Chem Homework

joe_chanboy

2009-07-22 8:08 pm

0.10 g of a salt substitute (a mixture of sodium chloride and
potassium chloride) was dissolved in water and titrated with 0.05 mol dm-3 silver nitrate solution using potassium chromate as indicator,
30.52 cm-3 of silver nitrate solution was required. Calculate the
percentage of sodium chloride and of potassium chloride in the
mixture.

!! 佢寫住potassium chloride 會contain less chloride than sodium
chloride !! 所以唔可以就咁計potassium 同 sodium 既比例 ...

唔該幫幫手! 感謝 !!

回答 (1)

老爺子

2009-07-22 9:41 pm

✔ 最佳答案

Consider the titration:
Ag+ + Cl- → AgCl
Mole ratio Ag+ : Cl- = 1 : 1

No. of moles of AgNO3 used = 0.05 x (30.52) = 0.001526 mol
No. of moles of Ag+ used = 0.001526 mol
No. of moles of Cl- in the nixture = 0.001526 mol
Mass of Cl- in the in the mixture = 0.001526 x 35.5 = 0.05417 g

Molar mass of Cl- ion = 35.5 g/mol
Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol
Molar mass of KCl = 39 + 35.5 = 74.5 g/mol
Mass fraction of Cl- in NaCl = 35.5/58.5
Mass fraction of Cl- in KCl = 35.5/74.5

Let the mass of NaCl be y g.
Then, the mass of KCl = (0.1 - y)g

Mass of Cl- in the mixture:
y(35.5/58.5) + (0.1 - y)(35.5/74.5) = 0.05417
y(35.5/58.5) - y(35.5/74.5) = 0.05417 - 0.1(35.5/74.5)
y = 0.05

The % by mass of NaCl in the mixture = (0.05/0.1) x 100% = 50%
The % by mass of KCl in the mixture = 1 - 50% = 50%

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