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# 第三章 电解质溶液和离子平衡 - PowerPoint PPT Presentation

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### 第三章电解质溶液和离子平衡

3.1 一元弱酸、弱碱的离解平衡
• 3.1.1 离解常数
• 以一元弱酸HA为例，其离解平衡式及其标准离解常数为：
• HA H+ + A-

KaΘ = {c(H+)/ cΘ}·{c(A-)/ cΘ}

c(HA)/cΘ = c‘(H+) ·c’(A-)c‘(HA)若以BOH表示弱碱，则：BOH B+ + OH-KbΘ = {c(B+)/ cΘ}·{c(OH-)/ cΘ}c(BOH)/cΘ = c‘(B+) ·c’(OH-)c‘(BOH)KaΘ可用来表示酸的强弱.KaΘ越大，该弱酸的酸性越强，HA Ka HCOOH 1.77×10-4 HAC 1.75×10-5 HClO 2.80×10-8 HCN 6.20×10-10

3.1.2 离解度对于弱电解质来说,除了用离解常数表示电解质的强弱外,还可用离解度(α)来表示其离解的程度:α= 已离解的弱电解质浓度 ×100% 弱电解质的起始浓度 以一元弱酸HA为例,其离解度α, 离解常数KaΘ和浓度c之间的关系推导如下: HA H+ + A-

KaΘ = c‘(H+) ·c’(A-) = c‘α ·c’ α =c‘α 2c‘(HA) c‘(1-α) 1-α

1. 精确公式:也就是说, c‘α 2 + KaΘα - KaΘ = 0,

α =-KaΘ + √ (KaΘ)2 + 4c’KaΘ

2c’

c(H+) = cα = c·-KaΘ + √(KaΘ)2 + 4c’KaΘ

2c’

= －KaΘ + √(KaΘ)2 + 4c’KaΘ ·cΘ

2

2. 近似公式:

α = √KaΘ/c’c'(H+) =√KaΘ·c’

α = √KbΘ/c’c’(OH-) = √KbΘ·c’

3.1.3 一元弱酸、弱碱溶液中离子浓度及pH的计算

(2)如将此溶液稀释至0.010 mol·L-1, 求此时溶液的H+浓度、pH及离解度。

HA H+ + Ac-

KaΘ = c‘(H+) ·c’(Ac-) = x·x

c‘(HAc) 0.10 - x

∵c/ KaΘ = 0.10÷(1.75×10-5) >500, ∴可用近似公式

c'(H+) = √KaΘ·c’

X =c’(H+)= c’(Ac-)=√KaΘ·c’=√1.75×10-5×0.10

=1.3×10-3

pH = -lgc’(H+) =-lg 1.3×10-3 = 2.88

α = (1.3×10-3/0.10) ×100% = 1.3%

(2)∵c/ KaΘ = 0.010÷(1.75×10-5) >500,

∴仍可用近似公式

x = c’(H+) = √KaΘ·c’ =√1.75×10-5×0.010

= 4.2×10-4

pH = -lgc’(H+) = -lg 4.2×10-4 = 3.38

α = (4.2×10-4/0.010) ×100% = 4.2%

c(OH-) = 6.0×10-4 mol·L-1

NH3+ H2O NH4+ + OH-

KbΘ（NH3）= c'(NH4+) ·c’(OH-) = (6.0×10-4)2 = 1.8 ×10-5

c'(NH3) 0.020

α = c’(OH-) ×100% = 6.0×10-4×100% = 3.0%

c'(NH3) 0.020

3.2 多元弱酸的离解平衡

1．碳酸分两步离解：

Ka1Θ（H2CO3）= c'(H+) ·c’(HCO3-) = 4.4 ×10-7

c'(H2CO3)

Ka2Θ（H2CO3）= c'(H+) ·c’(CO32-) = 4.7 ×10-11

c'(HCO3-)

2．磷酸分三步离解：

Ka1Θ（H3PO4）= c'(H+) ·c’(H2PO4-) = 7.1 ×10-3

c'(H3PO4)

Ka2Θ（H3PO4）= c'(H+) ·c’(HPO42-) = 6.3 ×10-8

c'(H2PO4-)

Ka3Θ（H3PO4）= c'(H+) ·c’(PO43-) = 4.2 ×10-13

c'(HPO42-)

3．对于多元弱酸的分步离解，因为Ka1Θ >> Ka2Θ, 其氢离子浓度主要来自第一级电离,当求氢离子浓度时可当作一元弱酸来处理.

c’(H+)= c’(HCO3-) = √KaΘ·c’ =√4.4×10-7×0.040

= 1.3×10-4

Ka2Θ（H2CO3）=c'(H+) ·c’(CO32-)

c'(HCO3-)

=c’(CO32-) = 4.7 ×10-11