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Topic: H2SO4 Help me!  (Read 9934 times)

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Offline Saint

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H2SO4 Help me!
« on: March 20, 2008, 02:45:37 PM »
Well i have this problem which it gives me a real headache...

We have the solution A which has H2SO4 in it and has PH=4
Find the Molarity of the acid c=?M
Ka2=10-2

then we mix solution A with solution B and we create a solution C

In solution B there is NaOH with PH=10


Find the PH of solution C

I did the first but on the other question i got messed up...
Is now ok?
« Last Edit: March 21, 2008, 01:09:24 PM by Saint »
E=mc2

Offline Borek

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Re: H2SO4
« Reply #1 on: March 20, 2008, 02:57:42 PM »
Question - as worded - doesn't make sense. Please elaborate or try to reword it.
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Offline Saint

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Re: H2SO4 Help me!
« Reply #2 on: March 20, 2008, 06:36:22 PM »
I modified it can you make sense now????
E=mc2

Offline Borek

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Re: H2SO4 Help me!
« Reply #3 on: March 20, 2008, 06:39:50 PM »
It still doesn't make sense, although I think I understand what you need. Thing is, NaOH solution can't have pH=4 - it must have pH >= 7.
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Offline Saint

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Re: H2SO4 Help me!
« Reply #4 on: March 21, 2008, 01:08:55 PM »
Holy s#*$ yep right sorry sorry... NaOH's PH=10 I was just thinking of its C=10-4 and that 4 made me write it wrong
E=mc2

Offline Borek

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Re: H2SO4 Help me!
« Reply #5 on: March 21, 2008, 03:00:18 PM »
Everything depends on amount of solutions A and B.
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Offline Saint

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Re: H2SO4 Help me!
« Reply #6 on: March 21, 2008, 04:21:48 PM »
equal Volumes V
the same amount of solutions...
Sorry for popping up each one of them but i have lost the page with the exercise and it's difficult to recall....
So now any hint?
 
E=mc2

Offline Borek

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Re: H2SO4 Help me!
« Reply #7 on: March 21, 2008, 06:04:28 PM »
Unfortunately, now all depends on your level. As first approximation - you are mixing identical volumes of solutions containing identical concentrations of H+ and OH-. What reaction takes place?
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Offline Saint

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Re: H2SO4 Help me!
« Reply #8 on: March 22, 2008, 07:45:50 AM »
well let's concentrate on the first query
H2SO4 + H20 -> HSO4- + H30+

Then
HSO4- + H2O <-> SO42- + H30+

But here Ka2= 10-2 do the approximations work here?
If not (i think they don't) how will i get C? too complicated ???

As for the other I think it's:
H2SO4 + NaOH -> Na2SO4 + H20

And here i have other queries which i am going to say after we finish the first question...
E=mc2

Offline Borek

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Re: H2SO4 Help me!
« Reply #9 on: March 22, 2008, 10:01:08 AM »
well let's concentrate on the first query
H2SO4 + H20 -> HSO4- + H30+

Then
HSO4- + H2O <-> SO42- + H30+

But here Ka2= 10-2 do the approximations work here?

At pH 4 you are two pH units from pKa, look at the Henderson-Hasselbalch equation - what does it tell you about how much HSO4- is left?

Quote
H2SO4 + NaOH -> Na2SO4 + H20

Try net ionic.
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Offline Saint

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Re: H2SO4 Help me!
« Reply #10 on: March 22, 2008, 10:59:22 AM »
well it doesn't tell anything about that.
It gives only PH of the H2SO4 solution equal with 4 (PH=4) and Ka2=10-2. My problem is if approximations work...

And about the second i have no clue what you're talking. Sorry. I tried making the concentration from the first question i found (?) into moles and the concentration of NaOH into moles and do the reaction but then I had a solution with bot H2SO4 and Na2SO4
E=mc2

Offline Borek

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Re: H2SO4 Help me!
« Reply #11 on: March 22, 2008, 11:07:02 AM »
well it doesn't tell anything about that.

Yes it does.

pH = pKa + log(A-/HA)

or

log(A-/HA) = pH - pKa

Your pH is 4, your pKa is 2. What is log of ratio, what is ratio?

Quote
It gives only PH of the H2SO4 solution equal with 4 (PH=4) and Ka2=10-2. My problem is if approximations work...

See above.

Quote
And about the second i have no clue what you're talking. Sorry. I tried making the concentration from the first question i found (?) into moles and the concentration of NaOH into moles and do the reaction but then I had a solution with bot H2SO4 and Na2SO4

I told you: start with net ionic reaction equation.
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Offline Saint

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Re: H2SO4 Help me!
« Reply #12 on: March 22, 2008, 02:10:39 PM »
pH = pKa + log(A-/HA)
Doesn't that typo concerns only ph indicators???
Is this of use for every acid or base???
I used simple ionization for the first one.
E=mc2

Offline Borek

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Re: H2SO4 Help me!
« Reply #13 on: March 22, 2008, 02:32:02 PM »
It works for every acid, in fact it is just a rearranged dissociation constant definition, see Henderson-Hasselbalch equation derivation and discussion.
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Re: H2SO4 Help me!
« Reply #14 on: March 24, 2008, 01:24:05 PM »
Ok i see now i just didn't understand the typo.
But taking that doesn't mean that approximities exist... If i take approximately the number we have C=x which is impossible since 2nd ionization of H2SO4 is not strong.
For the first question i just took the given things...
C+x=10-4
and
10-2=x10-4/C-x
E=mc2

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