Question

At \(25^{\circ} \mathrm{C}\), the reaction $$\mathrm{CaCrO}_{4}(s) \rightleftharpoons \mathrm{Ca}^{2+}(a q)+\mathrm{CrO}_{4}^{2-}(a q)$$ has an equilibrium constant \(K_{c}=7.1 \times 10^{-4}\). What are the equilibrium concentrations of \(\mathrm{Ca}^{2+}\) and \(\mathrm{CrO}_{4}{ }^{2-}\) in a saturated solution of \(\mathrm{CaCrO}_{4} ?\)

Short answer

The equilibrium concentrations of \(\mathrm{Ca}^{2+}\) and \(\mathrm{CrO}_{4}^{2-}\) in a saturated solution of \(\mathrm{CaCrO}_{4}\) are both \(7.1 \times 10^{-4} \, \text{mol/L}\).

Step-by-step solution

Set up the ICE table

Set up an ICE table to represent the initial concentrations of the reactants and products, the changes in their concentrations as the reaction proceeds, and their concentrations at equilibrium. In this case, since we are dealing with a saturated solution of \(\mathrm{CaCrO}_{4}\), the initial concentration of \(\mathrm{CaCrO}_{4}\) can be represented as "s", and the concentrations of \(\mathrm{Ca}^{2+}\) and \(\mathrm{CrO}_{4}^{2-}\) as 0. ICE Table: | | Initial | Change | Equilibrium | |---------------|------------|--------------|-------------------| | \(\mathrm{CaCrO}_{4}\) | s | -x | s - x | | \(\mathrm{Ca}^{2+}\) | 0 | +x | x | | \(\mathrm{CrO}_{4}^{2-}\) | 0 | +x | x |

Write the equilibrium constant expression

Write the equilibrium constant expression, \(K_c\), for the reaction in terms of the concentrations of the reactants and products at equilibrium. \[K_c = \frac{[\mathrm{Ca}^{2+}][\mathrm{CrO}_{4}^{2-}]}{[\mathrm{CaCrO}_{4}]}\]

Substitute equilibrium concentrations into the equilibrium constant expression

Substitute the equilibrium concentrations of the reactants and products from the ICE table into the equilibrium constant expression. \[K_c = \frac{(x)(x)}{(s - x)}\] Since the dissolution of a solid is negligible compared to the concentration of the aqueous ions, we can assume that the change in concentration of \(\mathrm{CaCrO}_4\) is negligible. Thus, \((s - x) \approx s\).

Solve for x

Given that the equilibrium constant \(K_c = 7.1 \times 10^{-4}\), substitute this value into the equilibrium constant expression and solve for the unknown x value. \[7.1 \times 10^{-4} = \frac{x^2}{s}\] Now we need to express 's' in terms of 'x'. Since the solution is saturated, the concentration of \(\mathrm{Ca}^{2+}\) and \(\mathrm{CrO}_{4}^{2-}\) are stoichiometrically equal, hence we can write: \[s = x\] Substituting s as x in the equilibrium expression, we get: \[7.1 \times 10^{-4} = \frac{x^2}{x}\]

Calculate the equilibrium concentrations

Solve for x to find the equilibrium concentrations of \(\mathrm{Ca}^{2+}\) and \(\mathrm{CrO}_{4}^{2-}\). \[x = 7.1 \times 10^{-4} \, \text{mol/L}\] Therefore, the equilibrium concentrations of \(\mathrm{Ca}^{2+}\) and \(\mathrm{CrO}_{4}^{2-}\) in the saturated solution are both \(7.1 \times 10^{-4} \, \text{mol/L}\).

Key concepts

Equilibrium Constant

The equilibrium constant, denoted as \( K_c \), is a crucial concept in chemistry that helps us understand how a chemical reaction behaves when it has reached a state of balance. It represents the ratio of the concentrations of the products to the reactants, each raised to the power of their stoichiometric coefficients, at equilibrium.

For the reaction \( \mathrm{CaCrO}_{4}(s) \rightleftharpoons \mathrm{Ca}^{2+}(a q)+\mathrm{CrO}_{4}^{2-}(a q) \), the equilibrium constant \( K_c \) at 25°C is given as \( 7.1 \times 10^{-4} \).

• This low \( K_c \) value indicates that at equilibrium, the reaction favors the reactant, meaning less product is formed.
• In essence, only a small amount of \( \mathrm{Ca}^{2+} \) and \( \mathrm{CrO}_{4}^{2-} \) ions are present in the solution at equilibrium.

The expression for \( K_c \) in this reaction is \( K_c = \frac{[\mathrm{Ca}^{2+}][\mathrm{CrO}_{4}^{2-}]}{[\mathrm{CaCrO}_{4}]} \), but since \( \mathrm{CaCrO}_{4} \) is a solid, its concentration is constant and does not appear in the expression.

ICE Table

An ICE table is a systematic method used to keep track of changes in concentrations of substances in a chemical reaction as it reaches equilibrium. "ICE" stands for Initial, Change, and Equilibrium, representing the three stages of a reaction.

To create an ICE table:

• Identify the initial concentrations of reactants and products. For a solid like \( \mathrm{CaCrO}_{4} \), this is its solubility "\( s \)" in the solvent.
• Determine the change in concentration as the reaction moves toward equilibrium. This is typically represented by "\( x \)" for variables changing due to the reaction.
• Calculate the equilibrium concentration by adding the change to the initial concentration.

In the given reaction, the initial concentrations of \( \mathrm{Ca}^{2+} \) and \( \mathrm{CrO}_{4}^{2-} \) are 0, since the starting material is a solid; hence they increase by the same amount \( +x \) to reach equilibrium.

Saturated Solution

A saturated solution occurs when a solution contains the maximum amount of solute that can dissolve in the solvent at a given temperature. Beyond this point, any added solute remains undissolved.

In the case of \( \mathrm{CaCrO}_{4}(s) \) dissolving to form \( \mathrm{Ca}^{2+} \) and \( \mathrm{CrO}_{4}^{2-} \), it means that the solution has reached a point where no more \( \mathrm{CaCrO}_{4} \) can dissolve beyond the equilibrium concentration.

• A saturated solution is in dynamic equilibrium, where the rate of dissolution is equal to the rate of precipitation.
• This equilibrium state is represented by a constant concentration of \( \mathrm{Ca}^{2+} \) and \( \mathrm{CrO}_{4}^{2-} \), given by the equilibrium constant \( K_c \).

Stoichiometry

Stoichiometry involves using balanced chemical equations to calculate the quantities of reactants and products involved in a chemical reaction. The concept of stoichiometry is crucial in determining the relationship between the concentrations of products and reactants.

In the equation \( \mathrm{CaCrO}_{4}(s) \rightleftharpoons \mathrm{Ca}^{2+}(a q) + \mathrm{CrO}_{4}^{2-}(a q) \), the stoichiometry ratio is 1:1:1.

• This means that one mole of \( \mathrm{CaCrO}_{4} \) produces one mole each of \( \mathrm{Ca}^{2+} \) and \( \mathrm{CrO}_{4}^{2-} \).
• In calculations, this ratio allows us to equate \( x \) to the concentrations of \( \mathrm{Ca}^{2+} \) and \( \mathrm{CrO}_{4}^{2-} \) at equilibrium.

With the help of stoichiometry, you can deduce the final concentrations of ions in the solution from the value of \( x \) calculated using the equilibrium constant.