Question

A \(250 \mathrm{mL}\) sample of saturated \(\mathrm{CaC}_{2} \mathrm{O}_{4}(\mathrm{aq})\) requires \(4.8 \mathrm{mL}\) of \(0.00134 \mathrm{M} \mathrm{KMnO}_{4}(\mathrm{aq})\) for its titration in an acidic solution. What is the value of \(K_{\mathrm{sp}}\) for \(\mathrm{CaC}_{2} \mathrm{O}_{4}\) obtained with these data? In the titration reaction, \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\) is oxidized to \(\mathrm{CO}_{2}\) and \(\mathrm{MnO}_{4}^{-}\) is reduced to \(\mathrm{Mn}^{2+}\).

Short answer

The value of Ksp for CaC2O4 obtained from the provided data is \(4.14 × 10^{-9}\)

Step-by-step solution

Establish the balanced redox reaction

The chemical reaction involved in this problem is: \(2MnO_4^- + 5C_2O_4^{2-} + 16H^+ \rightarrow 2Mn^{2+} + 10CO_2 + 8 H_2O\). The stoichiometry of the reaction is essential to find out the concentration of oxalate ions.

Calculate the moles of KMnO4 used

The molarity (M) and volume (V) of KMnO4 solution used in the titration can be leveraged to determine the number of moles. The mentioned volume should first be converted to liters. Apply the formula: n (moles) = Molarity × Volume \(= 0.00134 M × 4.8 ml × (1 L/1000 ml) = 6.432 × 10^{-6} mol \)

Determine moles of C2O4^2-

From the balanced reaction equation, 5 moles of C2O42- react with 2 moles of MnO4-. Hence, divide the moles of MnO4- by 2 and multiply by 5 to get the number of moles of C2O42- (\(6.432 × 10^{-6} mol × 5/2 = 1.608×10^{-5}\) mol) for the 250 mL of saturated solution.

Calculate the concentration of C2O42-

The concentration (molarity) of C2O42- can be obtained by dividing the number of moles by the volume in liters. \(1.608×10^{-5} mol ÷ 0.250 L = 6.432 × 10^{-5} M\). Since CaC2O4↔Ca2+ + C2O42-; the concentration of Ca2+ = concentration of C2O4^-

Determine Ksp for CaC2O4

The solubility product constant, Ksp is defined as Ksp = [Ca2+][C2O42-] = \(6.432 × 10^{-5}\) * \(6.432 × 10^{-5}\) = \(4.14 × 10^{-9}\)

Key concepts

Understanding Redox Titration

Redox titration is a process used to determine the concentration of a given substance by exploiting its ability to undergo a redox reaction. In a redox titration, one substance is oxidized while another is reduced, meaning electrons are transferred between molecules. This concept is key in understanding how substances react chemically in the presence of others.
In the exercise, permanganate ions ( MnO_4^- ) are the titrant responsible for oxidizing oxalate ions ( C_2O_4^{2-} ), turning them into carbon dioxide ( CO_2 ). The reaction involves a clear color change, as permanganate ions are reduced to manganese ( Mn^{2+} ). This color change helps to indicate the endpoint of the titration. Knowing the stoichiometry of the reaction is crucial.
It allows us to calculate exactly how much of the substance being investigated is present by relating the amounts of titrant and analyte used in the titration:

• Use a balanced chemical equation to determine stoichiometry.
• Relate the moles of titrant to the moles of the substance being titrated.

By analyzing the given data, we can comprehend how redox reactions facilitate identifying molarity and other chemical properties.

Performing Molarity Calculations

Molarity is a central concept in chemistry. It is defined as the number of moles of a solute in one liter of solution, serving as a measure of the concentration of a substance in a solution. Calculating molarity accurately is essential for solving many types of chemical equations.
In the exercise, we calculate the moles of the titrant (KMnO_4) used by multiplying its molarity by the volume in liters. This provides the number of moles of permanganate ions, which is the starting point for all subsequent calculations. Formula: n = M imes V where M is molarity and V is volume in liters.
This leads us to calculate the moles of oxalate ions ( C_2O_4^{2-} ). Using the stoichiometry from the balanced chemical equation, we can find how many moles of a substance react with another. Finally, with known moles and volume in liters, we calculate the molarity using:

• Convert volume from milliliters to liters.
• Apply the molarity formula.

This systematic approach facilitates determining the concentration of an ion or compound in question.

Concepts of Chemical Equilibrium

Chemical equilibrium refers to a state where the concentrations of all reactants and products remain constant over time. It signifies a balance between the forward and reverse reactions, resulting in no net change in the concentration of reactants and products.
For the reaction involving CaC_2O_4 in the exercise, equilibrium is important. At equilibrium, the solid salt dissolves and dissociates into calcium ions ( Ca^{2+} ) and oxalate ions ( C_2O_4^{2-} ). The solubility product constant ( K_{sp} ) is used to describe this equilibrium condition. It is the product of the concentrations of the ions raised to the power of their respective coefficients in the balanced equation.
The key relationships at equilibrium:

• K_{sp} = [Ca^{2+}][C_2O_4^{2-}] .
• This constant can help predict whether a precipitate will form under certain conditions.

Understanding equilibrium allows us to determine how changes in concentration, temperature, or pressure can affect the system, reinforcing the importance of dynamic balance in chemical processes.