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

Solution

Chemistry ma’am · 3 steps

Silver iodide precipitates almost completely, so ask which ions are left behind and how much of each.

1. Count millimoles. Ag⁺ added is 25 × 1 = 25 mmol. I⁻ present is 25 × 1.05 = 26.25 mmol. K⁺ and NO₃⁻ are spectator ions and stay fully in solution.
2. Precipitation. Ag⁺ + I⁻ → AgI v goes almost to completion because Kₛₚ of AgI is tiny, about 10⁻¹⁷. All 25 mmol of Ag⁺ uses up 25 mmol of I⁻, leaving only 1.25 mmol of I⁻ in the 50 mL mixture.

   Ion	Millimoles left	Concentration in 50 mL	Amount
   K⁺	26.25	0.525 M	large
   NO₃⁻	25	0.5 M	large
   I⁻	1.25	0.025 M	very small
   Ag⁺	almost 0	Kₛₚ/[I⁻] ∼ 10⁻¹⁵ M	very small

3. So. The spectator ions stay at about 0.5 M, while I⁻ is twenty times less and Ag⁺ is only what Kₛₚ allows. Both Ag⁺ and I⁻ are present in very small quantity, which is option (c).

The idea to keep. When two solutions mix and a precipitate forms, the spectator ions stay in full while the precipitating ions remain only as the small excess or as what Kₛₚ allows.

Answer from the NTA final answer key. Solution by PrepWiser.