How do i calculate the pH of a 500mL solution of 1M lysine....?

Question

given that lysine has the following pKa values: pK(1)= 2.2; pK(2)= 9.2; pK(3)= 10.8

Answer 1

you use only the first Ka

if the pKa is 2.2, then the Ka1 = 10 ^ -2.2  (aka  6.3e-3)

K = [H+] [lys-] / [Hlys]

6.3e-3 = [x] [x] / [1-x]

6.3e-3  [1-x] = [x] [x]

6.3e-3 - 6.3e-3 x  = x2

x2 + 6.3e-3x - 6.3e-3 = 0

http://www.1728.com/quadratc.htm

x = [H+] = 0.076285

pH = 1.1176

your answer is pH = 1.1

you may wish to have more sig figs in the pH, but the "1M" has only 1 sigfig