In: Chemistry
if you have 100 ml of HA with a pKa of 2.91, and 30.00 mL of 0.200M NaOH was needed to reach the equivalence point at pH 7.7. What was the initial pH right before you started the titration?
concentration of acid :
at equivalence point
millmoles of acid = millmoles of base
100 x C = 30 x 0.2
C = 0.06 M
molarity of acid = 0.06 M
Ka = 10^-pKa
Ka = 1.23 x 10^-3
HA ------------------------> H+ + A-
0.06 0 0 -------------> initial
0.06-x x x ------------- > equilibrium
Ka =[H+][A-]/[HA]
1.23 x 10^-3 = x^2 / 0.06 -x
x^2 + 1.23 x 10^-3 x - 7.38 x 10^-5 = 0
x = 8.0 x 10^-3
[H+] = 8.0 x 10^-3 M
pH = -log [H+]
pH = -log (8 x 10^-3)
pH = 2.10 -----------------------------------> before start titration