[amorite0428]

Calculate the percentage of malic acid molecules, malate 1- ions and malate 2- ions in pH 5.1 buffer. 1st pKa is 3.5 and 2nd pKa is 5.1. The answer is 1.225 % unionised molecules, 50  % malate 1- ions and 48.775 % malate 2- ions. I couldnt get the 1.225 % and shouldn't the percentage of 1- and 2- ions be the same as the 2nd pKa is the same as the pH of buffer which means half of the malate 1- ions dissociate to form 2- ions?

[Borek]

Definitely at pH=pKa2 [A^2-]=[HA^-].

You have a set of three equations in three unknowns - two HH equations for two pKa values and sum of fractions of all forms equals 1. Doesn't have to be easy to solve (you will probably end with a 3rd degree polynomial), but 1.2% looks more or less OK - at least after checking with the pH calculator.

[amorite0428]

So the percentage at of HA- and A- is the same ?

[amorite0428]

And I would like to know how to get 1.225 % as well... Is it by using this formula , pH-pKa= log ( HA-/H2A)

[Borek]

And I would like to know how to get 1.225 % as well...

I have explained you how to solve the problem - have you tried?

[Big-Daddy]

I don't think cubic equations should be needed. You can just work out [HA^-]/[H₂A] from K_a1/[H+], and [A^2-]/[H₂A] from K_a1*K_a2/[H⁺]² or from (as Borek noted) [A^2-]/[HA^-]=K_a2/[H⁺]. Now if f[H2A]+f[A^2-]+f[HA^-]=1 then f[H2A]+f[H2A]*ratio+f[H2A]*ratio=1 so f[H2A]=1/(1+2*ratio) to find the fraction of ions which are H₂A (the same ratios apply to fractions of concentrations as they do to concentrations themselves). As Borek said the answer is 1.24%. I don't think I have made any approximations concentration-wise ...

[Borek]

Yeah, I just realized it is trivial, see equations 9.11-9.13 here: http://www.chembuddy.com/?left=pH-calculation&right=pH-polyprotic-acid-base

[amorite0428]

Yup, I can get 1.2 something... so since Ka2= pH, just to confirm my basic idea, the % of HA- AND A2- should be the same right?

[Borek]

so since Ka2= pH, just to confirm my basic idea, the % of HA- AND A2- should be the same right?

Have you read my first post? This problem was already addressed.

And it is not Ka2=pH, but pKa2=pH.

[Big-Daddy]

Yup, I can get 1.2 something... so since Ka2= pH, just to confirm my basic idea, the % of HA- AND A2- should be the same right?

No point in memorizing these ideas rather than understanding the calculation that produces them. Rearrange the expression for Ka2 and you get what? K_a2/[H+] = [A^2-]/[HA^-]. Now what happens if K_a2 = [H+]?