Question

The y-carboxyl of Glu has a pKa of 4.3.

a) What percentage of these groups will be unprotonated (i.e. –COO- rather than –COOH) in a solution of Glu at pH=7.0

Answer 1

Apply buffer equation ( HEnderson hasselbach equations)

pH = pKa + log(COO-/COOH)

7.0 = 4.3 + log(COO-/COOH)

(COO-/COOH) = 10^(7-4.3) = 501.1

(COO-/COOH) = 501.1

total:

COO- + COOH = 100

(COO-/COOH) = 501.1

COO- = 501.1* COOH

substitute:

COO- + COOH = 100

501.1* COOH + COOH = 100

COOH = 100/(501.1+1) = 0.199

(COO-/COOH) = 501.1

(COO-/0.199 ) = 501.1

COO- = 501.5*0.199 = 99.79

Now...

% COO- = COO- / total * 100% =(99.79) / (99.79+0.199 ) * 100% = 99.79 % is COO-