Solid AgNO₃ is slowly added to a solution containing each
of 0.01 M NaCl and 0.001 M NaBr. What will be the
concentration of Cl⁻ ions in solution when AgBr will just
start to precipitate ?
Kₛₚ (AgBr) = 3.6 × 10⁻¹³, Kₛₚ (AgCl) = 1.8 × 10⁻¹⁰.
(a) 1.8 × 10⁻⁷ (b) 3.6 × 10⁻¹⁰
(c) 0.01 (d) 2 × 10⁻⁴
Answers
Answer:
Your answer is option (d) 2 × 10⁻⁴
Please Note - The correct set of options for the question is :
(a) 1.8 × 10⁻⁷ M
(b) 3.6 × 10⁻¹⁰ M
(c) 0.01 M
(d) 2 × 10⁻⁵ M
The concentration of Cl⁻ ions in the solution when AgBr will just start to precipitate is 2 × 10⁻⁵ M. Therefore, option (d) is the correct answer.
• The reaction between NaCl and AgNO₃ proceeds as follows :
NaCl + AgNO₃ → AgCl + NaNO₃
• Initial concentration of NaCl = 0.01 M
=> Initial concentration of Cl⁻ ion in NaCl = 0.01 M
=> Final concentration of Cl⁻ ion in AgCl = 0.01 M
• The solubility product of AgCl = Concentration of Ag × Concentration of Cl
∴ Kₛₚ (AgCl) = [Ag⁺] × [Cl⁻-
=> 1.8 × 10⁻¹⁰ M² = [Ag⁺] × 0.01 M
=> [Ag⁺] = (1.8 × 10⁻¹⁰ M²) / 0.01 M
=> [Ag⁺] = 1.8 × 10⁻⁸ M
• Now, the reaction between NaBr and AgNO₃ can be represented as follows :
NaBr + AgNO₃ → AgBr + NaNO₃
• Initial concentration of NaBr = 0.001 M
=> Initial concentration of Br⁻ ion in NaBr = 0.001 M
=> Final concentration of Br⁻ ion AgBr = 0.001 M
• The solubility product of AgCl = Concentration of Ag × Concentration of Br
∴ Kₛₚ (AgBr) = [Ag⁺] × [Br⁻]
=> 3.6 × 10⁻¹³ M² = [Ag⁺] × 0.001 M
=> [Ag⁺] = (3.6 × 10⁻¹³ M²) / 0.001 M
=> [Ag⁺] = 3.6 × 10⁻¹⁰ M
• Thus, we see that the concentration of Ag⁺ ions in AgBr is lower than the concentration of Ag⁺ ions in AgCl. This indicates that AgBr will precipitate before AgCl.
• Therefore, for AgCl to precipitate, the concentration of Ag⁺ ions in the solution should be equal to the concentration of Ag⁺ ions in AgCl, i.e., 1.8 × 10⁻⁸.
• Now , the concentration of Cl⁻ ions in the solution when AgBr just begins to precipitate is :
[Cl⁻] = Kₛₚ (AgBr) / [Ag⁺]
=> [Cl⁻] = (3.6 × 10⁻¹³) / (1.8 × 10⁻⁸)
=> [Cl⁻] = 2 × 10⁻⁵ M