Question

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

• A. Ag+
• B. I-
• C. Ag+ and I-
• D. K+ and NO3-

Answer 1

Answer: (C)

Ag+ and I-

Answer 2

The reaction between silver nitrate ($AgNO_3$) and potassium iodide ($KI$) forms silver iodide precipitate ($AgI$) and potassium nitrate ($KNO_3$). The balanced chemical equation is: $AgNO_3(aq) + KI(aq) \rightarrow AgI(s) + KNO_3(aq)$ The stoichiometry is 1:1.

Moles of $AgNO_3 = 1 \text{ mol/L} \times 0.025 \text{ L} = 0.025 \text{ mol}$ Moles of $KI = 1.05 \text{ mol/L} \times 0.025 \text{ L} = 0.02625 \text{ mol}$

$AgNO_3$ is the limiting reactant. After the reaction, $Ag^+$ is almost entirely consumed to form $AgI$. Its concentration is extremely low, determined by the solubility product ($K_{sp}$) of $AgI$. $KI$ is in excess. Moles of $I^-$ remaining = $0.02625 \text{ mol} - 0.025 \text{ mol} = 0.00125 \text{ mol}$. The total volume is $50 \text{ mL} = 0.050 \text{ L}$. Concentration of excess $I^- = 0.00125 \text{ mol} / 0.050 \text{ L} = 0.025 \text{ M}$. Concentration of $Ag^+ \approx (K_{sp} \text{ of } AgI) / [I^-] \approx (8.5 \times 10^{-17}) / 0.025 \approx 3.4 \times 10^{-15} \text{ M}$. Spectator ions $K^+$ and $NO_3^-$ are present in higher concentrations: $[NO_3^-] = 0.025 \text{ mol} / 0.050 \text{ L} = 0.5 \text{ M}$. $[K^+] = 0.02625 \text{ mol} / 0.050 \text{ L} = 0.525 \text{ M}$.

Both $Ag^+$ (due to $K_{sp}$) and $I^-$ (as the excess reactive ion, significantly less concentrated than spectator ions) are present in very small quantities compared to the spectator ions.