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How do i calculate the pH of a 500mL solution of 1M lysine....?
given that lysine has the following pKa values: pK(1)= 2.2; pK(2)= 9.2; pK(3)= 10.8
1 Answer
- Steve OLv 71 decade agoFavorite Answer
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
