第9章 重量分析法

1.41× 10−2 + Ka2
∴ s=2.65 × 10－8 mol ⋅ L−1 , BaSO4损失mg 数为： 2.65 × 10－8 × 100 × 233.4 = 6.2 × 10−4 mg
16．解：
（1）
NH 4 HF2 ⎯⎯ → NH 4 F + HF HF +
0.005−[ H ]
2＋ 2− 4
4
（2）
BaSO4在0.1mol ⋅ L−1 BaCl2中，I =
ο ο
1 ∑ CiZi = 0.30 2
2− 查表得 a ( Ba 2＋ ) = 5, a ( SO4 )=4
− lg γ Ba 2＋ = 0.512 ×102 − lg γ Ba 2＋ = 0.512 ×102
0.30 = 0.5909, γ Ba 2＋ = 0.26 1 + 0.328 × 5 × 0.30 0.30 = 0.6526, γ SO2− = 0.22 4 1 + 0.328 × 4 × 0.30
4
2− ∴ Ksp 0 = 1.1× 10−10 = [ Ba 2＋ ] × γ Ba2＋ × [ SO4 ] × γ SO 2− = ( s + 0.10) × 0.26 × s × 0.22
s = 1.92 × 10 −8 mol ⋅ L−1
5．解：
I= 1 ∑ ciZi 2 = 0.10, 2
0.05 ＝8.26 0.5
pOH = 5.74,[OH ] = 1.82 × 10−6 mol ⋅ L−1 ∴[ Mg 2+ ][OH − ]2 = 0.005 × (1.82 × 10−6 ) 2 = 1.66 × 10−14 < KspMg (OH )2 ∴ 无沉淀生成 19．解：
s = [ Zn 2+ ] + [ ZnOH − ] + [ Zn(OH ) 2 ]＋[ Zn(OH )－ ＋[ Zn(OH )2－ 3 ] 4 ] = [ Zn 2+ ]{1 + β1[OH − ] + β 2 [OH − ]2 + β 3 [OH − ]3 + β 4 [OH − ]4 } = Ksp × {1 + β1[OH − ] + β 2 [OH − ]2 + β3 [OH − ]3 + β 4 [OH − ]4 } − 2 [OH ]
第9章
重量分析法
1．解： Q S0=[CaSO4]=β[Ca2+][SO42-]=β×Ksp=200×9.1×10-6=1.82×10-3mol/L
[CaSO 4 ]水 1.82 × 10−3 = 0 = 37.6% S S Ksp ＋ 2+ 非离解形式 Ca 的百分数为
3．解：
BaSO4在0.1mol / LNaCl中, I =
= 2.5 ×10-7 mol ⋅ L−1
主要状态可由数值得 22．解：
F= M (Cr2O3 ) = 0.2351 2 M ( PbCrO4 ) 2 M ( MgSO4 ⋅ 7 H 2O ) = 2.215 M ( Mg 2 P2O7 ) M [Ca3 ( PO4 ) 2 ] = 0.08266 2 M [(NH 4 )3 PO4 ⋅12 MoO3 ] M ( P2O5 ) = 0.03783 2 M 2 M [(NH 4 )3 PO4 ⋅12 MoO3 ]
Ksp = [Ca 2＋ ][ HCO3− ][OH − ] ⋅
[CO32− ] [ H + ] Ksp ⋅ Kw ⋅ = Ka2 [CO32− ] [ H + ]
s=
3
Ksp ⋅ Kw 3 2.9 × 10−9 × 10−14 = Ka2 5.6 × 10−11
s = 8.02 × 10−5 mol ⋅ L−1
设混合后[Ag+]＝x/ mol·L-1 Ag+ + 2NH3 Ag(NH3)2+ 0.0175-x
x 1.5-2×(0.0175-x) ≈1.5
≈0.0175
0.0175 x(1.5) 2 ＝ß ＝107.40 2
x＝3.1×10-10
Ksp AgI
[Ag+][I-]＝3.1××0.025＝7.8×10-12 > 有 AgI 沉淀生成。
∴ Ag +浓度由AgCl ↓ 决定
s = [ Ag + ] = Ksp AgCl = 1.8 ×10−10 = 1.34 × 10−5 mol ⋅ L−1
9．解：
已知CaCO3沉淀在水中的主要离解平衡为：
CaCO3 ↓ + H 2O Ca 2＋ + HCO3 − + OH −
Ksp = [Ca 2＋ ][ HCO3− ][OH − ] = s 3
(2)
α Ag（NH ） ＝
3
c Ag [ Ag + ]
= 1 + 103.24 × 0.5 + 107.0 (0.5) 2 = 2.8 × 106
0.05 × 0.5 = 8.9 × 10−9 > Ksp AgCl 2.8 × 106 ∴ 有沉淀生成 [ Ag + ][Cl − ] =
(3)
pH = 9.26 + lg
ο ο
2− 查表得 a ( Ba 2＋ ) = 5, a ( SO4 ) = 4 ， γ Ba2＋ = 0.38, γ SO42− = 0.355
δ SO =
2− 4
1.0 ×10−2 = 0.125 0.07 + 1.0 ×10−2
4 4
Ksp 0 = 1.1× 10−10 = aBa ⋅ aSO4 = (0.01 + S ) × γ Ba 2＋ × s × γ SO 2− × δ SO 2−
[ SO4 2− ] = 0.01×
−3 −3 2+ 剩余的 Ba = (4.9 × 10 − 3.3 × ) ×150 × 137.33 = 3.3mg
100 mL 纯水洗涤时损失的 BaSO4 :
Q s = [ Ba 2+ ] = Ksp = 1.05 ×10−5 mol ⋅ L−1
∴为1.05 ×10−5 ×100 × 233.4＝0.245mg
0.3801 = 0.006806 55.85 y 0.01020 3 ∴ = = x 0.006806 2 则x =
⎞ ⎟ ⎠
∴ 为Fe 2 O3
50 AgCl: 10 × 143.3 ＝0.035(mol·L-1)
34．解：
cNH3＝3/2＝1.5(mol·L-1)
0.035 [Ag+]原＝ 2 ＝0.0175(mol·L-1) 0.05 [I-]原＝ 2 ＝0.025(mol·L-1)
[OH − ] = s = 8.02 × 10−5 mol ⋅ L−1
pOH = 4.1, pH = 9.9
11．解：
α Ag (S O ) =
2 2 3
cAg 2 3 = 1 + 10 8.82 × 0.01＋1013.46 × （10－2） ＋1014.15 × （10－3） = 3.02 × 10 9 ＝10 9.48 + [Ag ]
−1 100 mL0.010 mol ⋅ L H 2 SO4 洗涤时
Q 0.010mol ⋅ L−1 H 2 SO4的[H + ] = 1.41× 10 −2 mol ⋅ L−1
Ksp = 1.1× 10 −10 = [ Ba 2+ ][ SO4 2− ] = s × ( s + 0.01) × δ SO 2− = s × 0.01×
ο ο
（1）
1 ∑ ciZi 2 = 0.10, 2
2− 查表得 a ( Ba 2＋ ) = 5, a ( SO4 ) = 4, γ Ba 2＋ = 0.38, γ SO 2− = 0.355
4
Ksp 0 ∴ s = [ Ba ] = [ SO ] = Ksp = = 2.86 × 10−5 ( Ksp 0 = 1.1× 10−10 ) γ Ba 2＋ ⋅ γ SO2−
s × 0.010 9.48 ＝[Ag ][I ]＝ 10
+ －
KSP＝9.3×10
－17
−1 s=2.81×10－5 mol ⋅ L
13．解：
[ Ba 2+ ] = 0.1×
混合后，
1000 1 × = 4.9 ×10−3 mol ⋅ L−1 150 M ( Ba )
50 = 3.3 × 10 −3 mol ⋅ L−1 150
0.5805 × 28．解：
解得Na = 22.988865
M ( AgCl ) 107.868 + 35.453 = 0.5805 × = 1.4236 M ( NaCl ) Na + 35.453
31．解：
设为
Fe x O y
⎛ x × 55.85 + y × 16 = 0.5434 ⎜ x × 55.85 = 0.3801 ⎝
s = 6.44 × 10 −7 mol ⋅ L−1
7．解：
AgCl ↓ AgBr ↓ Ag + + Cl − , Ksp = 1.8 × 10−10 Ag + + Br − , Ksp = 5 × 10−13
+
在同一溶液中， Ag 只有一种浓度
Q Ksp AgCl > Ksp AgBr , AgCl ↓ 的溶解度大得多
(1)
F=
(2)
F=
(3)
F=
(4)
25．解：
∴x⋅
设 CaC2O4 为 x ,MgC2O4 y =0.6240- x