0.1mol溴化银能溶在1L多少浓度的硫代硫酸钠溶液中？（如果是碘化银呢）

Question

如题所述

Answer 1

AgBr + 2S2O3(2-) = [Ag(S2O3)](3-) + Br(-)
x 0 0
x-0.1*2 0.1 0.1
K = [ [Ag(S2O3)](3-) ]*[Br(-)]*[Ag(+)] /{[S2O3(2-)]^2 * [Ag(-)]
= Kf * KspAgBr = 2.9*10^13 * 5.35*10(-13) = 15.5
= 0.1^2/(x - 0.2)^2
x ~ = 0.225 mol/L

AgI + 2S2O3(2-) = [Ag(S2O3)](3-) + I(-)
y 0 0
y-0.1*2 0.1 0.1
K' = Kf * KspAgI = 2.9*10^13 * 8.51*10(-17) = 0.00247
0.1^2/(y - 0.2)^2 = 0.00247
y ~ = 2.21 mol/L本回答被提问者采纳