Question

Consider the following reaction, and assume that its equilibrium constant is \(1.00 \times 10^{14}\). $$ 2 \mathrm{CrO}_{4}^{2-}(a q)+2 \mathrm{H}^{+}(a q) \rightleftharpoons \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) $$ (a) Write the equilibrium equation for the reaction, and explain why \(\mathrm{CrO}_{4}^{2-}\) ions predominate in basic solutions and \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) ions predominate in acidic solutions. (b) Calculate the \(\mathrm{CrO}_{4}^{2-}\) and \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) concentrations in a solution that has a total chromium concentration of \(0.100 \mathrm{M}\) and a pH of \(4.000\) (c) What are the \(\mathrm{CrO}_{4}^{2-}\) and \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) concentrations if the \(\mathrm{pH}\) is \(2.000 ?\)

Short answer

In acidic solutions, \(\mathrm{Cr}_2\mathrm{O}_7^{2-}\) predominates, while in basic, \(\mathrm{CrO}_{4}^{2-}\) predominates. At pH 4.0, \([\mathrm{Cr}_2\mathrm{O}_7^{2-}]\approx 0.01 \text{ M}\), \([\mathrm{CrO}_4^{2-}]\approx 0.080 \text{ M}\). More \(\mathrm{Cr}_2\mathrm{O}_7^{2-}\) at pH 2.0.

Step-by-step solution

Writing the Equilibrium Expression

For the reaction \(2 \mathrm{CrO}_{4}^{2-}(aq) + 2 \mathrm{H}^{+}(aq) \rightleftharpoons \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(aq) + \mathrm{H}_{2} \mathrm{O}(l)\), the equilibrium expression, according to the law of mass action, is given by: \[ K = \frac{[\mathrm{Cr}_2\mathrm{O}_7^{2-}]}{[\mathrm{CrO}_4^{2-}]^2 [\mathrm{H}^+]^2} \]\ where \(K = 1.00 \times 10^{14}\). The high value of \(K\) suggests the formation of \(\mathrm{Cr}_2\mathrm{O}_7^{2-}\) is heavily favored either in systems where the concentration of \(\mathrm{H}^+\) is high or that are acidic.

Explanation of Ion Prevalence in Acid and Base

In acidic solutions (low pH), the concentration of \(\mathrm{H}^+\) ions is high, which favors the rightward shift of the equilibrium, resulting in more \(\mathrm{Cr}_2\mathrm{O}_7^{2-}\) ions. In basic solutions (high pH), the concentration of \(\mathrm{H}^+\) ions is low, causing the equilibrium to shift leftwards, favoring the presence of \(\mathrm{CrO}_{4}^{2-}\) ions.

Calculation for pH 4.0

Given the total chromium concentration \([\text{Cr}]_{total} = 0.100 \text{ M}\) and \(\text{pH} = 4.0\), we first convert this pH to \([\mathrm{H}^+]\) concentration: \([\mathrm{H}^+] = 10^{-4} \text{ M}\). Let \(x\) be the concentration of \(\mathrm{Cr}_2\mathrm{O}_7^{2-}\). Therefore, \([\mathrm{CrO}_4^{2-}] = 0.100 - 2x\). Substitute into the equilibrium expression: \[1.00 \times 10^{14} = \frac{x}{(0.100 - 2x)^2 (10^{-4})^2}\] Solve for \(x\). Assuming \(2x\) is small due to large \(K\), \[1.00 \times 10^{14} \approx \frac{x}{0.01}\] thus \(x = 0.01 \text{ M}\), hence, \([\mathrm{CrO}_4^{2-}] \approx 0.080 \text{ M}\).

Calculation for pH 2.0

For \(\text{pH} = 2.0\), \([\mathrm{H}^+] = 10^{-2} \text{ M}\). Using the same set for equations: \(\) \([\text{Cr}]_{total} = [\mathrm{Cr}_2\mathrm{O}_7^{2-}] + 2[\mathrm{CrO}_4^{2-}]= 0.100\text{ M}\). Substituting this and solving \[1.00 \times 10^{14} = \frac{x}{(0.100 - 2x)^2 (10^{-2})^2}\], and simplifying as in the previous step, solve to find \([\mathrm{Cr}_2\mathrm{O}_7^{2-}]\) and \([\mathrm{CrO}_4^{2-}]\). With \(\text{[CrO}_4^{2-}]\) dominating due to high \([\mathrm{H}^+]\), more \(\mathrm{Cr}_2\mathrm{O}_7^{2-}\) is expected.

Key concepts

Chromate-Dichromate Equilibrium

At the heart of understanding chemical equilibria between chromate (\( \text{CrO}_4^{2-} \)) and dichromate (\( \text{Cr}_2\text{O}_7^{2-} \)) lies the reversible nature of their reaction. In an aqueous solution, these ions exist in a dynamic balance described by the equation:\[2 \text{CrO}_4^{2-}(aq) + 2 \text{H}^+(aq) \rightleftharpoons \text{Cr}_2\text{O}_7^{2-}(aq) + \text{H}_2\text{O}(l)\] The equilibrium between these ions shifts depending on the pH of the solution. In essence, chromate and dichromate are interconvertible forms affected by hydrogen ion concentration.

• At a lower pH (acidic conditions), the solution has more hydrogen ions, driving the equilibrium towards the formation of dichromate ions.
• In a basic environment, where hydrogen ions are scarce, chromate ions predominate.

Grasping this equilibrium is crucial as it illustrates how chemical systems respond to changes in their environment.

pH Effect on Equilibrium

The pH level of a solution is a critical factor in shifting the equilibrium between chromate and dichromate ions. It determines which ion is more prevalent:Firstly, pH is a measure of hydrogen ion concentration in a solution. For the chromate-dichromate equilibrium:

• Acidic solutions (low pH) have high \( \text{H}^+ \) concentration.
  • This abundance of hydrogen ions encourages the equilibrium to produce more dichromate ions.
• Basic solutions (high pH) feature low \( \text{H}^+ \) concentration.
  • Here, the system shifts to favor chromate ions.

Thus, by adjusting the pH, we can control the dominant species in the solution. This effect is pivotal for many applications in chemistry and industrial processes.

Le Chatelier's Principle

Le Chatelier's Principle is a foundational concept in understanding how equilibrium responds to changes. According to this principle, if a system at equilibrium experiences a change in concentration, temperature, or pressure, the system will adjust to counteract the effect of the change and re-establish equilibrium.In the context of chromate-dichromate equilibrium:

• When extra \( \text{H}^+ \) ions are introduced, the equilibrium shifts to the right, producing more dichromate ions to reduce the increase in hydrogen ion concentration.
• Conversely, reducing \( \text{H}^+ \) shifts the equilibrium to favor chromate ions.

This principle is invaluable for predicting the direction of equilibrium shifts in response to different conditions, proving essential in chemical synthesis and reactions.

Equilibrium Constant Calculations

The equilibrium constant (\( K \)) is a key parameter that quantifies the position of equilibrium in a chemical reaction. For the chromate-dichromate equilibrium, it is expressed as: \[K = \frac{[\text{Cr}_2\text{O}_7^{2-}]}{[\text{CrO}_4^{2-}]^2 [\text{H}^+]^2}\]A large equilibrium constant (like \( 1.00 \times 10^{14} \) in this case) indicates that the product side is heavily favored. It means the reaction heavily leans toward the formation of dichromate ions in this instance.To solve problems using \( K \):

• Set up the expression using the known concentrations of reactants and products.
• Solve for the unknown concentration, often using assumptions or iterative approaches if the equilibrium constant is large or small.
• Analyze how changes in conditions like pH or concentration affect the equilibrium concentrations.

Calculating \( K \) not only helps in predicting system behavior but is also crucial for designing processes in chemical industries.