Question

25 mL of silver nitrate solution (1M) is added dropwise to 25 mL of potassium iodide (1.05 M) solution. The ions(s) present in very small quantity in the solution is/are:

• A. NO₃⁻ only
• B. Ag⁺ and I⁻ both
• C. Ag⁺ only
• D. I⁻ only

Hint:

When mixing solutions of ionic compounds, check the solubility products (Ksp) to predict whether the ions will precipitate or remain in solution.

Answer 1

The Correct Option is C

To solve this problem, we need to understand the chemical reaction that occurs when silver nitrate (\( \text{AgNO}_3 \)) is added to potassium iodide (\( \text{KI} \)).

Step 1: Determine the initial concentrations.

• Volume of \( \text{AgNO}_3 \) solution = 25 mL = 0.025 L
• Molarity of \( \text{AgNO}_3 \) = 1 M
• Moles of \( \text{Ag}^+ \) = \( 1 \, \text{mol/L} \times 0.025 \, \text{L} = 0.025 \, \text{mol} \)
• Volume of \( \text{KI} \) solution = 25 mL = 0.025 L
• Molarity of \( \text{KI} \) = 1.05 M
• Moles of \( \text{I}^- \) = \( 1.05 \, \text{mol/L} \times 0.025 \, \text{L} = 0.02625 \, \text{mol} \)

Step 2: Write the balanced chemical equation.

The reaction that occurs is:

\[\text{Ag}^+ + \text{I}^- \rightarrow \text{AgI} \, \text{(precipitate)}\]

Step 3: Determine the limiting reactant.

• Moles of \( \text{Ag}^+ \) = 0.025 mol
• Moles of \( \text{I}^- \) = 0.02625 mol

Since the moles of \( \text{Ag}^+ \) are less than the moles of \( \text{I}^- \), \( \text{Ag}^+ \) is the limiting reactant.

Step 4: Calculate the ions left in the solution after the reaction.

• All \( \text{Ag}^+ \) will react with \( \text{I}^- \) to form \( \text{AgI} \) precipitate.
• Moles of \( \text{I}^- \) left = \( 0.02625 \, \text{mol} - 0.025 \, \text{mol} = 0.00125 \, \text{mol} \)

Now, we consider other ions starting with \( \text{NO}_3^- \).

• \( \text{NO}_3^- \) does not participate in the reaction and remains in the solution.

Conclusion:

Since \( \text{I}^- \) and \( \text{NO}_3^- \) are in excess, they will remain in significant quantities compared to \( \text{Ag}^+ \). Since all \( \text{Ag}^+ \) is reacting to form \( \text{AgI} \), which precipitates, any leftover amount of \( \text{Ag}^+ \) would be very small, hence the answer is:

Ag⁺ only