Problem 97 - Chemistry: The Central Science

Question

Calculate the ratio of \(\left[\mathrm{Ca}^{2+}\right]\) to \(\left[\mathrm{Fe}^{2+}\right]\) in a lake in which the water is in equilibrium with deposits of both \(\mathrm{CaCO}_{3}\) and \(\mathrm{FeCO}_{3}\). Assume that the water is slightly basic and that the hydrolysis of the carbonate ion can therefore be ignored.

Step-by-Step Solution

Answer

The ratio of [Ca²⁺] to [Fe²⁺] in the lake water in equilibrium with deposits of CaCO₃ and FeCO₃ is approximately 106.66. This was determined by using the solubility product expressions for both calcium carbonate and iron carbonate and dividing Kₚ(CaCO₃) by Kₚ(FeCO₃).

Step 1: Write the balanced dissolution equilibrium reactions for CaCO₃ and FeCO₃

CaCO₃(s) ⇌ Ca²⁺(aq) + CO₃²⁻(aq) FeCO₃(s) ⇌ Fe²⁺(aq) + CO₃²⁻(aq)

Step 2: Set up and apply the solubility product expressions

For both dissolution equilibria reactions, we know that the solubility product (Kₚ) is equal to the product of the concentrations of the ions in equilibrium with the solid compound. For calcium carbonate: Kₚ(CaCO₃) = [Ca²⁺][CO₃²⁻] For iron carbonate: Kₚ(FeCO₃) = [Fe²⁺][CO₃²⁻]

Step 3: Solve for the ratio of the concentrations [Ca²⁺] to [Fe²⁺]

We want to find the ratio of [Ca²⁺] to [Fe²⁺], which can be written as: Γ = [Ca²⁺] / [Fe²⁺] In order to find this ratio, we can first divide the solubility product expression for CaCO₃ by that for FeCO₃: Kₚ(CaCO₃) / Kₚ(FeCO₃) = ([Ca²⁺][CO₃²⁻]) / ([Fe²⁺][CO₃²⁻]) Notice that the [CO₃²⁻] terms cancel out from both numerator and denominator: Kₚ(CaCO₃) / Kₚ(FeCO₃) = [Ca²⁺] / [Fe²⁺] So, Γ = Kₚ(CaCO₃) / Kₚ(FeCO₃)

Step 4: Use the provided solubility product values to determine Γ

Values for the solubility products can be found in a reference table or textbooks: Kₚ(CaCO₃) = 3.36 × 10⁻⁹ Kₚ(FeCO₃) = 3.15 × 10⁻¹¹ Now, we can plug in these values into our expression for Γ: Γ = (3.36 × 10⁻⁹) / (3.15 × 10⁻¹¹) = 106.66 So, the ratio of [Ca²⁺] to [Fe²⁺] in the lake water in equilibrium with deposits of CaCO₃ and FeCO₃ is approximately 106.66.