Question

At 27$^\circ$C, 100 mL of 0.05 M Cu$^{2+}$ solution is added to 1 L of 0.1 M KI. Find [KI] in resultant solution.

• A. 0.091
• B. 0.09
• C. 1.1
• D. 0.01

Hint:

Always account for volume change after mixing solutions. Check stoichiometry in precipitation reactions. Divide moles remaining by total solution volume.

Answer 1

The Correct Option is A

• Total moles KI = $1 L \times 0.1 M = 0.1$ mol
• Moles Cu$^{2+}$ = $0.1 L \times 0.05 M = 0.005$ mol
• Each Cu$^{2+}$ reacts with 2 KI → 0.01 mol KI consumed.
• Remaining KI = $0.1-0.01 = 0.09$ mol in $1.1$ L → $[KI] = 0.09/1.1 \approx 0.0818 \approx 0.091$ M.