Question

The concentration of \(\mathrm{Ag}^{+}\) in a solution saturated with \(\mathrm{Ag}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(s)\) is \(2.2 \times 10^{-4} M .\) Calculate \(K_{\text {sp }}\) for \(\mathrm{Ag}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\)

Short answer

The solubility product \(K_\text{sp}\) for \(\mathrm{Ag}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) in the saturated solution is \(5.324 \times 10^{-12}\).

Step-by-step solution

Write the dissociation equation

Write the chemical equation for the dissociation of \(\mathrm{Ag}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) in water: \[ \mathrm{Ag}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(s) \rightleftharpoons 2 \mathrm{Ag}^{+}(aq) +\mathrm{C}_{2} \mathrm{O}_{4}^{2-}(aq) \]

Determine the concentration of ions

The concentration of \(\mathrm{Ag}^{+}\) ions in the solution is given as \(2.2 \times 10^{-4} M\). Since the stoichiometry between \(\mathrm{Ag}^{+}\) and \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\) is 2:1, the concentration of \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\) ions in the solution is half of the concentration of \(\mathrm{Ag}^{+}\) ions: \[ [\mathrm{C}_{2} \mathrm{O}_{4}^{2-}] = \frac{1}{2} [\mathrm{Ag}^+] = \frac{1}{2} (2.2 \times 10^{-4} M) = 1.1 \times 10^{-4} M \]

Calculate the \(K_\text{sp}\)

Now we have the concentrations of both \(\mathrm{Ag}^+\) and \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\) ions, so we can calculate the solubility product \(K_\text{sp}\) for \(\mathrm{Ag}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) using the expression: \[ K_\text{sp} = [\mathrm{Ag}^+]^2 \cdot [\mathrm{C}_{2} \mathrm{O}_{4}^{2-}] \] Plug in the values for the concentrations: \[ K_\text{sp} = (2.2 \times 10^{-4})^2 \cdot (1.1 \times 10^{-4}) = 5.324 \times 10^{-12} \]

Present the final answer

The solubility product \(K_\text{sp}\) for \(\mathrm{Ag}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) in the saturated solution is \(5.324 \times 10^{-12}\).

Key concepts

Chemical Equilibrium

Chemical equilibrium is a state in a chemical reaction where the rate of the forward reaction equals the rate of the reverse reaction. This balance means that the concentrations of reactants and products remain constant over time, although they are not necessarily equal. Consider our saturated solution of silver oxalate, \(\mathrm{Ag}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(s)\). In this state, the solid salt dissolves at the same rate that the ions combine to form the solid, resulting in no net change in the concentration of ions in solution.

In calculating the solubility product constant \(K_{sp}\), the key concept is the understanding of the dynamic but stable nature of chemical equilibrium. This constant provides a quantitative measure of the solubility of a compound under equilibrium conditions and can be used to predict whether a precipitate will form when solutions containing different ions are mixed.

Dissociation Reactions

A dissociation reaction involves the separation of a substance into two or more products. For salts and other ionic compounds dissolved in water, dissociation refers to the breaking apart of the solid into its constituent ions. In our exercise, \(\mathrm{Ag}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(s)\) dissociates into two \(\mathrm{Ag}^{+}\) ions and one \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\) ion.

Understanding Stoichiometry in Dissociation

Stoichiometry, or the ratio in which the components dissociate, is essential for solving chemistry problems. Here, it is important to note that one mole of solid \(\mathrm{Ag}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) produces two moles of \(\mathrm{Ag}^{+}\) and one mole of \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\), thus influencing the respective concentrations of these ions in solution.

Ionic Product

The ionic product is the product of the molar concentrations of the ions involved in a dissociation reaction, each raised to the power of its stoichiometric coefficient. The formula for the ionic product reflects the proportions of each ion present in solution. When the ionic product equals the solubility product constant \(K_{sp}\), the solution is saturated and at equilibrium.

The calculated ionic product can also determine the solubility of a compound. If the ionic product is less than \(K_{sp}\), more solid can dissolve, while if the ionic product is greater than \(K_{sp}\), the solution is supersaturated, and a precipitate may form. These calculations are vital when predicting reaction outcomes in various conditions.

Concentration Calculation

Calculating the concentration of ions in a solution is integral in determining the solubility product constant \(K_{sp}\) of an ionic compound. To achieve this, you need to understand the mathematical relationship between the ions that result from the dissociation. In our example, for every mole of \(\mathrm{Ag}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) that dissolves, there are twice as many moles of \(\mathrm{Ag}^{+}\) as there are moles of \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\).

The concentration of \(\mathrm{Ag}^{+}\) ions is given, so you can calculate the concentration of \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\) ions accordingly. With both concentrations known, you apply them to the formula for \(K_{sp}\) to find the solubility product constant, which is a crucial step in solving problems involving solubility and precipitation reactions.