Buffer Question- TBR Mistake?

[discowisco]

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What is the pH of a solution made by madding 0.839 grams NaHCO3 (MW = 83.9) to 100 ml 0.10 M H2CO3? pka1 = 6.4

Now i understand that the moles of NaHCO3 = 0.01 and the moles of H2CO3 = 0.01

so you use pka + log base/acid

Now my question is why is it 6.4 + log 0.01/0.01 ?

Shouldnt it be 6.4+ log 0.01/0 ?

Since the amount of base added to the solution transforms that much solution into base. By adding 0.01 of base to the solution you transform that much into base and leave the rest as acid.


Its like that in every other problem I have done

for example:

What is the ph of a solution made by mixing 10 ml of .10 M HCOOH with 4 ml of .10 M KOH? pka = 3.64

moles of HCOOH = 1
moles of KOH = .4

So for this the setup is 3.64 + log .4/.6

The .4 of base makes .4 of the 1 mole of HCOOH into base and leaves 0.6 as acid.

Why isnt the 1st problem solved like this?!?!

 

[FeinMS]

TBR is correct. The first problem is a buffer problem (acid+conjugate base,) so it's correct to use HHB equation. The conjugate base doesn't deprotonate the acid as in the second problem probably because the base isn't strong enough. (Don't quote me on the reason because I'm not too sure.)

The second problem gives you an acid with strong base. This strong base will deprotonate the acid to convert the acid to conjugate base. Again, when KOH is all used up, you will have HCOOH (acid) with HCOO- (conjugate base) This is a buffer and you use HHB.

To make the first problem analogous to the second problem, it can be written like:
You have 0.02moles of H2CO3 and you add 0.01 KOH. What's the pH?
Since KOH will deprotonate H2CO3 to convert half of it to HCO3-, the final solution will be composed of 0.01 moles of H2CO3 + 0.01 moles of HCO3-. From there, HHB equation. or more simply since [acid]=[conjugate base], pH=pKa

 

[JingleChips]

What is the pH of a solution made by madding 0.839 grams NaHCO3 (MW = 83.9) to 100 ml 0.10 M H2CO3? pka1 = 6.4

Now i understand that the moles of NaHCO3 = 0.01 and the moles of H2CO3 = 0.01

so you use pka + log base/acid

Now my question is why is it 6.4 + log 0.01/0.01 ?

Shouldnt it be 6.4+ log 0.01/0 ?

Since the amount of base added to the solution transforms that much solution into base. By adding 0.01 of base to the solution you transform that much into base and leave the rest as acid.


Its like that in every other problem I have done

for example:

What is the ph of a solution made by mixing 10 ml of .10 M HCOOH with 4 ml of .10 M KOH? pka = 3.64

moles of HCOOH = 1
moles of KOH = .4

So for this the setup is 3.64 + log .4/.6

The .4 of base makes .4 of the 1 mole of HCOOH into base and leaves 0.6 as acid.

Why isnt the 1st problem solved like this?!?!

Pretty much what the poster above me said.

The conjugate base doesn't deprotonate the acid as in the second problem probably because the base isn't strong enough. (Don't quote me on the reason because I'm not too sure.)

Although this is true that NaHCO3- isn't a strong base, it doesn't matter because even if it was - is a conjugate pair with H2CO3.

So you have .01moles H2CO3 chilling in solution. If you assume that the NaHCO3- is a strong base, and you add .01moles NaHCO3- to the solution, look what happens:

.01moles HCO3- acting as a strong base would deprotonate .01moles of H2CO3. The original base would become .01moles of H2CO3 (from gaining the proton) and the original .01moles of H2CO3 would become .01moles of HCO3- (from losing the proton) - so you're back exactly where you started.

 

[discowisco]

Your right, thanks. Great explanation didn't think of starting back where you started from