How do I calculate the concentration for Fe2+ in the solution??

[匿名]

2016-10-26 5:17 pm

This is a question I got in one of my lectures, and couldn't get the answer. Please help

Students are given 250cm^3 of a Fe2+ salt solution. 25.00cm^3 of this solution is titrated to the equivalence point with 16.55cm^3 of 0.02400M KMnO4.

MnO4- (aq) + 5Fe2+(aq) + 8H+(aq) --> 5Fe3+(aq) + Mn2+(aq) + 4H2O(l)

a) What is the concentration of the Fe2+ solution?

b) The Fe2+ salt solution for this titration was made by dissolving 3.950g of FeCl2* xH2O in enough water to make the 250cm^3 solution. What is the value of x?

Please, even if you can just help with a, it would be greatly appreciated!

回答 (1)

不用客氣

2016-10-26 5:57 pm

✔ 最佳答案

(a)
MnO₄⁻(aq) + 5Fe²⁺(aq) + 8H⁺(aq) → 5Fe³⁺(aq) + Mn²⁺(aq) + 4H₂O(l)

No. of moles of KMnO₄ = (0.02400 mol/dm³) × (16.55/1000 dm³) = 0.0003972 mol
No. of moles of MnO₄⁻ = 0.0003972 mol

According to the equation, mole ratio MnO₄⁻ : Fe²⁺ = 1 : 5
No. of moles of Fe²⁺ = (0.0003972 mol) × 5 = 0.001986 mol
Volume of Fe²⁺ solution = 25.00 cm³ = 0.02500 dm³
Concentration of Fe²⁺ solution = (0.001986 mol) / (0.02500 dm³) = 0.07944 M

(b)
Molar mass of FeCl₂•xH₂O
= [55.85 + 35.45×2 + x(1.01×2 + 16.00)] g/mol
= (126.75 + 18.02x) g/mol

In the 250 cm³ of Fe²⁺ solution :
No. of moles of FeCl₂•xH₂O = No. of moles of Fe²⁺
(3.950 g) / [(126.75 + 18.02x) g/mol] = (0.07944 mol/dm³) × (250/1000 dm³)
126.75 + 18.02x = 3.950 / (0.07944 × 0.250)
126.75 + 18.02x = 198.89
18.02x = 72.14
x = 4

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