重量分析法和沉淀滴定法1

−11 0
sp
(实际上,该题中离子强度都较大,因
,[Ca 2+ ] = s,[ F - ] = 2 s
Ka 6.6 × 10−4 = −2 = 6.19 × 10−2 δF = + −4 [ H ] + K a 10 + 6.6 × 10 [Ca 2+ ] = S , [ F − ] = δ F [ F − ] = δ F ⋅ 2s
-15 -1 2+
2. 已知某金属氢氧化物 M[OH]2 的 Ksp=4×10 ,向 0.10mol.L M 溶液中加入 NaOH,忽略体 积变化和各种氢氧基络合物,计算下列不同情况生成沉淀时的 pH. 2+ 2+ 2+ a.M 有 1%沉淀; b.M 有 50%沉淀; c.M 有 99%的沉淀. 解: [OH ] =
第 5 章 重量分析和沉淀滴定法
参考答案
1. 已知 β =
[CaSO4 ]水
2+ [Ca 2+ ][ SO4 ]
= 200 , 忽略离子强度的影响, 计算 CaSO4 的固有溶解度,并计算
2+
饱和 CaSO4 溶液中,非离解形式 Ca 的百分数. 解:
2− s 0 = [CaSO4 ][ 水 ] = 200[Ca 2+ ][ SO4 ]
= 4 × 10−2
K sp = [Cu 2+ ][ S 2− ] = s ⋅ δ S 2− ⋅ s = s 2 ⋅ δ S 2− ∴s = K sp
δS
2−
6 ×10−36 = = 1.2 × 10−14 mol ⋅ L−1 −8 4 ×10
2+
(b) MnS + H 2O ⇔ Mn
+ HS − + OH −
1 b. I = (cBa 2+ × 22 + ccl −1 ×12 ) = 0.3 2
由 −1gγ i = 0.512 Z i
I 1 + Ba 0 I
计算得: γ Ba 2+ = 0.256
γ SO
4
2−
= 0.223
Ksp =
K sp 0
γ Ba γ
2+
SO4
2−
1.1× 10−10 = = 1.9 × 10−9 0.256 × 0.223 1.9 × 10−9 = 1.9 × 10−8 mol ⋅ L−1 0.10
= 200 K sp = 200 × 9.1× 10−6 = 1.82 × 10−3 mol ⋅ L−1 [Ca 2+ ] = K sp = 3.02 × 10−3 mol ⋅ L−1
[CaSO4 ][水 ] [CaSO4 ][水 ]+ [Ca 2+ ]
× 100% =
1.82 ×10−3 × 100% = 37.5% 1.82 × 10−3 + 3.02 ×10−3
2+
−
K sp [M ]
2+
pOH =
1 ( pK sp + lg[ M 2+ ]) 2
pK sp = 15 − lg 4 = 14.4
pOH=12(14.4+lg0.099)=6.7 pH=7.3 a.[M ]=0.10(1-1%)=0.099 2+ b.[M ]=0.10(1-50%)=0.050 pOH=12(14.4+lg0.05)=6.55 pH=7.45 2+ c.[M ]=0.10(1-99%)=0.0010 pOH=12(14.4+lg0.0010)=5.7 pH=8.3 3.考虑盐效应,计算下列微溶化物的溶解度. -1 -1 a.BaSO4 在 0.1mol.L NaCl 溶液中; b.BaSO4 在 0.1mol.L BaCl2 溶液中. 解: a. I =
−1 −1
5.计算 BaSO4 在 0.010mol ⋅L BaCl2 − 0.070mol ⋅ L HCl 的溶解度. 解: 此题需同时考虑盐效应、酸效应和同离子效应： [ SO4 ] = s
2− /
cBa2+ = 0.010 cCl − = 0.070 + 0.020 = 0.090 cH + = 0.070
∴s =
3
2 3 K sp = [Ca 2+ ][ F − ]2 = s (δ F − 2 s ) 2 = 4δ F −s
K sp
2 4δ F −
=
3
2.7 ×10−11 = 1.2 × 10−3 mol ⋅ L−1 4 × (6.19 × 10−2 ) 2
-1
（2）BaSO4 在 2.0mol.L HCl 中.
Ka Ka [ H 2S ] [ H + ]2 [ S 2− ] = K a1 K a2 [H + ] = 10−0.5 [ S 2− ] = 1 2 + 2 = 9.23 × 10−22 ∵ [H ] [ H2S ] s = [Cu 2+ ] = K sp [ S 2− ] = 6 × 10−36 = 6.5 × 10−15 mol ⋅ L−1 9.23 × 10−22
4 4
2− K sp = [ Pb 2+ ][ SO4 ] = s.δ SO2− .s = s 2 .δ SO2−
s=
K sp
δ SO
=
4
2−
1.6 ×10−8 = 4.2 ×10−4 mol ⋅ L−1 0.091
-1
(4) CuS 在 pH=0.5 的饱和 H2S 溶液中 ( H 2 S ≈0.1mol.L )
I = 1/ 2(cBa 2+ × 22 + cCl − × 12 + cH + ) = 0.10
由德一休公式计算得： γ Ba 2+ = 0.38
γ SO = 0.355
4 2−
δ SO =
2− 4
K a2 [ H + ] + K a2
2+ 2− 4
=
1.0 ×10−2 = 0.125 0.070 + 1.0 × 10−2
s = [ Mn 2+ ] = [ HS − ] = [OH − ]
K = [ Mn 2+ ][ HS − ][OH − ] = s 3 =
K sp ⋅ K w Ka 2
s=3 K =
3
2.0 ×10−10 ×1.0 × 10−14 = 6.5 × 10−4 mol ⋅ L−1 −15 7.1×10
1 (cNa + × 12 + CCl − ×12 ) = 0.10 2 I 计算得: γ Ba 2+ = 0.38 1 + Ba 0 I
由
− lg γ i = 0.5122Z 2i
K sp 0
γ SO
4
2−
= 0.355
s = K sp =
γ Ba + γ SO
2+
4
2−
−5 1.1×10−10 = = 2.8 × 10 mol ⋅ L−1 0.38 × 0.355
δ SO
=
4
2−
1.1×10−10 = 1.5 × 10−4 mol ⋅ L−1 5.0 ×10−3
-1
(3) PbSO4 在 0.10mol.L HNO3 中, [ Pb ] = s [ SO4
2+
2− /
]= s
δ SO =
4 2−
Ka2 1.0 ×10−2 = = 0.091 [ H + ] + K a 2 0.10 + 1.0 ×10−2
0 k sp
K sp = [ Ba ][ SO ] =
γ Ba + γ SO
2+
= 8.15 × 10−10
2−
4
[ SO4 2− ] = δ SO2− ⋅ s = K sp [ Ba 2+ ]
4
∴s =6.K 源自p [ Ba 2+ ]δ SO 2−
4
=
8.15 ×10−10 = 6.5 ×10−7 mol ⋅ L−1 0.010 × 0.125
2− ] = K sp [ Ba 2+ ] = s = [ SO4
4. 考虑酸效应,计算下列微溶化合物的溶解度. a.CaF2 在 pH=2.0 的溶液中; b.BaSO4 在 2.0mol.L-1HCl 中; -1 c .PbSO4 在 0.10mol.L-1HNO3 中; d.CuS 在 pH=0.5 的饱和 H2S 溶液中([H2S]≈0.1mol.L ). 解: 按题意,只考虑酸效应,忽略盐效应,取 K sp ≈ K 此计算结果将会与实验有一定出入). (1) CaF2 在 pH=2.0 的溶液中, K sp = 2.7 × 10
δ SO
4
2−
=
K a2 [H + ] + Ka2
=
1.0 ×10−2 = 5.0 × 10−3 −2 2.0 + 1.0 ×10
4 4 4
[ SO4 2− ] = s, [ SO4 2− ] = δ SO 2− s, K sp = s ⋅ δ SO 2− s = s 2δ SO 2− ∴s = K sp
a.CuS; b. MnS;
考虑 S 的水解，计算下列硫化物在水中的溶解度。
2+ 2− −
2-
解： （a） s = [Cu ] = cS 2− = [ S ] + [ HS ] + [ H 2 S ]
δS =
2−
K a1 ka2 [ H ] + K a1 [ H + ] + K a1 K a2