一、滴定分析概论 - 31 -
28. 10 分 (0173)
有关反应和滴定反应:
MnO+2Na2O2+H2O = MnO42-+2OH-+4Na+ 3MnO42-+4H+ = 2MnO4-+MnO2↓+2H2O MnO2+2Fe2++4H+ = 2Fe3++Mn2++2H2O MnO4-+5Fe2++8H+ = Mn2++5Fe3++4H2O Cr2O72-+6Fe2++14H+ = 2Cr3++6Fe3++7H2O
即 3MnO?3MnO42-?1MnO2?2Fe2+, 1 MnO4-?5Fe2+
(3/2)×(0.1000×10.00-5×0.01000×8.24)×70.94
w(MnO)= ───────────────────────×100% = 3.13% 2.000×1000 因为 1 Cr2O3?2 CrO42-?1 Cr2O72-?6 Fe2+ 1 MnO2?2 MnO4-
1/6(0.1000×50.00-5×0.01000×18.40-5×0.5880)×152.0
w(Cr2O3) = ───────────────────────────×100% 2.000×1000 = 1.44% 29. 5 分 (0174)
用KMnO4溶液处理SO2水溶液时的反应如下: 2MnO4-+5H2SO3 = 2Mn2++4H++5SO42-+3H2O
由反应中有关物质的物质的量之比可知,KMnO4溶液的浓度为: 2×0.1000×20.00
c(KMnO4) = ────────── = 0.05000 (mol/L) 4×20.00
在酸性条件下,KMnO4和Fe2+反应时n(KMnO4):n(Fe2+)=1:5 则消耗Fe2+溶液的体积V为: 5×0.05000×20.00
V(Fe2+) = ────────── = 25.00 (mL) 0.2000 30. 5 分 (0175)
(0.05000×20.00-5×0.0106×5.55)×52.00/3
w(Cr) = ──────────────────────×100% = 1.25% 1.000×1000 31. 5 分 (0176)
设K2Cr2O7 x g, KMnO4为(0.2400-x) g
(0.2400-x)×5 6x 0.2000×30.00 ─────── + ─── = ──────── 158.0 294.2 1000 x = 0.1418 g
w(K2Cr2O7) = (0.1418/0.2400)×100% = 59.07% w(KMnO4) = (100-59.07)% = 40.93% 32. 5 分 (0177)
因为n(MnO4-):n(H2C2O4) = 2:5
0.006305×1000 2
所以c(MnO4-)= ─────────×── = 0.02000 (mol/L) 126.1 5 因为2BaSO4?FeSO4(NH4)2SO4·6H2O?1/5 MnO4-
一、滴定分析概论 - 32 -
0.2334×1000
所以0.02000 V(MnO4-)= ──────── 233.4×2×5 解得V(MnO4-) = 5.00 mL 33. 5 分 (2150)
w(H2O) = [(0.9738-0.9656)/0.9738]×100% = 0.84% 干燥试样中
w(Fe) = 51.69%×100/(100-0.84) = 52.13% 34. 5 分 (2151)
m(H2O) = 5.30×5.00%+5.30×95%×4.00% = 0.466 (g) w(H2O) = [0.466/5.30]×100% = 8.80%
(或w(H2O) = 5.00%+4.00%×(100-5)/100 = 8.79%) 35. 5 分 (2152)
湿存水的百分含量:[0.9901-(14.0819-13.1038)/0.9901]×100% = 1.21%
结晶水的百分含量:[(14.0819-13.8809)/(14.0819-13.1038)]×100% = 20.55% w(CaSO4·2H2O) = 20.55%×(136.14+36.04)/36.04 = 98.2% 36. 5 分 (2153)
1. V(AgNO3) = 0.2×1000×0.01/(0.1×166.0) = 0.12 (mL) 2. V(Na2S2O3) = 0.2×1000×0.01/(0.02×166.0/6) = 3.6 (mL) 37. 5 分 (2154)
1. c(KCN) = 0.1245×18.56/25.00 = 0.0924 (mol/L) 2. V(AgNO3) = 18.56/2 = 9.28 (mL) 38. 5 分 (2155)
m = [(0.001×32.06/233.4)/0.1000%]×100% = 0.1374 (g) 39. 5 分 (2156)
m = [(5/2×0.02000×x×56.08)/(x%×1000)]×100% = 0.2804 (g) 40. 5 分 (2157)
m = [(0.1×30×52/3)/(5%×1000)]×100% = 1 (g) 41. 5 分 (2158)
m = [(0.05×30×63.55)/(20%×1000)]×100% = 0.47≈0.5 (g) 42. 5 分 (2159)
m = [(6×0.02000×x×55.85)/(x%×1000)]×100% = 0.6072 (g) 43. 5 分 (2160)
m(Zn) = 0.02×20×65.38/1000≈0.03 (g)
称取0.3g金属锌,溶解后定容于250mL,移取25mL作标定。 44. 5 分 (2161)
m(K2Cr2O7) = (0.02×20××294.2/(6×1000) = 0.02 (g)
称取0.02gK2Cr2O7溶于250mL容量瓶,移取250mL作标定。 45. 5 分 (2162)
m = 0.1×30×134.0/2 = 0.2 (g) 46. 5 分 (2163)
按x = (3×c×x/2×159.7)/(0.5000×1000)×100 解得c = 0.02087 (mol/L) 47. 5 分 (2164)
(1) m = 0.020×25×5/2×56/50% = 0.14g, 称取约0.14 (g)
一、滴定分析概论 - 33 -
(2) m = [(0.02010×x×5/2×56.08×100%)]/(x%×1000) = 0.2818 (g), 准确称取0.2818g(注意有效数字) 48. 5 分 (2165)
(40.00-3.05×22.06/20.00)c(HCl) = 0.3500/(100.1/2)×1000 解出c(HCl) = 0.1909mol/L
c(NaOH) = 0.1909×22.06/20.00 = 0.2106 (mol/L) 49. 5 分 (2166)
? =(0.002210×7.76×28.01×5/2)/3.21 =0. 374 (mg/L) 50. 10 分 (2167)
(1) w(KI) = [(0.05000×20.00-0.1023×25.34/6)×5×166.0/(0.6125×1000)]×100% = 76.96%
(2) V(Na2S2O3) = (0.05000×20.00 ? 0.1023×25.34/6)×6/0.1023 = 33.31(mL) 51. 5 分 (2168)
设耗HCl V(mL), 则
0.1015×1.01×Mr(P)/24 = 0.1075×V×Mr(P)/2 V = 0.08mL 52. 5 分 (2169)
m(KHC2O4·H2C2O4·2H2O) = 0.1012×22.55×254.2×4/(3×1000) = 0.7735 (g) m(Na2C2O4) = 1.000-0.7734 = 0.2265 (g) 53. 5 分 (2170)
w[CO(NH2)2] = [1/2(0.2012×30.00 ? 0.1122×8.02)×60.06/(0.1582×1000)]×100% = 97.5% 54. 5 分 (2171)
c(Ag+) = 0.02000×23.70×2/25.00 = 0.03792 (mol/L) 55. 5 分 (2172)
试样中Fe2O3质量为:
0.03333×3×25.00×159.7/1000 = 0.3992 (g)
试样中FeO质量为:0.3992×2×71.85/159.7 = 0.3592 (g) 所以w(FeO) = [0.3592/1.000]×100% = 35.92%
w(Al2O3) = [(0.5000-0.3992)/1.000]×100% = 10.08% 56. 5 分 (2173)
w(S) = [(0.2671×4×32.06/159.7)/0.5080]×100% = 42.22% 57. 5 分 (2174)
w(CaO) = {[0.1502×(30.00-4.02/1.132)×56.08/2]/(0.2212×1000)}×100% = 50.36% 58. 5 分 (2175)
c(K+) = (0.05580×29.64×39.10/4)/250 = 0.06467 (mg/mL) 59. 5 分 (2176)
c(NaF) = [0.02012×(30.52-26.05)×41.99×2]×10/25.00 = 3.021 (g/L) 60. 5 分 (2177)
c(NaCN) = [0.05015×(24.70-9.32)×49.01×4]/25.00 = 6.048 (g/L) 61. 5 分 (2178)
Mr(CO2) = 100.1-56.08 = 44.02
100g试样中CaCO3质量为5.00/44.02×100.1 = 11.37 (g) 所以w(CaCO3) = 11.4% 62. 5 分 (4106)
一、滴定分析概论 - 34 -
H?O?2KF?HSiO?4HF??HFNaOHSiO2?K2SiO3????K2SiF6?????NaF?H2O232n?SiO2?:n?NaOH??1:4T?SiO2NaOH??0.1000?60.08?0.001502g/mL4?100063. 10 分 (4220)
解： x=2.71%, s=0.070%, p=95% (?=0.05)
??x?t0.05,3sn?2.71?3.18?0.074??2.71?0.11??%?
2.60% ~2.82%, 95%的把握该区间不包含真值在内。 p=99% (?=0.01)
??x?t0.01,3sn?2.71?5.84?0.074??2.71?0.20??%?
2.51% ~2.91%, 99%的把握该区间包含真值在内。 64. 5 分 (4130)
2K4Fe(CN)6 + 3 Zn2+ = K2Zn3[Fe(CN)6]2? + 6K+
c?K4Fe?CN?6??65. 5 分 (4136)
0.1000?25.00?2?0.0826?mol/L?20.18?3m (H2C2O4?2H2O) = 0.1070 ? 34.40 ? 10-3 ? 126.07/2 = 0.2320 g
m (Na2C2O4) = [50.00 ? 0.1250 -(2.59 + 34.40) ? 0.1070] ? 10-3 ? 134.0/2 = 0.1536 g w (Na2C2O4) = 0.1536 / 0.7050 × 100% = 21.79% w (H2C2O4?2H2O) = 0.2320 / 0.7050 × 100% = 32.91% 66. 5 分 (4137)
c (HAc) = 0.4034 ? 30.24 ? 20.00 = 0.6099 (mol/L) w(HAc) = [ (0.6099 × 60.05) / (1.055 × 1000) ]× 100%=3.472% 67. 5 分 (4138)
设KHC2O4?H2C2O4?2H2O的质量为mg，则
m?3?2m?4?0.5000?m??2??254.2254.2134.0m?0.3274?g?0.3274w?KHC2O4?H2C2O4?2H2O???100%?65.48%0.5000w?Na2C2O4??34.52h. 5 分 (4139)
设Na2C2O4的质量为m、KHC2O4?H2C2O4的质量为am :
m?2am?4am?3???3.62134.0218.2218.2 a = 0.475
m (Na2C2O4) : m (KHC2O4?H2C2O4) = 1 : 0.475 69. 5 分 (4140)
500.0? 0.2000 = (0.005000 ? 1000 ? 2/56.08) ? (500.0 +V )
一、滴定分析概论 - 35 -
V = 60.8 mL 70. 5 分 (4142)
设Na2C2O4质量为m g
m?2?0.2608?m??20.0800?10.00?5??134.0128.131000m?0.1036?g??0.2608?0.1036??1000?7.67?mL?V?NaOH??128.13?0.160071. 5 分 (4144)
0.5000?286.94w?MnO2????0.01964?0.01650?5??100%?84.7%???134.02?0.3000??
72. 5 分 (4147)
2?S?H2S?I2?2S2O3???m?S???10.0?0.00500?2?0.0200?2.6??32.06?0.77?mg?2
73. 5 分 (4148)
w?H2O2???5?0.0215?25.00?0.112?5.10??34.01?100%?6.13%2?0.587?1000
四、问答题 ( 共 4题 ) 1. 5 分 (2110) 因为： (1) 生成BaSO4的反应不很完全,在重量法中可加过量试剂使其完全,而容量法基于计量反应,不能多加试剂;
(2) 生成BaSO4反应要达到完全,速度较慢, 易过饱和,不宜用于滴定, 而在重量法中可采用陈化等措施。 2. 2 分 (2126)
酸碱反应的完全程度不如EDTA络合反应高,若浓度太稀,终点误差大。EDTA络合反应完全度高,可以稀一些,而且EDTA溶解度小,也难以配成0.1 mol/L溶液。 3. 10 分 (2127)
上述说法不正确。
1. 此时必须增大试样量,使消耗体积与原来相近,这时过量1滴(0.04mL)所造成的误差为0.04×c/V×c = 0.04/V, 它仅与总消耗体积有关,而与浓度无关;
2. 对终点误差而言,则是浓度大些好,以NaOH滴定HCl为例,若是以酚酞为指示剂, Et = 10-5×V/c×V = 10-5/c,即浓度大则终点误差小; 3. 浓度不能过大,原因是防止造成浪费。 4. 5 分 (4135)
w?C8H10N4O2???c1V1?c2V2??M?C8H10N4O2?ms?1000?100%