Question

Determine whether a precipitate will form if 1.70 \(\mathrm{g}\) of solid AgNO \(_{3}\) and 14.5 \(\mathrm{g}\) of solid NaCl are dissolved in \(200 . \mathrm{mL}\) of water to form a solution at \(25^{\circ} \mathrm{C}\)

Short answer

A precipitate will form because \(Q = 0.062\) is much greater than \(K_{sp} = 1.77 \times 10^{-10}\).

Step-by-step solution

- Write the Dissolution Reactions

First, write the balanced chemical equations for the dissolution of \(\text{AgNO}_3\) and \(\text{NaCl}\): \[ \text{AgNO}_3 (s) \rightarrow \text{Ag}^+ (aq) + \text{NO}_3^- (aq) \] \[ \text{NaCl} (s) \rightarrow \text{Na}^+ (aq) + \text{Cl}^- (aq) \]

- Calculate Moles of AgNO3 and NaCl

Use the molar masses to calculate the moles of each salt. \(\text{Molar mass of AgNO}_3 = 169.87\text{ g/mol}\) and \(\text{Molar mass of NaCl} = 58.44\text {g/mol}\). Moles of AgNO3: \(\text{moles of AgNO}_3 = \frac{1.70\text{ g}}{169.87\text {g/mol}} = 0.0100\text{ mol}\). Moles of NaCl: \(\text{moles of NaCl} = \frac{14.5\text{ g}}{58.44\text{ g/mol}} = 0.248\text{ mol}\)

- Determine Ion Concentrations

Since both salts are dissolved in 200 mL (0.200 L) of water, calculate the concentration of \(\text{Ag}^+\) and \(\text{Cl}^-\) ions in the solution. \[ \text{[Ag}^+] = \frac{0.0100\text { mol}}{0.200 \text{ L}} = 0.050 \text{ M} \] \[ \text{[Cl}^-] = \frac{0.248 \text {mol}}{0.200 \text {L}} = 1.24 \text{ M} \]

- Write the Precipitation Reaction

The precipitation reaction between \(\text{Ag}^+\) and \(\text{Cl}^-\) ions is: \[ \text{Ag}^+ (aq) + \text{Cl}^- (aq) \rightarrow \text{AgCl} (s) \]

- Calculate the Ion Product (Q)

The ion product (Q) can be determined using the formula: \[ Q = [\text{Ag}^+][\text{Cl}^-] = (0.050\text{ M})(1.24\text{ M}) = 0.062 \]

- Compare Q to Ksp

Compare the ion product (Q) to the solubility product constant \(K_{sp}\) for \(\text{AgCl}\). The \(K_{sp}\) for \(\text{AgCl}\) at 25°C is \(1.77 \times 10^{-10}\). Since \(\text{Q} = 0.062\) is much greater than \(K_{sp} \), a precipitate will form.

Key concepts

Solubility Product Constant (Ksp)

The Solubility Product Constant (Ksp) is a key concept in understanding precipitation reactions. It represents the maximum product of ion concentrations in a saturated solution before a precipitate forms. For example, the Ksp for AgCl is given as \(1.77 \times 10^{-10}\) at 25°C. This extremely low value indicates that AgCl is not very soluble in water. Knowing the Ksp value allows us to predict whether a precipitate will form in a solution by comparing it to the ion product (Q). If Q exceeds Ksp, the solution is supersaturated, and a precipitate will form.

Ion Concentration Calculations

Ion concentration calculations are essential for determining whether a precipitate will form. To calculate ion concentrations, you first need to determine how many moles of each ion are present. In the given problem, we calculated the moles of AgNO3 and NaCl:

• Moles of AgNO3: \(\frac{1.70 \text{ g}}{169.87 \text{ g/mol}} = 0.0100 \text{ mol} \)
• Moles of NaCl: \(\frac{14.5 \text{ g}}{58.44 \text{ g/mol}} = 0.248 \text{ mol} \)

We next determined the concentrations of Ag+ and Cl- ions by dividing the number of moles by the volume of the solution in liters (0.200 L):

• \([\text{Ag}^+] = \frac{0.0100 \text{ mol}}{0.200 \text{ L}} = 0.050 \text{ M} \)
• \([\text{Cl}^-] = \frac{0.248 \text{ mol}}{0.200 \text{ L}} = 1.24 \text{ M}\)

These concentrations are then used to determine the ion product (Q).

Dissolution Reactions

Dissolution reactions help us understand how compounds break down into ions in a solution. For example, AgNO3 and NaCl dissolve in water to form ions as follows:

• \(\text{AgNO}_3 (s) \rightarrow \text{Ag}^+ (aq) + \text{NO}_3^- (aq)\)
• \(\text{NaCl} (s) \rightarrow \text{Na}^+ (aq) + \text{Cl}^- (aq)\)

When these salts dissolve, they release ions into the solution. In our problem, solid AgNO3 produces Ag+ and NO3- ions, while NaCl produces Na+ and Cl- ions. These dissolution reactions are the first step in analyzing whether a precipitate will form.

Chemical Equilibrium

Understanding chemical equilibrium is essential for solving precipitation problems. When a solution reaches equilibrium, the rate of the forward reaction (dissolution) equals the rate of the reverse reaction (precipitation). At this point, the concentration of ions in the solution reaches a constant value.

In our example, the equilibrium can be reached if the ion product (Q) is equal to the solubility product constant \(K_{sp}\). If Q is less than Ksp, more dissolution occurs until equilibrium is reached. Conversely, if Q exceeds Ksp, excess ions combine to form a precipitate, which continues until the system returns to equilibrium. In the problem, Q was significantly larger than Ksp for AgCl, indicating that a precipitate (AgCl) forms as the system seeks equilibrium.