分析化學-滴定問題計算

[匿名]

2011-10-17 9:35 pm

A 100.0 mL sample of a spring water was analyzed for its iron content by acidifying and reducing all the iron present to Fe2+ (AW = 55.847). A 25.00 mL aliquot of a 0.002107M K2Cr2O7 (FW = 294.185) was added, which resulted in the reaction:

6(Fe2+) + (Cr2O7)(2-) + 14H+ → 6(Fe3+) + 2(Cr3+) + 7H2O

The excess K2Cr2O7 was back-titrated with 10.00mL of 0.01M Fe2+. Calculate the concentration (in ppm) of the iron in the sample.

基本題型

求解 !! ~~ 謝謝 ^_^

回答 (2)

Uncle Michael

2011-10-17 10:27 pm

✔ 最佳答案

6Fe²⁺ + Cr2O7²⁻ + 14H⁺ → 6Fe³⁺ + 2Cr³⁺ + 7H2O

No. of moles of K2Cr2O7 reacted = 0.002107 ´ (25/1000) = 0.000052675 mol

No. of moles of Cr2O7²⁻ reacted = 0.000052675 mol

Total no. of moles of Fe²⁺ used = 0.000052675 ´ 6 = 0.00031605 mol

No. of moles of Fe²⁺ used in back titration= 0.01 ´ (10/1000) = 0.0001 mol

No. of moles of Fe²⁺ in the sample =0.00031605 - 0.0001 = 0.00021605 mol

Mass of Fe²⁺ in the sample =0.00021605 ´ 55.847 = 0.01207 g

Mass of the sample ≈ (100 mL) ´ (1 g/mL) = 100 g

Conc. of iron in the sample = (0.01207/100) ´ (1000000 ppm) = 120.7 ppm

參考: Uncle Michael

👍 0 👎 0

[匿名]

2011-10-17 10:31 pm

120.63 ppm

👍 0 👎 0