无机及分析化学第七章酸碱平衡与酸碱滴定法习题答案

无机及分析化学第七章酸碱平衡与酸碱滴定法习题答案

第七章 酸碱平衡与酸碱滴定法

7-3 (1) Kb = (3) Kb = (5) Kb1 =

KwKa 1 10

14 10

6.2 101 10

14

=1.6×10-5； (2) Kb =

KwKa

1 10

14 4

=5.6×10-11

1.8 101 10

14

KwKaKw

1.1 10

10

=9.1×10-5； (4) Kb = =2.3×10； (6) Kb2 =

Kac

-2

KwKaKwKa1

5.8 10 1 10

10

=1.7×10-5 =1.7×10-13

1 10

14 13

14 2

Ka34.4 105.9 10

7-4 (1) ∵ = 4.2% < 5% ∴

c(HAc)c(NaAc)

n(HAc)

Ka c

2

= 1.76×10-5 c(H+) =c = 0.010×0.042 = 4.2×10-4 mol.L-1

c(H)c(HAc)

(2) c(H+)=Ka

Ka

n(NaAc)

1.76 10

5

0.010 0.5

2.182

=3.4×10-6mol.L-1

100

3.4 10

6

100

0.010

= 0.034%

7-5 (1) c(H+) = c(OH-) + c(H2PO4) + 2c(HPO4) + 3c(PO4)； (2) c(H+) + c(H3PO4) = c(OH-) + c(HPO4) + 2c(PO4) (3) c(H+) + c(HS-) + 2c(H2S) = c(OH-) + c(NH3)； (4) c(H+) + c(H2CO3) = c(OH-) + c(CO3) + c(NH3) (5) c(H+) + c(H2PO4) + 2c(HPO4) + 3c(PO4) = c(OH-)

(6) c(Na+) + c(NH4) + c(H+) = c(OH-)； (7) c(H+) = c(OH-) + c(Ac-) + c(H2BO3) (8) c(H+) = c(OH-) + c(Ac-) + c(Cl-)； (9) c(H+) + c(HF) = c(OH-) + c(NH3) 7-6 c(NH3) = c×

c(NH

4

2 3 2 3

2

2 3

c(OHc(OH

)

0.10

1 10

1 10

11

11

5

) Kb

1.8 10

= 5.6×10-8 mol.L-1

)

= c - c(NH3) = 0.10 - 5.6×10-8 ≈0.10 mol.L-1

-5

7-7 (1) ∵Kac = 1.8×10×0.100>>20Kw ∴c(H+) =

5

cKa

0.1001.8 10

5

= 5556 > 500

Kac

.8 10 0.100

=1.342×10-3 mol.L-1 pH= -lgc(H+) =2.87

cKa

0.1505.0 10

2

(2) ∵Kac = 5.0×10-2×0.150>>20Kw ∴c(H+) = (3) Ka =

KwKb(NH3)

2

= 3 < 500

12

( Ka Ka 4Kac)

14 5

12

( 0.050 0.05

2

4 0.050 0.150)

=0.06514 mol.L-1 pH = -lgc(H+) = 1.19

cKa

0.1005.6 10

10

1 10

1.8 10

= 5.6 ×10-10 ∵Kac = 5.6×10-10×0.100 > 20Kw

10

= 1.8×108 > 500

∴c(H+) =

Kac

5.6 10 0.100

=7.483×10-5 mol.L-1 pH= -lgc(H+) =5.13

cKa1

0.4007.6 10

3

(4) ∵Ka1 >> Ka2 >> Ka3 Ka1c = 7.6×10-3×0.400>>20Kw ∴c(H+) = (5) Kb =

= 53 < 500

12

( Ka1 Ka1 4Ka1c)

14 10

2

12

( 0.0076 0.0076

2

4 0.0076 0.400)

= 0.05147 mol.L pH = -lgc(H+) = 1.29

-1

KwKa(HCN)

Kbc

1 10

6.2 10

= 1.6 ×10-5 ∵Kbc = 1.6×10×0.100 >> 20Kw

-5

-1

cKb

0.1001.6 10

5

= 6250 > 500

∴c(OH-) =

.6 10

5

0.100

= 1.265×10-3 mol.L pOH= -lgc(OH-) = 2.90 pH = 11.10