[o1ocups]

"What amount of solid NaOH must be added to 1.0 L of a 0.12 M H2CO3 solution to produce a solution with [H+] = 3.5×10^-11 ? There is no significant volume change as the result of the addition of the solid."

How do I approach this problem? I know it has something to do with working backwards with the value of Ka (I looked it up, it's 4.3*10^-7 for H2CO3) using [H+], but I don't know. I think I am thinking about buffer questions. Can anyone give me a hint on how to start solving this problem? Thanks!!

[Borek]

http://www.chemicalforums.com/index.php?topic=31862

[o1ocups]

Think what reactions take place when you add NaOH.

That's what I couldn't figure out. I can write a buffer equation for H2CO3 though:
H2CO3 + H2O <-> HCO3- + H3O+   

[Borek]

Will you be able to solve the question if you were asked to prepare carbonate buffer mixing NaHCO₃ with Na₂CO₃?

[o1ocups]

OK, with some help from my friend, here is what I did:

So apparently there are two deprotenations, I am not sure but I calculated the second ones:
pH = pka2 + log ([CO3 2-]/[HCO3-])
-log(3.5*10^-11) = 10.25 + log ([CO3 2-]/[HCO3-])
10.46-10.25 = log ([CO3 2-]/[HCO3-])
10^0.21 = [CO3 2-]/[HCO3-] = 1.622
1.622 = (CO3 2-) / 0.1-(CO3 2-)
CO3 2- = 0.06164
multiply that by 1 L, moles CO3 2- = 0.06164
0.06164 + 0.1 = 0.16164
0.16164 * 39.99 = 6.47 grams NaOH

Is this right?  I entered 6.46 before and it was wrong.
And as you can see I don't really understand what I did myself, so any clarification would be appreciated..

Edit: yeah and I don't know where she got the second value of pka either. That was really confusing to me.

[o1ocups]

Uh nevermind I think I realized what I did. I used my friend's number (0.1) instead of mine! (0.12)

[Borek]

You have not shown the final result, so it is impossible to check, but approach seems correct.