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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

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) + H_{2}SO_{4}(aq) :rarrow: BaSO_{4}(s) + 2H_{2}O(l);
2) 2KOH(aq) + H_{2}SO_{4}(aq) :rarrow: K_{2}SO_{4}(aq) + 2H_{2}O(l);
3) 2NaCl(aq) + H_{2}SO_{4}(aq) :rarrow: Na_{2}SO_{4}(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) + H_{2}O(l).
b) 1) n(BaSO_{4}) = 0,45 g/233 g/mol = 0,00193 mol
n(BaSO_{4}) = 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?

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 K_{2}SO_{4}. Right? But I can't really put it together just yet...

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 K_{2}SO_{4}, dissolved NaCl doesn't react with sulfuric acid.

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'

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 K_{2}SO_{4}, dissolved NaCl doesn't react with sulfuric acid.
Oh, you're right, I missed it somehow...

(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).

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

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

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.

Borek is right. I glanced at the solution for this.
This reaction: 2NaCl(aq) + H_{2}SO_{4}(aq) :rarrow: Na_{2}SO_{4}(aq) + HCl(aq) is not happening, because you can't get two aquas, so NaOH is neutralizing the excess of H_{2}SO_{4}.

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 K_{2}SO_{4}, dissolved NaCl doesn't react with sulfuric acid.
Wait, what about NaOH?

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 K_{2}SO_{4}, dissolved NaCl doesn't react with sulfuric acid.
Wait, what about NaOH?
Yes, NaOH reacts with sulfuric acid  how much NaOH did you use?

Yes, NaOH reacts with sulfuric acid  how much NaOH did you use?
0.00226 mol or 0.0904 g

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?

H2SO4 + 2NaOH. Ratio 1:2

So n(H2SO4)=0.00113 mol
m(H2SO4)=0.11074 g

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.

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.

Hybrid, give me your answers and I'll later check them in the solution with answers that I have. (After I solve this myself)

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.

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?

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)?

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 H_{2}SO_{4}, how many moles of KOH would you need? Alternatively, if you needed 34 mol of KOH, how much acid did you have?

2 mol. And 17 mol.

2 mol. And 17 mol.
So what is the relationship between the n of KOH and the n of H_{2}SO_{4} ?

2 mol. And 17 mol.
So what is the relationship between the n of KOH and the n of H_{2}SO_{4} ?
The coefficients in the balanced equation?

So can you answer your previous question?
How do I know if n of KOH is 1/2n(H2SO4) or 2n(H2SO4)?

Well if we have an equation 2KOH + H_{2}SO_{4} :rarrow: K_{2}SO_{4} + 2H_{2}O we can see that 2 moles of KOH react with 1 mole of H_{2}SO_{4}, so 2n(H_{2}SO_{4})=n(KOH), 1/2n(KOH)=n(H_{2}SO_{4})