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Evaldas

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« on: December 28, 2010, 01:04:20 PM »
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)  BaSO4(s) + 2H2O(l);
2) 2KOH(aq) + H2SO4(aq)  K2SO4(aq) + 2H2O(l);
3) 2NaCl(aq) + H2SO4(aq)  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)  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

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Re: Olympiad question. (Salts, acids, bases...)
« Reply #1 on: December 28, 2010, 01:08:58 PM »
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

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Re: Olympiad question. (Salts, acids, bases...)
« Reply #2 on: December 28, 2010, 01:59:09 PM »
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.
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Hybrid

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Re: Olympiad question. (Salts, acids, bases...)
« Reply #3 on: December 28, 2010, 02:18:10 PM »
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

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Re: Olympiad question. (Salts, acids, bases...)
« Reply #4 on: December 28, 2010, 02:21:46 PM »
Not "count", but "calculate". <...>
Oh, thanks , I'll keep that in mind.

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.
Oh, you're right, I missed it somehow...

Borek

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Re: Olympiad question. (Salts, acids, bases...)
« Reply #5 on: December 28, 2010, 04:09:04 PM »
(0.01mol 'H2SO4'- 0.00193mol 'BaOH'- 1/2 mol KOH - 1/2 mol NaCl) + mol HCl = 0.00226 mol  'NaOH'

Since when NaCl takes part in the neutralization?

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Hybrid

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Re: Olympiad question. (Salts, acids, bases...)
« Reply #6 on: December 28, 2010, 04:46:43 PM »

Since when NaCl takes part in the neutralization?

Note: it is not your job to solve the question for someone. If you want to help, give hints. Please read forum rules.

and science when H2SO4 can't displace NaCl, before suggesting your own thoughts , think about it

Hybrid

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Re: Olympiad question. (Salts, acids, bases...)
« Reply #7 on: December 28, 2010, 04:53:46 PM »
and b y the way try to understand the above equation correctly then you can discuss it.

Borek

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Re: Olympiad question. (Salts, acids, bases...)
« Reply #8 on: December 28, 2010, 07:21:16 PM »
and science when H2SO4 can't displace NaCl, before suggesting your own thoughts , think about it

Concentrated sulfuric acid with solid NaCl - yes, that's a laboratory way of producing gaseous HCl. But that's not the case here, we are dealing with diluted (0.1M) sulfuric acid, so NaCl will be completely inert.

and b y the way try to understand the above equation correctly then you can discuss it.

Feel free to try to explain it. It is wrong. You are trying to balance all acids and bases that were neutralized. Idea is good, but execution is lousy.
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Evaldas

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Re: Olympiad question. (Salts, acids, bases...)
« Reply #9 on: December 29, 2010, 06:05:04 AM »
Borek is right. I glanced at the solution for this.
This reaction: 2NaCl(aq) + H2SO4(aq)  Na2SO4(aq) + HCl(aq) is not happening, because you can't get two aquas, so NaOH is neutralizing the excess of H2SO4.

Evaldas

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Re: Olympiad question. (Salts, acids, bases...)
« Reply #10 on: December 29, 2010, 09:46:30 AM »
Ahhh, I still don't know how to solve this. I don't want to jump straight to the answers that I have, I want to try to solve it.

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.
« Last Edit: December 29, 2010, 10:22:54 AM by Evaldas »

sjb

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Re: Olympiad question. (Salts, acids, bases...)
« Reply #11 on: December 29, 2010, 11:11:54 AM »
Ahhh, I still don't know how to solve this. I don't want to jump straight to the answers that I have, I want to try to solve it.

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.

Yes, NaOH reacts with sulfuric acid - how much NaOH did you use?

Evaldas

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Re: Olympiad question. (Salts, acids, bases...)
« Reply #12 on: December 29, 2010, 11:13:59 AM »
Yes, NaOH reacts with sulfuric acid - how much NaOH did you use?
0.00226 mol or 0.0904 g

sjb

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Re: Olympiad question. (Salts, acids, bases...)
« Reply #13 on: December 29, 2010, 11:46:07 AM »
Yes, NaOH reacts with sulfuric acid - how much NaOH did you use?
0.00226 mol or 0.0904 g

which reacts with how much acid?

Evaldas

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Re: Olympiad question. (Salts, acids, bases...)
« Reply #14 on: December 29, 2010, 12:04:55 PM »
H2SO4 + 2NaOH. Ratio 1:2