几道化学题
1.某溶液水电离出的c[H+]=1*10^-3 mol/L，该溶液PH=？
2.常温下PH=11某溶液，则水电离出的c[OH-]=？
3.往水中分别加入NaHCO3和NaHSO4会促进还是抑制水的电离？为什么？
4.0.1mol/L的Na2S溶液
Na+,S2-,HS-,H2S,H+,OH-的大小关系是？
需要思考过程。谢谢。
1. PH = -log[H+]= -log[0.001]= 3.0
2. [H+] = 10^(-11) = 1.0 * 10^-11
常温下，[OH-] = Kw / [H+] = 1.0*10^-14 / 1.0*10^-11 = 0.001 mol/L
3. HCO3- Ka = 4.8*10^-11 Kb = 2.08*10^-4
HSO4- Ka = 6.3*10^-8 Kb = 1.89*10^-7
因HCO3- HSO4- 均为弱酸，其Kb值大于 Ka，所以其在水中电离出的氢氧根离子将多余H+， 根据Le Chatelier's Principle, 大量增加的氢氧根离子会抑制水的电离。
4. 0.1mol/L的Na2S溶液 S2- + H20 = HS- + OH-
Ka1 = 8.9*10^-8 Ka2 = 1.2*10^-13
Kb1 = Kw / Ka2 = 1.0*10^-14 / 1.2*10^-13 = 0.083
Kb2 = Kw / Ka1 = 1.0*10^-14 / 8.9*10^-8 = 1.12*10^7
let [OH-] is X
Kb1 = [OH-][HS-] / [S2-]
= X*X / (0.1-X)
X^2 + 0.083X - 0.1*0.083 = 0
X = 0.0586M = [OH-] = [HS-]
HS- + H2O = H2S + OH-
let [OH-] is Y
Kb2 = [OH-][H2S] / [HS-]
= Y*Y / (0.0586 - Y)
Y = 8.11*10^-5M
[OH-] = X + Y = 0.0586M
[H+] = Kw/[OH-] =1.71*10^-13M
[S2-] = 01-0.0586 = 0.0414M
[HS-] = 0.586 - Y = 0.586M
[H2S] = Y = 8.11*10^-5M
[HS-] 大于 [OH-] 大于 [S2-] 大于 [H2S] 大于 [H+]
