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# 第 6 章 酸碱滴定法 3 - PowerPoint PPT Presentation

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### 第6章 酸碱滴定法3

6.3 酸碱溶液的H+浓度计算

1、强酸碱溶液:

[H+]＝ Ka[HA] + Kw

KW

Ka[HA]

[H+]= +

[H+]

[H+]

2、弱酸(碱)溶液:

[H+]＝ Ka (ca - [H+])

[H+]＝ Kaca

[HA]=ca-[A-]=ca-([H+]-[OH-])≈ ca-[H+]

[H+]＝ Kaca + Kw

[H+]＝ Ka (ca - [H+])

c/Ka= 0.20 / 10-1.26 =100.56＜ 400

pH=14-pOH

Kw

[B-] Kb

= +

[OH-]

[OH-] [OH-]

[OH-]=

[OH-]=

Kb cb + Kw

Kb (cb-[OH-])

Kb[B-]+Kw

KaKw

[H+]=

cb

[OH-]=

Kbcb

(1) Kbc > 20Kw

(2) c/Kb> 400 :

(3) Kbc >20Kw, c/Kb> 400 :

Ka1[H2A]

Kw

2Ka1Ka2[H2A]

[H+]= + +

[H+]

[H+]2

[H+]

2Ka2

[H+]= Ka1[H2A] (1+ ) + Kw

[H+]

[H+] = [HA-] + 2[A2-] + [OH-]

[H+]= Ka1[H2A]

2Ka2

2Ka2

[H+]

[H+]

[H+]＝ Ka1ca

2Ka2

2Ka2

[H+]= Ka1[H2A] (1+ ) + Kw

[H+]= Ka1[H2A] (1+ ) + Kw

[H+]

[H+]

Ka1ca >20Kw

≤0.05,

(忽略二级及以后各步离解)

ca/Ka1≥400

3、两性物质溶液

Ka1(Ka2[HA-]+Kw)

[H+]＝

Ka1+[HA-]

[H+][HA-]

Kw

Ka2[HA-]

[H+]+ = +

[H+]

[H+]

Ka1

Ka1(Ka2c＋Kw)

[H+]=

Ka1+ c

Ka1Ka2c

[H+]=

Ka1+ c

[H+]＝ Ka1Ka2

pH = 1/2(pKa1 + pKa2)

[H+]＝ KaKa’

Ka(Ka，c+Kw)

KaKa’c

[H+]＝

[H+]=

Ka+c

Ka+ c

[NH4+] ≈ [Ac-]≈c

Ka’ NH4+

Ka HAc

Ka’c >20Kw

c >20 Ka

Ka1Ka2c

[H+]＝

Ka1+ c

CH2ClCOOH: Ka=1.4×10-3

NH3: Kb=1.8×10-4

Ka’c ≥ 20Kw , c<20Ka

pH = 6.24

[H+]＝ Ka1Ka2

Ka1(Ka2c+Kw)

Ka1Ka2c

[H+]＝

[H+]＝

Ka1+c

Ka1+ c

PBE:[H+] + [+H3N-R-COOH] = [H2N-R-COO-] + [OH-]

Ka2c > 20Kw

c/Ka1> 20

Kaca

Kw

[H+]= cHCl + +

[H+]

Ka+[H+]

4、混合酸碱:

(近似式)

cHCl >20 [A-],忽略弱酸的离解: [H+] ≈ c HCl (最简式)

Kbcb

Kw

[OH-]=cNaOH + +

[OH-]

Kb+[OH-]

KHA[HA]

Kw

KHB[HB]

[H+]= + +

[H+]

[H+]

[H+]

[H+]＝ KHAcHA+KHBcHB

[H+]＝ KHAcHA

[HA]≈ cHA [HB]≈cHB

KHAcHA>>KHBcHB

[H+]的精确表达式

[H+]的近似计算式和最简式

6.4 对数图解法

1、强酸强碱的浓度对数图:

0.1mol/L HCl Cl- H+ OH-

lg [Cl-]= -1

lg [H+]= -pH

lg [OH-]= pH-14

+

c

c

Ka

[H+]

=

=

a

a

[Ac -]

[HAc]

+

+

+

+

[H+]

[H+]

K

K

a

a

0.01mol/L HAc

HAc Ac- H+ OH-

2、一元弱酸（碱）的浓度对数图：

1确定体系点 S （pKa，lgca）

• 一元弱酸（碱）的浓度对数图绘制：

2过S，画斜率为0，1的三条直线

3S附近lgc 与pH的曲线关系

0.01mol/L H2A ( pKa1=4,pKa2=8)

H2A HA- A2- H+ OH-

3、多元弱酸（碱）的浓度对数图：

4、对数图解法的应用：

1 ） 计算pH值

2 ） 计算各种分布形式的平衡浓度及分布分数

P

0.01mol/L HAc

[H+]=[Ac-]+[OH-]

1）pH值计算：

0.01mol/L NaAc

[H+] + [HAc] = [OH-]

2）平衡浓度及分布分数的计算

cH2A＝ 0.01mol/L

pH＝9.0

log[H2A]=-8.2

log[HA-]=-3.2

[A2-]=cH2A-[H2A]-[HA-]