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### Topic: Olympiad question. (Salts, acids, bases...)  (Read 16413 times)

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#### Evaldas

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« Reply #15 on: December 29, 2010, 01:48:37 PM »
So n(H2SO4)=0.00113 mol
m(H2SO4)=0.11074 g

#### Hybrid

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« Reply #16 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.

#### Borek ##### Re: Olympiad question. (Salts, acids, bases...)
« Reply #17 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.
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#### Evaldas

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« Reply #18 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)

#### Borek ##### Re: Olympiad question. (Salts, acids, bases...)
« Reply #19 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.
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#### Evaldas

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« Reply #20 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?

#### Evaldas

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« Reply #21 on: December 30, 2010, 09:47:17 AM »
I solved it.
A question, tho:
2KOH + H2SO4 K2SO4 + 2H2O
n(H2SO4)=0.00694 mol
How do I know if n of KOH is 1/2n(H2SO4) or 2n(H2SO4)?

#### sjb

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« Reply #22 on: December 30, 2010, 12:03:33 PM »
I solved it.
A question, tho:
2KOH + H2SO4 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?

#### Evaldas

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« Reply #23 on: December 30, 2010, 12:19:05 PM »
2 mol. And 17 mol.

#### sjb

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« Reply #24 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 ?

#### Evaldas

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« Reply #25 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?

#### sjb

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« Reply #26 on: December 31, 2010, 07:50:48 AM »
How do I know if n of KOH is 1/2n(H2SO4) or 2n(H2SO4)?

#### Evaldas

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« Reply #27 on: December 31, 2010, 09:26:42 AM »
Well if we have an equation 2KOH + H2SO4 K2SO4 + 2H2O we can see that 2 moles of KOH react with 1 mole of H2SO4, so 2n(H2SO4)=n(KOH), 1/2n(KOH)=n(H2SO4)