分析化学有效数字运算规则习题及答案 《分析化学习题答案及详解》高等教育出版社(36)
∴ ω feo = 6 c cr
2 2o 7
v cr
2 2o 7
m feo
m样
=
× 100 % =
6 × 0 . 03333 × 25 . 00 × 10 3 × 71 . 84 × 100 % = 35 . 92 % 1 . 00
又 m fe
2o 3
1 ×6×c 2 × v cr 2 o 2 × m cr 2 o 7 7 2
fe 2 o 3
ω al o
2
3
(m =
共
mfe 2o 3 m样
) × 100% = ω
al 2 o 3
=
0.5000
1 × 6 × 0.03333 × 25.00 × 10 3 × 159.69 2 × 100% = 10.08% 1.000
15.称取含有as2o3与as2o5的试样1.500 g,处理为含aso33-和aso43-的溶液。将溶液调节为弱碱性, 以0.05000 mol.l-1碘溶液滴定至终点,消耗30.00 ml。将此溶液用盐酸调节至酸性并加入过量ki溶液, 释放出的i2再用0.3000 mol.l-1 na2s2o3溶液滴定至终点,消耗30.00ml。计算试样中as2o3与as2o5的质 量分数。 解:有关反应: h3aso3 + i2 + h2o = h3aso4 + 2 i + 2 h+ (弱碱介质中) (1) + h3aso4 + 2 i + 2 h = h3aso3 + i2 + h2o (酸性介质中) (2) 2 2 i2 + 2 s2o3 = 2 i + s4o6 (3) 故: as2o3 ~ 2h3aso3 ~ 2 i2
∴ ω as 2 o 3 1 1 c i 2 v i 2 .m as 2 o 3 × 0 . 05000 × 30 . 00 × 10 3 × 197 . 84 = 2 × 100 % == 2 × 100 % = 9 . 89 % m样 1 . 500
又: as2o5 ~ 2 h3aso4 ~ 2 i2 ~ 4 s2o32 参与(2)式反应的h3aso4也包括(1)生成的h3aso4
∴ ω as 2 o 5
1 1 c s 2 o 32 v s 2 o 32 c i 2 v i 2 × × m as 2 o 5 2 2 = × 100 % m样
1 3 3 1 × 0.3000 × 30 .00 × 10 0 .0500 × 30 .00 × 10 × × 229 .84 2 2 = × 100 % = 22 .98 % 1 .500
16.漂白粉中的“有效氯”可用亚砷酸钠法测定,现有含“有效氯”29.00%的试样0.3000g,用25.00ml na3aso3溶液恰能与之作用。每毫升na3aso3溶液含多少克的砷又同样质量的试样用碘法测定,需 要na2s2o3标准溶液(1 ml相当于0.01250 g cuso4.5h2o)多少毫升 解:(1) ca(ocl)cl + na3aso3 = cacl2 + na3aso4 ca(ocl)cl + 2 h+ = cl2 + ca2+ + h2o 故: ca(ocl)cl ~ cl2~ na3aso3
∴每毫升na3aso3含砷的克数 = nna aso
3
3
m as
vna3aso3
=
ncl2 m as vna 3aso3
=
0 .3000 × 29 .00 % × 74 .92 = 0 .003677 ( g ml1 ) 70 .90 × 25 .00
(2)
cl2 + 2 i = 2 cl + i2 i2 + 2 s2o32 = 2 i + s4o62 2cu2+ + 4 i = 2 cui + i2 故: 2cu2+ ~ i2 ~ 2 s2o32 2×249.69 g 2 mol 0.01250 v1 × c (s2o32- )
早上好