Question

You make 100ml 0.10M solution of histidine at pH of 5.4 You then add 80ml of 0.10M HCl. What is the resulting pH? Assume the pKa of histidine to be 6.0. The answer we were given was about 1.8-1.9 and I cannot get that answer

Answer 1

Note that histidine has 2 pKa values for acids, and 1 for a base:

pKa1 = 1.82 pKa2 = 6.00 and finally, pKa3 (the base) = 9.17

Note that initially:

mmol of buffer = 100*0.1 = 10 mmol of histidine (acidic1/acidic2)

pH can be modeld with buffer equation

pH = pKa + log(Acid2/Acid1)

pKa for acid 2 --> 6.00

5.40 = 6.00 + log(Acid2/Acid1)

find ratio

acid2/acid1 = 10^(5.4-6.0) = 0.25118

acid2 = 0.25118*acid1

and we know

acid2 + acid1= 10

0.25118*acid1 + acid1 = 10

acid1 = (10)/(1+0.25118) = 7.992;

acid2 = 10-7.992 = 2.008

Now...

after adding

mmol of HCl = MV = 80*0.1 = 8 mmol

COO- group + H+ = COOH

note that all the acid 2 group (COO-) will become COOH

then, H+ left = 8-2 = 6 mmol

the new pH must be then analysed from the acid 1 pKa1

pH = pKa1 + log(COO-/COOH)

pH = 1.82+ log(7/6) = 1.88

pH = 1.88 approx

as expected (1.8-1.90)