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Chemistry Forums for Students => High School Chemistry Forum => Topic started by: Saint on March 20, 2008, 02:45:37 PM

Title: H2SO4 Help me!
Post by: Saint 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?
Title: Re: H2SO4
Post by: Borek on March 20, 2008, 02:57:42 PM
Question - as worded - doesn't make sense. Please elaborate or try to reword it.
Title: Re: H2SO4 Help me!
Post by: Saint on March 20, 2008, 06:36:22 PM
I modified it can you make sense now????
Title: Re: H2SO4 Help me!
Post by: Borek 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.
Title: Re: H2SO4 Help me!
Post by: Saint 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
Title: Re: H2SO4 Help me!
Post by: Borek on March 21, 2008, 03:00:18 PM
Everything depends on amount of solutions A and B.
Title: Re: H2SO4 Help me!
Post by: Saint 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?
 
Title: Re: H2SO4 Help me!
Post by: Borek 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?
Title: Re: H2SO4 Help me!
Post by: Saint 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...
Title: Re: H2SO4 Help me!
Post by: Borek on March 22, 2008, 10:01:08 AM
Quote from: Saint 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?

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.
Title: Re: H2SO4 Help me!
Post by: Saint 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
Title: Re: H2SO4 Help me!
Post by: Borek on March 22, 2008, 11:07:02 AM
Quote from: Saint on March 22, 2008, 10:59:22 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.
Title: Re: H2SO4 Help me!
Post by: Saint 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.
Title: Re: H2SO4 Help me!
Post by: Borek 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 (http://www.chembuddy.com/?left=pH-calculation&right=pH-buffers-henderson-hasselbalch).
Title: Re: H2SO4 Help me!
Post by: Saint 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
Title: Re: H2SO4 Help me!
Post by: Borek on March 24, 2008, 05:39:14 PM
No idea what you mean.
Title: Re: H2SO4 Help me!
Post by: Saint on March 24, 2008, 06:25:34 PM
I didn't finished my reply my internet was cut.
Now...
That typo you gave to me the Hederson one exists only when the approximations work.
Here after ionizations and with C H2SO4's concentration we have:
1) C+x=10-4 (from the ph)
2) 10-2=x*10-4/C-x  (from the second ionization)
Well that Hederson typo is the 2) but with the approximations:
3) 10-2=x*10-4/C

If we solve 1-3 we find that C=x which is impossible
If we solve the 1-2 we take "reasonable" results which I'll write down tomorrow...
I hope you understand... Goodnight...
Title: Re: H2SO4 Help me!
Post by: AWK on March 25, 2008, 03:20:25 AM
H2SO4 with pH 4 and NaOH with pH 10 practically forms Na2SO4 with concentration 2.5 10-5 M (error less then 0.5 %). I think this problem can be approximately treated as hydrolysis of Na2SO4 instead of buffer Na2SO4/NaHSO4 far outside of reasonable bufferring capacity range.

I think Borek can check both possibilities with his program