Problem 104 - Chemistry A Molecular Approach

Question

Predict whether a precipitate will form if you mix \(175.0 \mathrm{~mL}\) of a \(0.0055 \mathrm{M} \mathrm{KCl}\) solution with \(145.0 \mathrm{~mL}\) of a \(0.0015 \mathrm{M} \mathrm{AgNO}_{3}\) solution. Identify the precipitate, if any.

Step-by-Step Solution

Answer

Yes, a precipitate of silver chloride (AgCl) will form, as the calculated ion product (Q) is expected to exceed the solubility product (Ksp) for AgCl.

Step 1: Identify the potential precipitate

Write the potential double displacement reaction that occurs when mixing solutions of potassium chloride (KCl) and silver nitrate (AgNO3). In this reaction, the cations and anions switch partners, potentially forming silver chloride (AgCl) and potassium nitrate (KNO3).

Step 2: Write the ionic equation

The full ionic equation for this reaction is written as: \( K^+ + Cl^- + Ag^+ + NO_3^- \rightarrow AgCl(s) + K^+ + NO_3^- \). Simplifying by removing the spectator ions (ions that do not participate in the reaction), we get the net ionic equation: \( Ag^+ + Cl^- \rightarrow AgCl(s) \). This indicates that silver chloride may precipitate if the product of the concentrations of Ag+ and Cl- exceeds the solubility product constant for AgCl.

Step 3: Calculate the ion product (Q)

The ion product (Q) is calculated using the concentrations of the ions that could form a precipitate. Use the initial concentrations and the volumes of the two solutions to find the final concentrations of Ag+ and Cl- ions after they are mixed. Q = [Ag+] * [Cl-].

Step 4: Determine the final concentrations

The final concentration of each ion is found by applying the dilution formula: C1V1 = C2V2. For Ag+: Cfinal = (0.0015 M * 145.0 mL) / (175.0 mL + 145.0 mL). For Cl-: Cfinal = (0.0055 M * 175.0 mL) / (175.0 mL + 145.0 mL).

Step 5: Compare the ion product with the solubility product (Ksp)

Find the Ksp value for AgCl from a solubility table, which is approximately \(1.8 \times 10^{-10}\) at room temperature. If Q < Ksp, no precipitate forms; if Q > Ksp, a precipitate of AgCl will form.

Step 6: Predict whether a precipitate will form

Calculate the value of Q using the final concentrations found in Step 4. If it is greater than the Ksp value found in Step 5, a precipitate of AgCl will form. If not, no precipitate will form.

Key Concepts

Solubility Product Constant (Ksp)

Understanding the solubility product constant, commonly denoted as Ksp, is crucial when predicting the formation of a precipitate during chemical reactions. It is a unique value for each sparingly soluble compound, representing the limit at which a compound will dissolve in solution.

The solubility product constant is defined as the product of the concentrations of the ions of a compound, each raised to the power of its coefficient in the balanced chemical equation. For a generic salt AB that dissociates into A+ and B- ions, the Ksp expression is given by: \[ Ksp = [A^+]^{m} \times [B^-]^{n} \] Here, m and n are the stoichiometric coefficients of the ions in the balanced equation. In the case of silver chloride (AgCl), which has a 1:1 ratio of silver ions (Ag+) to chloride ions (Cl-), the Ksp is simply: \[ Ksp = [Ag^+] \times [Cl^-] \] The solubility product is affected by temperature and the presence of common ions. When two solutions are mixed, if the ion product (Q) - the product of the ionic concentrations at any moment - exceeds the Ksp, the solution is supersaturated, and a precipitate will form. Otherwise, the compound remains dissolved in equilibrium, and no precipitation occurs.

Net Ionic Equation

The net ionic equation is a succinct representation of the actual chemical reaction that occurs during precipitation. It excludes spectator ions, which are ions that remain aqueous and unchanged on both the reactant and product sides of the equation.

Writing the net ionic equation involves the following steps:

• Starting with the full ionic equation, which lists all the ions participating in the reaction.
• Identifying and canceling out the spectator ions.
• Writing the simplified equation that only shows the ions and compounds participating in the formation of the precipitate.

For our example with silver chloride precipitation, the net ionic equation from the problem is: \[ Ag^+ + Cl^- \rightarrow AgCl(s) \] This equation clearly shows that when silver ions and chloride ions are in the same solution and their concentration product is above the solubility product, they combine to form solid silver chloride, which precipitates out of solution.

Double Displacement Reaction

Also known as a metathesis reaction, the double displacement reaction involves the exchange of ions between two compounds. It is a chemical process where the cations and anions of two different molecules swap places, forming two entirely different compounds.

This swapping happens when two ionic compounds are dissolved in water, and it typically results in the formation of a precipitate, a gas, or a weak electrolyte. Double displacement reactions can be easily identified by their general formula: \[ AB + CD \rightarrow AD + CB \] In the textbook exercise, the reaction between potassium chloride (KCl) and silver nitrate (AgNO3) is a classic example of a double displacement reaction, which is expected to produce silver chloride (AgCl), a precipitate, and potassium nitrate (KNO3), which remains dissolved in solution. It is important to identify potential precipitates in these reactions by considering the solubility rules for ionic compounds, thereby enabling one to predict the products of the reaction correctly.