To solve this problem, we need to break it down into several steps. First, we need to understand the electrolysis of copper(II) sulfate (CuSO4) and how it affects the pH of the solution. Then, we will look at the reaction that occurs when iodine (I2) is formed after treating the solution with potassium iodide (KI) and how to calculate the volume of sodium persulfate (Na2S2O8) needed for titration.
Understanding the Electrolysis Process
When you electrolyze a solution of copper(II) sulfate using inert electrodes, copper ions (Cu²⁺) are reduced at the cathode, and sulfate ions (SO₄²⁻) are oxidized at the anode. This process leads to the deposition of copper metal and the formation of sulfuric acid, which can lower the pH of the solution. In your case, the pH dropped to 2, indicating a significant concentration of hydrogen ions (H⁺) in the solution.
Formation of Iodine
After electrolysis, when you treat the solution with excess potassium iodide (KI), iodine (I2) is produced. This occurs because the copper ions can oxidize iodide ions (I⁻) to form iodine. The reaction can be summarized as follows:
- 2 Cu²⁺ + 4 I⁻ → 2 CuI + I2
Here, the iodine produced will react with sodium persulfate during the titration.
Calculating the Amount of Iodine Formed
To find out how much sodium persulfate is needed, we first need to determine the amount of iodine produced. The stoichiometry of the reaction shows that 1 mole of I2 reacts with 1 mole of Na2S2O8. Therefore, we need to find out how many moles of I2 were produced from the electrolysis process.
Determining Moles of Iodine
Assuming that all the Cu²⁺ ions were converted to I2, we need to know the initial concentration of CuSO4. If we started with 100 mL of CuSO4 solution, we can calculate the moles of Cu²⁺ present. For example, if the concentration of CuSO4 was 0.1 M, then:
- Moles of Cu²⁺ = Concentration × Volume = 0.1 mol/L × 0.1 L = 0.01 moles
This means that 0.01 moles of I2 would be produced, as the stoichiometry of the reaction shows a 1:1 ratio.
Using Sodium Persulfate for Titration
Now, we can calculate the volume of 0.04 M Na2S2O8 needed to react with the iodine. Since 1 mole of I2 reacts with 1 mole of Na2S2O8, we can use the following formula:
- Volume (L) = Moles / Concentration
Substituting the values we have:
- Volume of Na2S2O8 = 0.01 moles / 0.04 mol/L = 0.25 L
This converts to 250 mL of Na2S2O8 solution required for the titration.
Final Summary
In summary, after electrolyzing 100 mL of CuSO4 solution and treating it with excess KI, you would produce iodine. The titration of this iodine with 0.04 M Na2S2O8 would require 250 mL of the sodium persulfate solution to fully react with the iodine formed. This process illustrates the interconnectedness of electrochemical reactions and stoichiometry in analytical chemistry.
