Michael S. answered • 13d
B.S. in Chemistry, Indiana University; Organic Chem Teaching Intern
The pKa values did not come through with your post. Pyridoxal-5-phosphate is tetraprotic, and the values tabulated in Harris are:
pK1 = 1.4 pK2 = 3.44 pK3 = 6.01 pK4 = 8.45
Substitute your own book's numbers if they differ - the method below is what matters, and the arithmetic is quick to redo.
Call the fully protonated form H4A. The five species are H4A, H3A-, H2A2-, HA3-, and A4-.
The general formula
Every fraction is one term over the same denominator D:
D = [H+]^4 + K1[H+]^3 + K1K2[H+]^2 + K1K2K3[H+] + K1K2K3K4
α(H4A) = [H+]^4 / D, α(H3A-) = K1[H+]^3 / D, and so on down the line, with α(A4-) = K1K2K3K4 / D.
Plug in pH 7.6
[H+] = 10^-7.6 = 2.512 x 10^-8
K1 = 3.98 x 10^-2, K2 = 3.63 x 10^-4, K3 = 9.77 x 10^-7, K4 = 3.55 x 10^-9
The five numerators:
[H+]^4 = 3.98 x 10^-31
K1[H+]^3 = 6.31 x 10^-25
K1K2[H+]^2 = 9.12 x 10^-21
K1K2K3[H+] = 3.548 x 10^-19
K1K2K3K4 = 5.01 x 10^-20
D = 4.141 x 10^-19
The answer
α(H4A) = 1 x 10^-12 (negligible)
α(H3A-) = 1.5 x 10^-6 (negligible)
α(H2A2-) = 0.022, about 2.2%
α(HA3-) = 0.857, about 85.7%
α(A4-) = 0.121, about 12.1%
They sum to 1.00, which is the check you should always run.
A faster way to see it, and to catch errors
pH 7.6 sits between pK3 (6.01) and pK4 (8.45), so HA3- has to be the dominant form - and it is. The two neighbors follow directly from Henderson-Hasselbalch:
α(A4-)/α(HA3-) = 10^(pH - pK4) = 10^-0.85 = 0.141
α(H2A2-)/α(HA3-) = 10^(pK3 - pH) = 10^-1.59 = 0.0257
Check against the numbers above: 0.121/0.857 = 0.141 and 0.022/0.857 = 0.0257. Both match, which confirms the big calculation.
The forms more than two protonation steps away are always negligible in a case like this. That is worth internalizing: at any given pH, a polyprotic acid is essentially a two- or three-species mixture, and the rest of the ladder can be ignored.
