Problem 15 - Analytical Chemistry

Question

A 50 -mL solution that is \(0.10 M\) in both chloride and iodide ions is titrated with \(0.10 M\) silver nitrate. (a) Calculate the percent iodide remaining unprecipitated when silver chloride begins to precipitate. (b) Calculate the potential of a silver electrode versus the SCE when silver chloride begins to precipitate and compare this with the theoretical potential corresponding to end point for the titration of iodide. (c) Calculate the potential at the end point for chloride. For simplicity, in lieu of activities, use concentrations in calculations.

Step-by-Step Solution

Answer

(a) Approximately 0.000047% iodide remains; (b) Potential is ~1.013 V for silver chloride, 1.327 V for iodide endpoint; (c) Potential at chloride endpoint is ~1.451 V.

Step 1: Determine the reaction equations

The titration involves the formation of precipitates as silver ions react with chloride and iodide ions. The primary reactions are:1. \( \text{Ag}^+ + \text{Cl}^- \rightarrow \text{AgCl(s)} \)2. \( \text{Ag}^+ + \text{I}^- \rightarrow \text{AgI(s)} \)Since the product of these reactions are solid precipitates, they are largely irreversible and drive the reaction forward.

Step 2: Establish solubility product constants

We now need to use the solubility product constants \( K_{sp} \) for silver chloride and silver iodide. These values are crucial as they determine when precipitation occurs:- \( K_{sp,AgCl} = 1.8 \times 10^{-10} \)- \( K_{sp,AgI} = 8.5 \times 10^{-17} \)

Step 3: Calculate concentration of iodide when chloride begins to precipitate

Chloride begins to precipitate when the concentration \([\text{Ag}^+]\) equals the corresponding \( K_{sp,AgCl} \) divided by \([\text{Cl}^-]\). Since initially both chloride and iodide are at 0.1 M, as the silver nitrate is added, chloride will start to precipitate when:\( [\text{Ag}^+] = \frac{K_{sp,AgCl}}{[\text{Cl}^-]} = \frac{1.8 \times 10^{-10}}{0.1} = 1.8 \times 10^{-9} \text{ M} \)Next, the concentration of iodide left is now calculated using the same silver ion concentration:\( K_{sp,AgI} = [\text{Ag}^+][\text{I}^-] = 8.5 \times 10^{-17} \)So \([\text{I}^-] = \frac{K_{sp,AgI}}{[\text{Ag}^+]} = \frac{8.5 \times 10^{-17}}{1.8 \times 10^{-9}} \approx 4.7 \times 10^{-8} \text{ M} \).

Step 4: Calculate percent iodide remaining unprecipitated

Initially, iodide was 0.10 M. Find out how much of this remains:\( \text{Percent Iodide remaining} = \frac{[\text{I}^-]_{remaining}}{[\text{I}^-]_{initial}} \times 100% = \frac{4.7 \times 10^{-8}}{0.10} \times 100\% \approx 0.000047\% \)

Step 5: Calculate the potential when silver chloride begins to precipitate

Use the Nernst equation:\[ E = E^0 - \frac{RT}{nF} \ln [\text{Ag}^+] \]Where:- \( E^0_{Ag/Ag^+} \) is the standard electrode potential of silver \( (+0.799 \text{ V}) \).- \( [\text{Ag}^+] = 1.8 \times 10^{-9} \text{ M} \)Assuming room temperature, substitute the values:\[ E = 0.799 - \frac{8.314 \times 298}{1 \times 96485} \ln(1.8 \times 10^{-9}) \]\[ E \approx 0.799 + 0.117 \text{ V} \times \, (-20.13) \approx 1.013 \text{ V} \]

Step 6: Theoretical endpoint potential for iodide

For the iodide ion at its endpoint (where iodide concentrations become negligible):\[ E = E^0 - \frac{RT}{nF} \ln \left( \frac{K_{sp, AgI}}{0.10} \right) \approx 0.799 - 0.059 \times (-8.93) \approx 1.327 \text{ V} \]

Step 7: Calculate endpoint potential for chloride

For chloride:\[ E = E^0 - \frac{RT}{nF} \ln \left( \frac{K_{sp, AgCl}}{0.10} \right) \approx 0.799 - 0.059 \times (-10.95) \approx 1.451 \text{ V} \]

Key Concepts

Titration

Titration is a common laboratory method used to determine the concentration of an unknown solution by reacting it with a solution of known concentration. In this process, the known solution, called the titrant, is added to the unknown solution until a reaction is complete, often signified by a noticeable change, such as a color change or a precipitate formation. For the exercise problem, silver nitrate serves as the titrant, while the solution with chloride and iodide ions is the analyte.
Titrations involving precipitate formation, such as the reaction between silver ions and chloride or iodide ions, are often referred to as precipitation titrations. As the titrant is added, it reacts with one ion until it exceeds a certain threshold concentration, causing a solid to form. This solid, or precipitate, helps indicate the point where a particular ion's reaction is complete.

• Precipitation begins once the concentrations meet the solubility product constant conditions.
• The titration endpoint is approached when no more analyte can react with the titrant.

In our specific example, as silver nitrate adds silver ions, they react first with the more reactive iodide ions before forming silver iodide precipitate, leaving chloride ions for continued titration until silver chloride formation begins.

Solubility Product Constant

The solubility product constant, denoted as \( K_{sp} \), is a value that represents the equilibrium point of a dissolution reaction. It indicates the extent to which a compound can dissolve in a solution. For sparingly soluble salts like silver chloride (AgCl) and silver iodide (AgI), the \( K_{sp} \) values are critical in determining when precipitation occurs.
In the titration exercise, as silver ions are added to the solution, their interaction with chloride and iodide ions reaches a level defined by the respective \( K_{sp} \) values. When the concentration of silver ions multiplied by that of the chloride ions in the solution equals \( K_{sp,AgCl} \), precipitation of AgCl starts. Similarly, the same relationship applies to iodide before AgI precipitates.

• For AgCl: \( K_{sp,AgCl} = 1.8 \times 10^{-10} \)
• For AgI: \( K_{sp,AgI} = 8.5 \times 10^{-17} \)

These constants help predict and calculate the concentrations at which salts will precipitate, thus aiding in understanding the titration's progress and endpoint.

Electrode Potential

Electrode potential is a measure of how likely an electrode can gain or lose electrons, thus driving an electrochemical reaction. For silver electrodes in this problem, the potential reflects the reaction between the silver ion \(( ext{Ag}^+)\) and the silver electrode. The Nernst equation helps us calculate this potential under non-standard conditions.
The Nernst equation is given by:\[ E = E^0 - \frac{RT}{nF} \ln Q \]Where:

• \( E \) is the electrode potential.
• \( E^0 \) is the standard electrode potential.
• \( R \) is the gas constant.
• \( T \) is the temperature in Kelvin.
• \( n \) is the number of moles of electrons transferred.
• \( F \) is the Faraday constant.
• \( Q \) is the reaction quotient, often expressed as the concentration of ions.

In the problem, as \( [ ext{Ag}^+] \) reaches the required concentrations to begin precipitation of chloride or iodide, we calculate the corresponding electrode potential using the Nernst equation. This potential informs us about the feasibility and progression of the titration, enabling insights into when the completion of both the iodide and chloride reactions are reached.