Question:
The solubility of AgCl in 0.2 M NaCl solution is ($Kₛp$ of $AgCl = 1.20 × 10^-10$).
The solubility of AgCl in 0.2 M NaCl solution is ($Kₛp$ of $AgCl = 1.20 × 10^-10$).
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The solubility of AgCl in 0.2 M NaCl solution is ($Ksp$ of $AgCl = 1.20 x 10$).
Updated On: Apr 15, 2026
- $6.0 \times 10^{-10} M$
- 0.2 M
- $1.2 \times 10^{-10} M$
- $0.2 \times 10^{-10} M$
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The Correct Option is A
Solution and Explanation
Step 1: Concept
Due to the common ion effect, the presence of $Cl^{-}$ from NaCl will reduce the solubility of AgCl.
Step 2: Equilibrium Equations
$AgCl (s) \rightleftharpoons Ag^{+} (aq) + Cl^{-} (aq)$
Let solubility be $x$. $[Ag^{+}] = x$ and $[Cl^{-}] = (x + 0.2)$.
Step 3: Calculation
$K_{sp} = [Ag^{+}][Cl^{-}] = x(x + 0.2)$
As $x$ is very small, $(x + 0.2) \approx 0.2$.
$1.2 \times 10^{-10} = 0.2x \Rightarrow x = 6.0 \times 10^{-10} M$.
Final Answer: (A)
Due to the common ion effect, the presence of $Cl^{-}$ from NaCl will reduce the solubility of AgCl.
Step 2: Equilibrium Equations
$AgCl (s) \rightleftharpoons Ag^{+} (aq) + Cl^{-} (aq)$
Let solubility be $x$. $[Ag^{+}] = x$ and $[Cl^{-}] = (x + 0.2)$.
Step 3: Calculation
$K_{sp} = [Ag^{+}][Cl^{-}] = x(x + 0.2)$
As $x$ is very small, $(x + 0.2) \approx 0.2$.
$1.2 \times 10^{-10} = 0.2x \Rightarrow x = 6.0 \times 10^{-10} M$.
Final Answer: (A)
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