# Chemical Forums

## Chemistry Forums for Students => High School Chemistry Forum => Chemistry Olympiad and other competitions => Topic started by: Evaldas on December 28, 2010, 01:04:20 PM

Title: Olympiad question. (Salts, acids, bases...)
Post by: Evaldas 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)  :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?
Title: Re: Olympiad question. (Salts, acids, bases...)
Post by: Evaldas 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...
Title: Re: Olympiad question. (Salts, acids, bases...)
Post by: Borek 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.
Title: Re: Olympiad question. (Salts, acids, bases...)
Post by: Hybrid 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'

Title: Re: Olympiad question. (Salts, acids, bases...)
Post by: Evaldas 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...
Title: Re: Olympiad question. (Salts, acids, bases...)
Post by: Borek 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?

Note: it is not your job to solve the question for someone. If you want to help, give hints. Please read forum rules (http://www.chemicalforums.com/index.php?topic=33740.0).
Title: Re: Olympiad question. (Salts, acids, bases...)
Post by: Hybrid 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 (http://www.chemicalforums.com/index.php?topic=33740.0).

and science when H2SO4 can't displace NaCl, before suggesting your own thoughts , think about it
Title: Re: Olympiad question. (Salts, acids, bases...)
Post by: Hybrid 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.
Title: Re: Olympiad question. (Salts, acids, bases...)
Post by: Borek 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.
Title: Re: Olympiad question. (Salts, acids, bases...)
Post by: Evaldas on December 29, 2010, 06:05:04 AM
Borek is right. I glanced at the solution for this.
This reaction: 2NaCl(aq) + H2SO4(aq)  :rarrow: Na2SO4(aq) + HCl(aq) is not happening, because you can't get two aquas, so NaOH is neutralizing the excess of H2SO4.
Title: Re: Olympiad question. (Salts, acids, bases...)
Post by: Evaldas 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.
Title: Re: Olympiad question. (Salts, acids, bases...)
Post by: sjb 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?
Title: Re: Olympiad question. (Salts, acids, bases...)
Post by: Evaldas 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
Title: Re: Olympiad question. (Salts, acids, bases...)
Post by: sjb 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?
Title: Re: Olympiad question. (Salts, acids, bases...)
Post by: Evaldas on December 29, 2010, 12:04:55 PM
H2SO4 + 2NaOH. Ratio 1:2
Title: Re: Olympiad question. (Salts, acids, bases...)
Post by: Evaldas on December 29, 2010, 01:48:37 PM
So n(H2SO4)=0.00113 mol
m(H2SO4)=0.11074 g
Title: Re: Olympiad question. (Salts, acids, bases...)
Post by: Hybrid on December 29, 2010, 02:31:02 PM
Evaldas, if you look in this equation i suppose you must get the answer.
(0.01mol 'H2SO4'- 0.00193mol 'BaOH'- 1/2 mol KOH - 1/2 mol NaCl) + mol HCl = 0.00226 mol  'NaOH'

N.B. about  H2SO4 / HCl it happens for sure but at the end all species are ionized " rule: strong acid can displace the weaker one from its salt" so it doesn't matter as 2H 'H2SO4' = 2H 'HCl' and the most important is that H+ is equivalent and present.

now back to the equation simply 1/2 mol NaCl =  mol HCl because any NaCl converted to HCl will consume 1/2 mol of H2SO4.

N.B. in the above equation 1/2 in KOH, NaCl correspond to the consumed H2SO4 in the equations you posted by yourself.

so finally you left with
0.01mol 'H2SO4'- 0.00193mol 'BaOH'- 1/2 mol KOH  = 0.00226 mol  'NaOH'
and from it " all mole are known except that of KOH" calculate moles of consumed H2SO4 for neutaralizing KOH and by multiplying it '2' you get the actual moles of KOH.
Title: Re: Olympiad question. (Salts, acids, bases...)
Post by: Borek on December 29, 2010, 06:46:34 PM
Evaldas, if you look in this equation i suppose you must get the answer.
(0.01mol 'H2SO4'- 0.00193mol 'BaOH'- 1/2 mol KOH - 1/2 mol NaCl) + mol HCl = 0.00226 mol  'NaOH'

Stop posting nonsense. NaCl is irrelevant.
Title: Re: Olympiad question. (Salts, acids, bases...)
Post by: Evaldas on December 30, 2010, 04:49:17 AM
Hybrid, give me your answers and I'll later check them in the solution with answers that I have. (After I solve this myself)
Title: Re: Olympiad question. (Salts, acids, bases...)
Post by: Borek on December 30, 2010, 05:15:48 AM
Hybrid is temporarily banned.

Evaldas, you were already on the right track. Do the analysis of what is happening with all acids and bases. You use NaOH to neutralize excess sulfuric acid - that should allow you to calculate sum of amounts of barium and potassium hydroxide present in the original mixture.
Title: Re: Olympiad question. (Salts, acids, bases...)
Post by: Evaldas on December 30, 2010, 06:11:31 AM
But according to my calculations 0.11074 g of H2SO4 reacted with NaOH, and we only have 0.098 g of H2SO4  ???

Edit: Wait, or is it 0.98 g?
Title: Re: Olympiad question. (Salts, acids, bases...)
Post by: Evaldas on December 30, 2010, 09:47:17 AM
I solved it.
A question, tho:
2KOH + H2SO4  :rarrow: K2SO4 + 2H2O
n(H2SO4)=0.00694 mol
How do I know if n of KOH is 1/2n(H2SO4) or 2n(H2SO4)?
Title: Re: Olympiad question. (Salts, acids, bases...)
Post by: sjb on December 30, 2010, 12:03:33 PM
I solved it.
A question, tho:
2KOH + H2SO4  :rarrow: K2SO4 + 2H2O
n(H2SO4)=0.00694 mol
How do I know if n of KOH is 1/2n(H2SO4) or 2n(H2SO4)?

If you have 1 mol of H2SO4, how many moles of KOH would you need? Alternatively, if you needed 34 mol of KOH, how much acid did you have?
Title: Re: Olympiad question. (Salts, acids, bases...)
Post by: Evaldas on December 30, 2010, 12:19:05 PM
2 mol. And 17 mol.
Title: Re: Olympiad question. (Salts, acids, bases...)
Post by: sjb on December 31, 2010, 05:01:58 AM
2 mol. And 17 mol.

So what is the relationship between the n of KOH and the n of H2SO4 ?
Title: Re: Olympiad question. (Salts, acids, bases...)
Post by: Evaldas on December 31, 2010, 05:11:32 AM
2 mol. And 17 mol.

So what is the relationship between the n of KOH and the n of H2SO4 ?
The coefficients in the balanced equation?
Title: Re: Olympiad question. (Salts, acids, bases...)
Post by: sjb on December 31, 2010, 07:50:48 AM