(20點)兩小題AL chem

[匿名]

2010-07-03 7:03 pm

1.Question
An experiment as described below was carried out to determine the total hardness in a sample of hard water:

' 50.0 cm^3 of the sample was allowed to pass through an ion-exchange column, in which the metal ions present in the sample were totally exchanged by hydrogen ions. The eluent collected required 15.0 cm^3 of 0.020M KOH(aq) for complete neutralization. '

Assuming that the metal ions present in the sample are Mg2+ and Ca2+ only, calculate the total hardness, in M, of the sample.

1.Answer
Amount of H+(aq) exchanged = 0.020 x 15 x 10^-3 = 3.0 x 10^-4 mol
Total hardness of the water sample = (0.5 x 3.0 x 10^-4) / (50 x 10^-3) = 3 x 10^-3 M

為什麼Total hardness of the water sample要乘0.5??total hardness不是要包括both Mg2+和Ca2+嗎?

2. Question
Write chemical equation for the following reaction:
the disproportionation of (MnO4)2-(aq) in water to give (MnO4)-(aq) and MnO2(s)

請說明balance方法。

回答 (1)

andrew

2010-07-03 11:59 pm

✔ 最佳答案

1.
Considering the charge balance during the process.
No. of moles Ca2+ : No. of moles of H⁺ ions exchanged with Ca²⁺ ions = 1 : 2
No. of moles of Ca²⁺ = 0.5 x (No. of moles of H⁺ ions exchanged with Ca²⁺ ions) … (1)

Similarly,
No. of moles of Mg²⁺ = 0.5 x (No. of moles of H⁺ ions exchanged with Mg²⁺ ions) … (2)

Combine (1) and (2):
Total no. of moles Ca²⁺ and Mg²⁺ = 0.5 x (Total no. of moles of H⁺ ions exchanged)

Total hardness of water sample
= Total concentration of Ca²⁺ and Mg²⁺
= (Total no. of moles of Ca²⁺ and Mg²⁺) / (Volume of water sample)
= 0.5 x (Total no. of moles of H⁺ ions exchanged) / (Volume of water sample)
= 0.5 x (3.0 x 10¯⁴) / (50 x 10¯³)
= 3 x 10¯³ M

2.
Half equation of oxidation:
MnO4²¯(aq) → MnO4¯(aq) + e¯ …… (1)

Half equation of reduction:
MnO4²¯(aq) + 2H2O(l) + 2e¯ → MnO2(s) + 4OH¯(aq) …… (2)

(1)x2 + (2), and elimination 2e¯ on the both sides. The overall equation is:
3MnO2(aq) + 2H2O(l) → MnO4¯(aq) + 2MnO2(s) + 4OH¯(aq)

參考: andrew

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