Question

Suppose that a 10-mL sample of a solution is to be tested for \(\mathrm{Cl}^{-}\) ion by addition of 1 drop \((0.2 \mathrm{~mL})\) of \(0.10 \mathrm{M}\) \(\mathrm{AgNO}_{3}\). What is the minimum number of grams of \(\mathrm{Cl}^{-}\) that must be present for \(\mathrm{AgCl}(\mathrm{s})\) to form?

Short answer

The minimum number of grams of Cl⁻ that must be present in the 10 mL solution for AgCl(s) to form upon the addition of 0.2 mL of 0.10 M AgNO₃ is approximately \(3.2\) × \(10^{-8}\) g.

Step-by-step solution

Write the balanced chemical equation

For the precipitation reaction between Ag⁺ and Cl⁻ ions, the balanced equation is: \[ \mathrm{Ag^+}(aq) + \mathrm{Cl^-}(aq) \longrightarrow \mathrm{AgCl}(s) \]

Find the solubility product constant

In this problem, the equilibrium constant we are interested in is the solubility product constant (Kₛₚ) for AgCl. To find this, we can refer to a table for the Kₛₚ values of various compounds. The Kₛₚ of AgCl is: \[ K_{sp}(\mathrm{AgCl}) = 1.77\, × \,10^{-10} \]

Calculate the concentration of Ag⁺ ions

Since we are given that a 0.2 mL droplet containing 0.10 M AgNO₃ is added to the 10 mL sample, we can compute the resulting concentration of Ag⁺ ions in the entire solution. First, we find the moles of Ag⁺ in the drop: Moles of Ag⁺ = volume (drop) × concentration of AgNO₃ Moles of Ag⁺ = (0.2 mL) × (0.10 mol/L) To convert the volume to liters, we divide by 1000: Moles of Ag⁺ = (0.0002 L) × (0.10 mol/L) = 2 × 10^{-5} mol Now compute the concentration of Ag⁺ ions in the entire solution: Concentration of Ag⁺ ions = moles of Ag⁺ / total volume (drop + solution) Volume of drop + solution = 10 mL + 0.2 mL = 10.2 mL Concentration of Ag⁺ = (2 × 10^{-5} mol) / (0.0102 L) = 1.96 × 10^{-3} M

Calculate the minimum concentration of Cl⁻ ions for precipitation

The equilibrium expression for the precipitation of AgCl is: \[ K_{sp}(\mathrm{AgCl}) = [\mathrm{Ag^+}][\mathrm{Cl^-}] \] We need to find the minimum concentration of Cl⁻ ions for precipitation to occur. Therefore, we can solve for this by rearranging the formula: Minimum [\(\mathrm{Cl^-}\)] = \(K_{sp}\,(\mathrm{AgCl})\) / [\(\mathrm{Ag+}\)] Substitute the values for Kₛₚ and [\(\mathrm{Ag+}\)]: Minimum [\(\mathrm{Cl^-}\)] = \(1.77 × 10^{-10} \) / \(1.96 × 10^{-3} \) = \(9.03 × 10^{-8}\) M

Find the minimum number of grams of Cl⁻ ions needed

Now that we have the minimum concentration of Cl⁻ ions required for precipitation, we can find the minimum mass of Cl⁻ ions that must be present in the 10 mL solution. First, we determine the moles of Cl⁻ ions in the 10 mL solution: Moles of Cl⁻ = concentration of Cl⁻ × volume (solution) Moles of Cl⁻ = \(9.03 × 10^{-8}\,M \) × \(0.010\,L\) = \(9.03 × 10^{-10}\,mol\) Now, we convert moles of Cl⁻ into grams, knowing that the molar mass of Cl⁻ ion is 35.45 g/mol: Minimum grams of Cl⁻ ions = moles of Cl⁻ × molar mass of Cl⁻ Minimum grams of Cl⁻ ions = \(9.03 × 10^{-10}\,mol\) × \(35.45\,g/mol\) ≈ \(3.20 × 10^{-8}\,g\) Thus, the minimum number of grams of Cl⁻ that must be present in the 10 mL solution for AgCl(s) to form upon the addition of 0.2 mL of 0.10 M AgNO₃ is approximately \(3.2\) × \(10^{-8}\) g.

Key concepts

Precipitation Reaction

A precipitation reaction occurs when two soluble substances in solution react to form an insoluble solid, called a precipitate. In this scenario, when silver nitrate (AgNO₃) is added to a solution containing chloride ions (Cl⁻), a chemical reaction takes place. The products of this reaction are silver chloride (AgCl), which is the precipitate, and nitrate ions (NO₃⁻), which remain in the solution.

Here is the balanced chemical equation for the reaction:

• \[ \mathrm{Ag^+} (aq) + \mathrm{Cl^-} (aq) \rightarrow \mathrm{AgCl} (s) \]

For precipitation to occur, the product of the ionic concentrations of the involved ions must exceed the solubility product constant (Kₛₚ) of the precipitate. Once the limit is reached or exceeded, the solid precipitate will form out of the solution.

AgNO₃ Solution

Silver nitrate (\(\mathrm{AgNO₃}\)) is a chemical compound often used in precipitation reactions due to its solubility in water. When dissolved, it dissociates fully to give silver ions (\(\mathrm{Ag^+}\)) and nitrate ions (\(\mathrm{NO₃^-}\)). This dissociation is critical for the precipitation reaction since the \(\mathrm{Ag^+}\) ions are the reactive component that will combine with \(\mathrm{Cl^-}\) ions to form \(\mathrm{AgCl(s)}\).

In this exercise, adding a 0.2 mL drop of a 0.10 M \(\mathrm{AgNO₃}\) solution introduces \(\mathrm{Ag^+}\) ions into the system, which can then react to form a precipitate. The concentration of \(\mathrm{Ag^+}\) ions and their interaction with \(\mathrm{Cl^-}\) ions determine if precipitation will occur.

It is important to note that \(\mathrm{AgNO₃}\) solutions should be handled with care, as they can stain skin and clothing upon exposure.

Concentration Calculation

Concentration calculations are essential for understanding how and when a precipitation reaction will occur. In this exercise, it involves determining the concentration of \(\mathrm{Ag^+}\) ions when a known volume of \(\mathrm{AgNO₃}\) is added to a solution.

First, you calculate the moles of \(\mathrm{Ag^+}\) in a given volume. Let's break it down:

• Moles of \(\mathrm{Ag^+} = \text{volume (in liters)} \times \text{concentration (molarity)}\)
• For 0.2 mL of a 0.10 M solution, convert to liters by dividing by 1000.
• Moles = \(0.0002 \, L \times 0.10 \, \mathrm{mol/L} = 2 \times 10^{-5} \, \mathrm{mol} \)

Next, compute the concentration of these ions in the new, total volume of solution (10.2 mL):

• Total volume = drop + solution
• \[ \mathrm{Concentration \ of \ Ag^+} = \frac{\mathrm{Moles \ of \ Ag^+}}{\mathrm{Volume \ in \ liters}} \approx 1.96 \times 10^{-3} \, \mathrm{M} \]

This concentration is then used to determine if the ion product exceeds the solubility product constant.

AgCl Formation

To form silver chloride (\(\mathrm{AgCl}\)), the concentration of chloride ions (\(\mathrm{Cl^-}\)) must be high enough in the solution. The reaction is driven by the solubility product constant (Kₛₚ) value of AgCl, which is a very low 1.77 x 10⁻¹⁰. This indicates that AgCl is not very soluble in water.

For precipitation to occur, the ionic product of \(\mathrm{Ag^+}\) and \(\mathrm{Cl^-}\) ions should be equal to or exceed the Kₛₚ. Using the formula below, you can determine the minimum concentration of \(\mathrm{Cl^-}\) required:

• \[ \mathrm{K_{sp} = [Ag^+][Cl^-]} \]
• Rearranging, \( \mathrm{[Cl^-]_{min} = \frac{K_{sp}}{[Ag^+]}} \)
• Plugging in, \( \mathrm{[Cl^-]_{min} = \frac{1.77 \times 10^{-10}}{1.96 \times 10^{-3}} \approx 9.03 \times 10^{-8} \, M} \)

Once you have the minimum concentration, calculate the mass of \(\mathrm{Cl^-}\) necessary in the solution to achieve this concentration, ensuring enough ions are present for AgCl precipitation.