✔ 最佳答案
1)
Let y M be the [Ag⁺] in the sat. AgCl solution.
y < 0.01 M
For the 0.01 M AgNO3 solution: E1 = E°+ 0.059 x log(0.01) ... [1]
For the sat. AgNO3 solution: E2 = E° + 0.059 x logy ... [2]
[1] - [2]:
Ecell = 0.059 x {log(0.01) - logy}
0.059 x (-2 - logy) = 0.17
-2 - logy = 0.17/0.059
logy = -2 - (0.17/0.059)
y = 1.31 x 10^-5
[Ag⁺] in the sat. AgCl solution = 1.31 x 10^-5 M
2)
AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq)
When the small volume of KCl solution is added, the concentration of Cl⁻ ions increases. This will shift the equilibrium position of the left, and thus the solubility of AgCl decreases. This is known as common ion effect.