发布时间 : 星期日 文章重量法沉淀滴定 练习更新完毕开始阅读
1009:0.2670; 1010:0.3621; 1011:0.1183; 1012:1.956; 1013:39.12%; 1014:5.044%;
1015: 设混合物中氯的质量为m g,则溴的质量为2m g,银的质量为m(Ag) g, 则混合物中
m(Ag)
w(Ag)= ───────────────×100% Mr(AgCl) Mr(AgBr)
────×m+ ─────×2m Ar(Cl) Ar(Br) Ar(Ag) Ar(Ag) ───×m+ 2m×──── Ar(Cl) Ar(Br)
= ──────────────×100% Mr(AgCl) Mr(AgBr) ────×m+ 2m×──── Ar(Cl) Ar(Br) 107.87 107.87
──── + ─────×2 35.45 79.90
= ──────────────×100% 143.32 187.77
──── + ──────×2 35.45 79.90 = 65.69% 1016
灼烧后试样的组成为Mn3O4和SiO2
Ar(Mn)
─────×80%×m试样 Mr(MnO2)
则 w(Mn)= ────────────────── Mr(Mn3O4)
[───── ×80%+SiO2%]×m试样 3Mr(MnO2)
54.94
──── × 80% 86.94
w(Mn)= ────────────×100% 228.81
100% × ─────×80%+15% 3×86.94 = 59% 1017
3Mr(Fe2O3) w(Fe)= 10.11%×[0.970 + 0.030×───── ] 2Mr(Fe3O4) 3×159.69 = 10.11%×(0.970 + 0.030×───── ) 2×231.54 = 10.12%
1018: 设BaSO4中BaS质量分数为x
Ar(Ba) Ar(Ba) Ar(Ba)
[(1-x)────── + ─────x]×0.980= ────── Mr(BaSO4) Mr(BaS) Mr(BaSO4) 137.3 137.3 137.3 [(1-x)──── + ────x]×0.980= ──── 233.4 169.4 233.4 解得 x=5.40% 1019
设BaS的质量为x g Mr(BaSO4)
0.5021 - ───── x = 0.5013 - x Mr(BaS) 233.4
0.5021 - ───── x = 0.5013 - x 169.4 x = 2.1×10-3(g) 2.1×10-3
w(BaS)= ──────×100% = 0.42% 0.5013 1020
107.87
m(Ag) = ─────×0.2067 = 0.1556(g) 143.32 0.1556
w(Ag) = ─────×100% = 77.80% 0.2000
已沉淀的m(PbCl2) = 0.2466 - 0.2067 = 0.0399 (g) 试样中的m(Pb) = 0.2000 - 0.1556 = 0.0444 (g) 278.10
若完全沉淀应有的m(PbCl2) = ─────×0.0444 = 0.0596 (g) 207.2
未被沉淀的m(PbCl2) = 0.0596 - 0.0399 = 0.0197 (g)
1021
纯MgSO4·7H2O中含MgSO4理论值为48.84%,含H2O 51.16%;试样中MgSO4质量为: 2Mr(MgSO4)
──────×0.3900 = 0.4219 (g) Mr(Mg2P2O7) Mr(MgSO4)
或 ────── ×0.8179 = 0.4219 (g) Mr(BaSO4)
0.4219
w(MgSO4)= ─────×100 = 52.74 0.8000 含水则为47.26%
由计算结果可知,泻盐试样不符合已知的化学式,原因是失去部分结晶水。 1022
设试样中NaCl的质量为x g, NaBr的质量为y g
1023:(D); 1024:(C); 1025:(D); 1026:(A); 1027
PbCrO4 PbClF s2 s3 Pb(IO3)2 Pb3(AsO4)2 4s3 108s5 1028: 1. > 2. > 3. < 4. > 5. < 1029: 由于CdCO3的溶解度较大,H2CO3的Ka2很小,设以HCO3-为主 CdCO3 + H2O===Cd2+ + HCO3-+ OH- Ksp·Kw K= ────── Ka2 10-11.28-14 = ────── 10-10.25 = 10-15.03
s = K1/3 = 10-15.03/3 = 10-5.01 (mol/L)
即pH9.0, 此时以HCO3-为主,以上假设合理,结果正确 1030
[Pb2+] = [SO42-] + [HSO4-] = 2.0×10-4 mol/L,
?SO2-4(-2+2
= 1+10=2 H) 2.0×10-4
[SO42-] = ────── = 1.0×10-4 (mol/L) 2
Ksp = [Pb2+][SO42-] = 2.0×10-4×10-4 = 2.0×10-8 1031
?F(H) = 1 + 10-1.00+3.18 = 102.18
×
K'sp =Ksp·(?F(H))2= 10-10.57+22.18 = 10-6.21
K'sp 10-6.21
s = [Ca2+] = ─── = ───── c2(F) 10-2.00 = 10-4.21
= 6.2×10-5 (mol/L) 1032
0.80×10-6
未沉淀的[Ca2+] = ───────×1000 = 2.0×10-7(mol/L) 100×40.08 10-1.22-4.19
x(C2O42-) = ────────────────
×
10-4.72+10-4.7-1.22+10-1.22-4.19 = 0.76
Ksp 2.0×10-9
[C2O42-] = ──── = ────── = 1.0×10-2 (mol/L) [Ca2+] 2.0×10-7
[C2O42-] 1.0×10-2
则c(H2C2O4) = ────── = ───── = 1.3×10-2(mol/L) x(C2O42-) 0.76 1033
?PO3-4(H)=1+10-8.00+12.36+10-16.00+19.56 +10-24.00+21.68
= 104.42 s
[PO43-] = ───── = 10-4.42s
?PO3-4(H)
[H+] 10-8.00
x(NH4+)= ───── = ──────── =10-0.023 [H+]+Ka 10-8.00+10-9.26
[NH4+] = cx(NH4+) = 0.2×0.95 = 0.19 = 10-0.72 [Mg2+]= s
s×10-0.72×10-4.42s =10-12.7 s =10-3.78 (mol/L) 1034
[Ca2+] = s c(H2C3O4) = 0.050 mol/L 5.9×10-2×6.4×10-5.0
x2= ────────────────────── = 0.86 (10-5.0)2+5.9×10-2×10-5.0+5.9×10-2×6.4×10-5.0 [C2O42-] = 0.050×0.87 = 0.043 (mol/L) [Ca2+][C2O42-] =s×0.043 = 2.0×10-9 s = 4.7×10-8 (mol/L) 1035
s = [Ag+] + [AgCl] + [AgCl2] = [Ag+] +?1[Cl-][Ag+] +?2[Cl-]2[Ag+]