Question

A 1.0 - \(\mathrm{L}\) saturated silver carbonate solution at \(5^{\circ} \mathrm{C}\) is filtered to remove undissolved solid and treated with enough hydrochloric acid to decompose the dissolved compound. The carbon dioxide generated is collected in a \(19-\mathrm{mL}\) vial and exerts a pressure of \(114 \mathrm{mmHg}\) at \(25^{\circ} \mathrm{C}\). What is the \(K_{\mathrm{sn}}\) of silver carbonate at \(5^{\circ} \mathrm{C} ?\)

Short answer

The \(K_{sp}\) of silver carbonate at \(5^{\circ} \mathrm{C}\) is calculated using the solubility derived from the ideal gas law.

Step-by-step solution

Write the Chemical Reaction

For dissolved silver carbonate (\(\text{Ag}_2\text{CO}_3\)), the reaction with hydrochloric acid (HCl) produces carbon dioxide \((\text{CO}_2)\), according to the equation: \(\text{Ag}_2\text{CO}_3 + 2\text{HCl} \rightarrow 2\text{AgCl} + \text{CO}_2 + \text{H}_2\text{O}\). The moles of \(\text{CO}_2\) produced are equivalent to the moles of dissolved \(\text{Ag}_2\text{CO}_3\).

Use Ideal Gas Law to Find Moles of CO2

Use the Ideal Gas Law \(PV = nRT\) to find the moles of \(\text{CO}_2\):\[n = \frac{PV}{RT}\] where \(P = 114 \text{ mmHg} = 0.150 \text{ atm}, V = 19 \times 10^{-3} \text{ L}, R = 0.0821 \text{ L atm/mol K}, T = 298 \text{ K}\). Calculate \(n = \frac{0.150 \times 0.019}{0.0821 \times 298}\).

Calculate Moles of Ag2CO3 Dissolved

Since 1 mole of \(\text{Ag}_2\text{CO}_3\) produces 1 mole of \(\text{CO}_2\), the moles of \(\text{Ag}_2\text{CO}_3\) that dissolved is equal to the moles of \(\text{CO}_2\) calculated in Step 2.

Express Solubility in Terms of Molarity

The moles of \(\text{Ag}_2\text{CO}_3\) calculated correspond to its solubility in the 1.0 L solution, which directly gives its molarity \([\text{Ag}_2\text{CO}_3]\).

Write the Ksp Expression

For the dissolution of silver carbonate: \(\text{Ag}_2\text{CO}_3 (s) \rightleftharpoons 2\text{Ag}^+ (aq) + \text{CO}_3^{2-} (aq)\), the solubility product constant \(K_{sp}\) is given by \[K_{sp} = [\text{Ag}^+]^2[\text{CO}_3^{2-}]\].

Determine Ion Concentrations

Based on the solution stoichiometry, \([\text{Ag}^+] = 2[\text{Ag}_2\text{CO}_3]\) and \([\text{CO}_3^{2-}] = [\text{Ag}_2\text{CO}_3]\).

Calculate the Ksp Value

Substitute the ion concentrations into the \(K_{sp}\) expression: \[K_{sp} = (2[\text{Ag}_2\text{CO}_3])^2[\text{Ag}_2\text{CO}_3] = 4[\text{Ag}_2\text{CO}_3]^3\]. Insert the solubility value from Step 4 and solve for \(K_{sp}\).

Key concepts

Ideal Gas Law

The ideal gas law is a crucial concept in chemistry that describes the behavior of gases under different conditions. It combines several gas laws into one concise equation: \(PV = nRT\). Here, \(P\) represents the pressure of the gas, \(V\) stands for the volume it occupies, \(n\) is the number of moles of gas, \(R\) is the ideal gas constant (which is 0.0821 L atm/mol K), and \(T\) is the temperature in Kelvin.

When dealing with gases, like the carbon dioxide in our exercise, we can use this equation to calculate the number of moles. This is particularly useful when the volume and pressure of a gas sample are known, as the temperature will allow us to solve for \(n\), the number of moles. For instance, in our exercise, we use the ideal gas law to determine the moles of carbon dioxide produced in a reaction, based on its pressure, volume, and the temperature at which it was collected.

Chemical Reactions

Chemical reactions involve the transformation of reactants into products. In the provided exercise, the reaction between silver carbonate and hydrochloric acid is fundamental. This reaction can be represented as: \(\text{Ag}_2\text{CO}_3 + 2\text{HCl} \rightarrow 2\text{AgCl} + \text{CO}_2 + \text{H}_2\text{O}\).

This equation shows how silver carbonate decomposes in the presence of hydrochloric acid to form silver chloride, carbon dioxide, and water. The reaction is a classic example of how a solid compound can break down into simpler substances, including gases, which can be collected and measured. Understanding these reactions is key for knowing how substances interact under certain conditions and help predict the products formed.

Dissolution

Dissolution is the process through which a solid becomes incorporated into a solvent, forming a solution. For silver carbonate, dissolution in water leads to the breakdown of the solid into its constituent ions in the solution. The equilibrium of this is represented as \(\text{Ag}_2\text{CO}_3 (s) \rightleftharpoons 2\text{Ag}^+ (aq) + \text{CO}_3^{2-} (aq)\).

This equilibrium is fundamental in determining the solubility of a compound, indicating how much can dissolve in a given amount of solvent at a certain temperature. In our exercise, the solubility of silver carbonate is assessed through its dissolution and measured by the concentration of ions in the solution. The solubility product constant, \(K_{sp}\), which we calculate, provides a quantitative measure of this solubility and helps predict precipitation of the compound from the solution.

Silver Carbonate

Silver carbonate is a compound with the formula \(\text{Ag}_2\text{CO}_3\). It is known to decompose readily in acidic solutions, as shown by its reaction with hydrochloric acid. This property is utilized in the exercise where it helps release carbon dioxide gas when reacted.

As with many carbonate salts, silver carbonate's solubility in water is relatively low, making it useful in certain experimental scenarios where limited ion presence is desired. Understanding its solubility behavior is crucial when exploring precipitation reactions and solubility equilibria. By calculating its solubility product constant \(K_{sp}\), we can gain insights into its behavior in different chemical environments and predict its presence in equilibrium reactions.