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Step-by-Step Solution
Step 1: Write the dissolution equilibrium for AgI
Silver iodide (AgI) dissociates in water according to the balanced equilibrium equation:
AgI
ā
Ag+
+
Iā
Step 2: Express the solubility product (Ksp)
The solubility product for AgI is given by:
$K_{sp} = [\text{Ag}^+] \, [\text{I}^-]$
Numerically, it is:
$K_{sp} = 1.0 \times 10^{-16} \, \text{mol}^2 \, \text{L}^{-2}$
Step 3: Determine the solubility in pure water
If the solubility of AgI in pure water is $s$ mol Lā1, then:
$K_{sp} = s \times s = s^2
\quad \Rightarrow \quad
1.0 \times 10^{-16} = s^2
\quad \Rightarrow \quad
s = 1.0 \times 10^{-8} \, \text{mol L}^{-1}$
Hence, in pure water:
$[\text{Ag}^+] = 1.0 \times 10^{-8} \, \text{mol L}^{-1}$
Step 4: Account for the common ion (Iā) from KI solution
The question provides a 10ā4 N (approx. 10ā4 M) solution of KI. In this solution:
$[\text{I}^-] \approx 10^{-4} \, \text{mol L}^{-1}$
Because adding $1.0 \times 10^{-8} \, \text{mol L}^{-1}$ of Iā (from dissolved AgI) is negligible compared to 10ā4, the iodide ion concentration essentially remains 10ā4 M.
Step 5: Apply the solubility product expression with the common ion
Let the new solubility of AgI in the KI solution be $s'$ mol Lā1. Then:
$K_{sp} = [\text{Ag}^+] \, [\text{I}^-]
\quad \Rightarrow \quad
1.0 \times 10^{-16} = s' \times (10^{-4})$
Solving for $s'$ gives:
$s' = \frac{1.0 \times 10^{-16}}{10^{-4}} = 1.0 \times 10^{-12} \, \text{mol L}^{-1}$
Final Answer
The solubility of AgI in 10ā4 N KI at 25Ā°C is
$1.0 \times 10^{-12} \, \text{mol L}^{-1}$.
