[HaroTaro]

You want to prepare a 1L buffer HA/A- with a total concentration of 0.1 M and a pH of 6.8. HA has a pKa of 6.2

HA: 1M
A- : 1M  <<< in aqueous solution
NaOH: 5M
HCl: 5M
water

Will you be able to prepare the desired buffer solution with the following recipe?
1. 50 mL HA + 2mL NaOH + 948mL water
2. 100 mL A- + 16mL HCl + 884 mL water.

Please show me the actual calculation. I'm really confused. It's hard to take the fixed amount of water into consideration. I got stuck after I obtained the A-/HA ratio from the Henderson-Hassalbeck equation. I can't seem to take the problem from there on. Please help~

[AWK]

You have a problem
x+y =0.1 from stoichiometry
x/y = (from H-H equation)
Having known moles you can calculate volumes
Finally add enough water to get 1000 mL. (practically for diluted solutions their volumes are additive)

[HaroTaro]

I'm sorry, I don't quite capture what you meant. Can you elaborate a bit?

THrough the H-H equation, I got [A-]/[HA]= 3.98.
That then give me ~80% A- and 20% HA. Would this then be 0.08 M A- and 0.02 M HA? I can't get pass here. 

[AWK]

Yes, but this result concerns a mixture of HA + A_-.
You should check your data

1. 50 mL HA + 2mL NaOH + 948mL water
2. 100 mL A- + 16mL HCl + 884 mL water

which combination of stoichiometry fulfills both equations. I can exclude one from this list. Check the second if it works.

[HaroTaro]

wait,do we even need the H-H equation.

50mL of 1M HA in 948mL of water.THis will give me 0.0527M
2mL of 5M HCL in 948 mL of water.THis will give me 0.0105M
0.0527M + 0.0105M = 0.06 M
Since this doesn't reach a total of 0.1 M, this recipe won't work?

But then why r we given a pKa of 6.8?

[HaroTaro]

ahhhh~I think I finally got it.

Thanks for all the help~~~