A+B=2 moles
Based upon your calculations, 2*0.316=0.632 moles of A
2-0.632=1.37 moles of B
pH=pka+log[moles of B/moles of A]=6.0+log[moles of B/moles of A]
****Check my thinking. Not sure about the problem since you have to take into account the isolectric points of the amino acids
You have a solution containing 1 mol each of two unknown amino acids in a mixture at pH6.5 (pKa=6.0). What would be the pH if you added 0.2 mol of NaOH?
I've used the Henderson-Hasselbalch to get this:
6.5 = 6.0 + log [B/A]
10^.5 = [B/A]
3.16/1 = [B/A]
What do I do next?
1 answer