[tjpj77]

Calculate the amount of glycine and HCl required to prepare a 1L solution of 0.1M buffer at a pH of 2.0. The pKa of the glycine carboxyl group is 2.4 and the pKa of the amine group is 9.6. The amu of glycine is 75.

Using pH = pKa + log ([A-]/[HA]), I get 2.0 = 6.0 + log ([A-]/[HA]). I get 6.0 because that is the average of the pKa values of glycine. But we can't have the log of a negative number, so I'm totally stuck here... 

Please *delete me*

Thanks...

[Borek]

As I told you you can't use average pKa. Besides, there is no negative number here, so I have no idea where is the problem.

[tjpj77]

so I use 2.4 = 2.0 + log ([A-]/[HA]
then 0.4 = log ([A-]/[HA]

but where is the buffer here? i thought a buffer involved a weak acid or base, not HCl...

[Borek]

HCl protonates glycine yielding a weak acid.

[tjpj77]

ok, then (0.1 mol/L)(75 g Gly/mol Gly) = 7.5 g Gly/L in the solution
but what about the HCl?
I think I"m still confused about what my A- is and what my HA is...

[Borek]

Forget about glycine for a moment.

What happens when ammonia reacts with hydrochloric acid? Write reaction equation.

[tjpj77]

ok, i have NH3 + HCl --> +NH4 + Cl-
so, i think that because of this analogy, that:
0.4 = log [Cl-]/[Gly+]
we know I have 7.5 g of Gly, so is that the right track?

[Borek]

No, Cl^- doesn't matter here, it is just a spectator. Write net ionic equation for the ammonia protonation. Use definition of the Bronsted acid/base to show the acid and the base. Then apply the same logic to glycine.

[tjpj77]

i get 0.4 = log [gly-]/[gly+]
but then i'm stuck. so stuck.

[Borek]

Why Gly^-/Gly⁺?

Write the net ionic reaction equation for the protonation process.

[bbassa9823]

Here is the calculation:
pH = pKa + Log [Salt]/[Acid]
2.0 = 2.4 + Log [Salt]/[Acid]
- 0.4 =  Log [Salt]/[Acid]
10 ^ (-0.4) = [Salt]/[Acid]
0.398 = [Salt]/[Acid]    Note:[Salt]/[Acid]  represents the ratio of salt and acid conc.               
Or
0.398/1 = [Salt]/[Acid]
Total moles = 0.1 * 1     1 = 1 liter volume
moles of salt = 0.1 * 0.398/1.398 = 0.028
moles of acid = 0.1 * 1/1.398 = 0. 0715 (or 0.1 - 0.028 = 0.072)