when PkaEPH + Pka Ep = 14
so, Pka EPH = 14 - Pka EP
∴ Pka EPH = 14 - 3.86 = 10.14
and when Pka EPH = -㏒ Ka(EPH)
∴ Ka(EPH) = 10^-10.14
= 7.24 * 10^-11
by using the ICE table :
EPH + H2O ↔ H3O+ + EP
initial 0.0131 0 0
change -X +X +X
Equ (0.0131-X) X X
when Ka = [H3O+][EP]/[EPH]
by substitution:
7.24 x 10^-11 = X^2 / (0.0131-X) by solving for X
∴X = 9.74 x 10^-7
∴[H3O+] = X = 9.74 X 10^-7
∴PH = -㏒[H3O+]
= -㏒ (9.74 X 10^-7)
= 6
