The solubility product of AgI at 25ºC is 1.0×10–16 mol2L–2. The solubility of AgI in 10–4N solution of KI at 25ºC is approximately (in mol L–1):
- 1.0×10–16
- 1.0×10–12
- 1.0×10–10
- 1.0×10–8
The Correct Option is B
Solution and Explanation
To find the solubility of AgI in a 10–4 N solution of KI, we need to consider the common ion effect. AgI is a sparingly soluble salt, and its solubility is reduced in the presence of a common ion (I– from KI).
The solubility equilibrium for AgI is:
\[ \text{AgI (s)} \leftrightarrow \text{Ag}^+ (aq) + \text{I}^- (aq) \]The solubility product (Ksp) of AgI is given as 1.0×10–16 mol2L–2. In an unsaturated or saturated solution of AgI, Ksp is expressed as:
\[ K_{sp} = [\text{Ag}^+][\text{I}^-] \]Let S be the solubility of AgI in mol L–1 in the KI solution. In this solution, the concentration of I– ions originally provided by KI is 10-4 mol L–1, and S is very small compared to 10-4. Therefore, the total concentration of I– can be approximated as 10-4 mol L–1.
Substitute the concentrations into the Ksp expression:
\[ K_{sp} = [\text{Ag}^+][\text{I}^-] = S \times 10^{-4} \]Setting up the equation with Ksp:
\[ 1.0 \times 10^{-16} = S \times 10^{-4} \]Solving for S:
\[ S = \frac{1.0 \times 10^{-16}}{10^{-4}} = 1.0 \times 10^{-12} \]Thus, the solubility of AgI in a 10-4N solution of KI is approximately 1.0 × 10-12 mol L-1.
Therefore, the correct answer is 1.0 × 10-12.
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