrO_4\) can be lowered to \(5.0 \times 10^{-8} mol/ L\) by using \(CrO_4^{2-}\) as the common ion if the approximate total concentration of \(CrO_4^{2-}\) is equal to or greater than \(5.0 \times 10^{-8} mol/L\). And the same condition applies to \(Ag+\) if it is used as a common ion. Thus, both can be used as common ions to achieve the desired solubility, as long as the ion concentrations are carefully managed.

Step-by-step solution

Find the Initial Concentration from Ksp

It is given that the solubility product constant (Ksp) for \(Ag_2CrO_4\) is \(1.9 \times 10^{-12}\). This sets up the following equilibrium: \(Ag_2CrO_4 (s) \leftrightarrow 2Ag+ (aq) + CrO_4^{2-} (aq)\). As the equation shows, for each formula unit of \(Ag_2CrO_4\) that dissolves, it produces two \(Ag+\) ions and one \(CrO_4^{2-}\) ion. So, if we let x represent the solubility of \(Ag_2CrO_4\) in moles per liter, then the concentration of \(Ag+\) will be 2x, and the concentration of \(CrO_4^{2-}\) will be x. Setting the solubility product expression equal to the Ksp value and solving for x gives us the initial concentrations. (Ksp = [2x]^2[x] = \(1.9 \times 10^{-12}\))

CrO4^2- as common ion

We are asked to discover if the solubility of silver chromate could be lowered to \(5.0 \times 10^{-8} mol/L\). We assume that this is the concentration of \(CrO_4^{2-}\). We know that the chromate ions originate from the dissolution of silver chromate and the external supply. In the case of the lowered solubility, the contribution from the silver chromate dissolution is going to be insignificant compared to external supply due to common ion effect, because it's much less than the chromate ions supplied (we can verify this by comparing \(x\) and \(5.0 \times 10^{-8} mol/L\)). So we can approximate the \(CrO_4^{2-}\) concentration as \(5.0 \times 10^{-8} mol/L\). Substituting this and the concentration of Ag+ from Step 1 into the expression for \(Ksp\), we can find if this chromate concentration will yield a product that is equal to or greater than the \(Ksp\) of \(Ag_2CrO_4\).

Ag+ as the common ion

This time we will use the silver ion \(Ag+\) as the common ion. Again, the supply from the dissolved silver chromate will be negligible to that of the common ion supply. If we assume that the additional \(Ag+\) ions do not come from the dissolution of \(Ag_2CrO_4\), then the chromate concentration would equal the reduced solubility(i.e. \(5.0 \times 10^{-8} mol/L\)). Substituting the concentration of chromate and the concentration of silver ions (from Step 1) into the expression \(Ksp=[Ag+]^2[CrO_4^{2-}]\), we will find out if these conditions would satisfy the solubility product of \(Ag_2CrO_4\).

Key concepts

Equilibrium Constant

The Equilibrium Constant, often represented as K, is a measure of the extent of a chemical reaction at equilibrium. For a dissolution equilibrium, it is given a specific name, the solubility product constant, represented as Ksp. The Ksp for a salt is the product of its constituent ions' concentrations, each raised to the power of their stoichiometric coefficients.

For instance, the Ksp equation for silver chromate, (Ag2CrO4), is derived from its dissociation into two silver (Ag+) ions and one chromate (CrO4^2-) ion:
\[ Ksp = [Ag+]^2[CrO_4^{2-}] \]
In this equation, [Ag+] and [CrO4^2-] represent the molar concentrations of these ions at equilibrium. The value of Ksp is crucial in solubility calculations as it allows us to compare the solubility of different salts under various conditions and predict whether a precipitate will form when solutions containing these ions are mixed.

Common Ion Effect

The common ion effect occurs when a salt, which contains an ion already present in the solution, is added to that solution, leading to a decrease in the solubility of the original salt. This is because the presence of the common ion shifts the equilibrium to favor the formation of the undissolved salt, thus reducing its solubility. In other words, adding more of an ion that is part of the equilibrium system reduces the solubility of the compound that produces that ion.

For the exercise regarding silver chromate (Ag2CrO4), if either Ag+ or CrO4^2- ions are added to the solution as a common ion, the solubility of Ag2CrO4 is expected to decrease according to Le Chatelier's principle, which states that the system will adjust to minimize the change. This will be particularly evident in a saturated solution of Ag2CrO4, where any addition of Ag+ or CrO4^2- will push the equilibrium back towards the solid, precipitating out of the solution and thus reducing solubility.

Solubility Calculations

Solubility calculations involve determining how much of a particular compound can dissolve in a given amount of solvent at a specific temperature. The solubility is related to the Ksp value of the salt. A higher Ksp value typically indicates a more soluble compound.

To calculate solubility, we first set up an equation representing the dissolution process and the corresponding equilibrium expression. By defining a variable, commonly x, to represent the solubility of the substance, we can relate the molar concentrations of the ions at equilibrium to x. Taking into account the stoichiometry of the dissolution, we can express the concentrations of each product ion in terms of x, substitute these expressions into the Ksp equation, and solve for x.

In this example, the solubility of silver chromate (x) can be decreased using a common ion. When external sources of CrO4^2- or Ag+ are introduced, the corresponding equilibrium shifts, affecting x. Given Ag2CrO4's Ksp, the solubility in the presence of additional CrO4^2- or Ag+ can be calculated, as done in the step-by-step solution, to determine if the solubility can be lowered to the desired levels.