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Olympiad question. (Salts, acids, bases...)

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Evaldas:
I'm preparing for a chemistry olympiad.
I'm stuck at this exercise:
2 g of a mixture, made up of barium hydroxide, potassium hydroxide and sodium chloride, was mixed with 0,1 liter, 0,1 mol/l sulfuric acid solution. White precipitate was formed. This precipitate was separated by filtering and drying. The mass of the dry precipitate 0,45 g. To neutralize the filtrate 0,0113 l of 0,2 mol/l sodium hydroxide solution was required.
a) write the equations for all the reactions.
b) count the mass of every substance in the original mixture.
c) count, how many percent of the original mixture were: i) barium ions; ii) potassium ions; iii) sodium ions.

My attempt:
a) 1) Ba(OH)2(aq) + H2SO4(aq)  :rarrow: BaSO4(s) + 2H2O(l);
2) 2KOH(aq) + H2SO4(aq)  :rarrow: K2SO4(aq) + 2H2O(l);
3) 2NaCl(aq) + H2SO4(aq)  :rarrow: Na2SO4(aq) + HCl(??) question here: in this (3) reaction HCl gas or aq? If aq then it wouldn't be happening, would it?
4) HCl(aq) + NaOH(aq)  :rarrow: NaCl(aq) + H2O(l).

b) 1) n(BaSO4) = 0,45 g/233 g/mol = 0,00193 mol
n(BaSO4) = n(Ba(OH)2) = 0,00193 mol
m(Ba(OH)2) = 0,00193 mol x 171 g/mol = 0,33 g.
2) c(NaOH) = 0,2 l/mol = n/0,0113 l => n(NaOH) = 0,2 l/mol x 0,0113 l = 0,00226 mol
I stop here.
What do I do next?
How do I find the mass of potassium sulfate?

Evaldas:
I'm thinking: chloride ions should help me get the mass of NaCl, and having that I would subtract masses of NaCl and Ba(OH)2 from 2 g and get the mass of K2SO4. Right? But I can't really put it together just yet...

Borek:
Not "count", but "calculate". When you want to know how many matches you have, you count them: one match, two matches, three matches and so on. That's counting. When you are multiplying, adding, taking square roots and so on - you are calculating.

It is simpler than you think. Precipitate gives you amount of barium hydroxide. Titration tells you how much bases in total were in the solution, as you already know how much Ba(OH)2 was present that gives you amount of KOH. Whatever is left was NaCl. You were not asked about and don't need mass of K2SO4, dissolved NaCl doesn't react with sulfuric acid.

Hybrid:
first we assume that there is xss H2SO4
so , (0.01mol 'H2SO4'- 0.00193mol 'BaOH'- 1/2 mol KOH - 1/2 mol NaCl) + mol HCl = 0.00226 mol  'NaOH'

Evaldas:

--- Quote from: Borek on December 28, 2010, 01:59:09 PM ---Not "count", but "calculate". <...>

--- End quote ---
Oh, thanks ;), I'll keep that in mind.


--- Quote from: Borek on December 28, 2010, 01:59:09 PM ---It is simpler than you think. Precipitate gives you amount of barium hydroxide. Titration tells you how much bases in total were in the solution, as you already know how much Ba(OH)2 was present that gives you amount of KOH. Whatever is left was NaCl. You were not asked about and don't need mass of K2SO4, dissolved NaCl doesn't react with sulfuric acid.

--- End quote ---
Oh, you're right, I missed it somehow...

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