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

We add 25 mL of 1 M AgNO$$_3$$ to 25 mL of 1.05 M KI and need to determine which ions are present in very small quantity in the resulting solution.

The initial moles are: Ag$$^+$$ from AgNO$$_3$$ = $$25 \times 1 = 25$$ mmol, I$$^-$$ from KI = $$25 \times 1.05 = 26.25$$ mmol, K$$^+$$ = 26.25 mmol, and NO$$_3^-$$ = 25 mmol.

The precipitation reaction is

Silver iodide is highly insoluble with $$K_{sp} = 8.5 \times 10^{-17}$$, so the reaction goes essentially to completion.

Since Ag$$^+$$ (25 mmol) and I$$^-$$ (26.25 mmol) react in a 1:1 ratio, Ag$$^+$$ is the limiting reagent. After precipitation, the remaining I$$^-$$ is $$26.25 - 25 = 1.25$$ mmol in 50 mL total, giving a concentration of $$\frac{1.25}{50} \times 1000 = 0.025$$ M.

Now, from the solubility equilibrium of AgI in the presence of excess I$$^-$$,

Comparing the concentrations of all ions in the 50 mL solution: K$$^+$$ = 0.525 M, NO$$_3^-$$ = 0.5 M, I$$^-$$ = 0.025 M, and Ag$$^+$$ = $$3.4 \times 10^{-15}$$ M. Both Ag$$^+$$ and I$$^-$$ are present in very small quantity relative to the spectator ions.

Hence, the correct answer is Option 4: Ag$$^+$$ and I$$^-$$ both.