[K']

The pH of a 0.10 M HA solution is 3.8.

How many mL of 0.10 M NaOH should be added to 20 mL 0.10 M HA to prepare a  buffer with a pH of 4.3?

[Borek]

You have to show your attempts at solving the question to receive help. This is a forum policy.

[K']

Here is my work:
[H+] = 10^(-4.3) = 5.01x10^-5 mol/L

.020 L (5.01x10^-5 mol HA/1 L)(1 L/ 0.10 mol NaOH) =  1.00 L

But I don't know it this is correct.

Thanks.

[Borek]

No idea what you are doing - it is definitely not correct.

Do you know any equations that are used for buffer pH calculations?

[K']

I took the anti log to get [H+] then preceded to get volume required.
I don't think that I need an equation to solve this problem, you guys are the experts here.

[Borek]

Sorry, but it looks like you have no idea how to do buffer problems. First you need acid Ka, then it is all bout calculating ratio of concentrations of acid and its conjugate base.

[K']

Actually I think I know about ka and pka etc.
my calculations showed that:
pka = 6.6
ka = 2.51x10^-7
ratio = 2.40x10^-5

But I still think that I don't need an equation to solve that problem.
Can you please give more details.

[Borek]

Ratio is incorrect. You have neutralized part of the acid - you can safely assume neutralization was stoichiometric and went to completion.

And while there are many ways of skinning that cat, the simplest approach is to calculate ratio and plug it into the Henderson-Hasselbalch equation.

[K']

I used that equation and I think did a mistake.
Thanks for pointing that out.

after few corrections, I arrived at 5.32 mL required.

I just need your confirmation that is the correct answer.

[Borek]

Your pKa value is OK, but the volume is wrong.

However, there is something wrong with the question IMHO. pKa 6.6 acid should not be used to prepare pH 4.3 buffer.

[K']

Thanks for the help