[o1ocups]

"Calculate the volume of 0.10 M HCl needed to produce a pH of 4.00 in the original buffered sample."
And the original buffer is a 20 mL solution that is 0.32 M in HBz and 0.28 M in Bz-. Ka for benzoic acid is 6.2*10^-5

OK I know I can calculate [H+] from the pH, but what to do next?
Sorry for not trying enough...my brain is half-dead, and the lab is tomorrow
Thanks..

[Loyal]

When you got a buffer what equation should you be using and also what other information do you need to obtain to use that equation?

[o1ocups]

Henderson-Hasselbalch equation?
pH=pKa + log ([base]/[acid])
It seems like I already have all the necessary information?
Do I calculate for ka?

[Borek]

Do I calculate for ka?

No. You need concentrations of acid and conjugate base. Think about stoichiometry of the reaction that takes place when you add strong acid (HCl) to the salt containing weak base (Bz^-).

[o1ocups]

OK So I just calculate for this whole thing ([base]/[acid]) ?

Think about stoichiometry of the reaction that takes place when you add strong acid (HCl) to the salt containing weak base (Bz-)

When you add HCl the amount of weak base decreases by the amount of HCl added because they have to neutralize the HCl right? Also the acid would increase by that same amount. I don't know how to work backwards though.

Edit: wait! I think I might know. So [base]/[acid] = 0.62
Amount of HCl added = 0.1M * x mL = 0.1 x mmol HCl
[Bz-]=5.6-0.1x mmol / 20+x mL
[HBz]=6.4+01.x mmol / 20+x mL
So you just set [Bz-]/[HBz] = 0.62?

[Borek]

Assume x is amount of strong acid added. Calculate amounts of HBz and Bz^- using initial amounts and x. Put it into the HH equation. SOlve for x.

[o1ocups]

Assume x is amount of strong acid added. Calculate amounts of HBz and Bz^- using initial amounts and x. Put it into the HH equation. SOlve for x.

How do you even put that into the HH equation? The base and acid are in concentrations so I also need to divide the (initial amount + or - x) by the volume right? That just gets really convoluted.

Edit: but I guess I should try first. let me try to do it now.

OK I got -0.02 for x. That's probably wrong right? I mean how can it be negative.
Do you think the approach that I described in my last post would work?

I used my approach and got 4.295 mL HCl, which I think is also wrong because it can absorb 64 mL of 0.10 M NaOH.

[Borek]

The base and acid are in concentrations so I also need to divide the (initial amount + or - x) by the volume right?

Yes. Note, that both volumes (the one used to calculate concentration of the acid after the reaction, and the one used to calculate concentration of the conjugate base after the reaction) are identical, so they simply cancel out. You may safely use just numbers of moles oin HH equation.

OK I got -0.02 for x. That's probably wrong right? I mean how can it be negative.

It probably means you have switched concentrations or sign somewhere.

Do you think the approach that I described in my last post would work?

You mean the edit? Honestly, I have missed that part.

But essentially it doesn't differ from the thing I have described - you have calculated concentrations of acid and conjugate base assuming complete reaction, and you have expressed both in terms of x - amount (volume) of HCl added. This is exactly approach I wrote about. And equations you wrote look OK to me (+/- one or two typos).

I used my approach and got 4.295 mL HCl, which I think is also wrong because it can absorb 64 mL of 0.10 M NaOH.

Check your math, I got different volume.

[o1ocups]

I did it again and got 10.06 mL, but I am really not sure because it can absorb 6 times as much base. Here is what I did:
pH = pKa + log ([base]/[acid])
4=-log(6.2*10^-5) + log([base]/[acid])
[base]/[acid]=0.62

(5.6 -0.1x)/(20+x) * (20+x)/(6.4+0.1x) = 0.62
(5.6 -0.1x)/(6.4+0.1x)=0.62
5.6 -0.1x = 3.97 + 0.062x
0.162x=1.63
x=10.06 mL

[Borek]

10 mL is what I got as well.

Not sure what you mean by "absorbing some amount of base". Please elaborate, I have a feeling that you are thinking about something completely irrelevant for the question.

[o1ocups]

There is another question "Calculate the total volume of 0.10 M NaOH that the buffered sampel can absorb" and I got 64 mL. here is what I did:
HBz=6.4 mmol
Can absorb 6.4 mmol of NaOH
Volume of NaOH = 0.0064 * 1L / 0.1 mol = 0.064 L = 64 mL
So my reasoning was that since the buffer can absorb that many milimeters of base before it even becomes basic, why is only 10 mL of acid able to change the pH to 4??

But again the pH of the original buffered solution is 4.15 so that's not that much of a difference. I guess this is right.
Thanks.

[Borek]

Question is ambiguous, as it doesn't precisely state what they mean by "buffered sample can absorb".

It is all in the pH change, but what change can be accepted? Sometimes it is 0.01 pH unit, sometimes 0.1, sometimes change by 1 pH unit doesn't matter much.

You have added 10 mL of HCl to change pH by 0.15 unit, 64 mL of NaOH to change pH by 1.25 unit. Doesn't make sense to compare these numbers.