pH = -log(H^+) = approxmately 1E-6
Solve for (H^+).
..........HX ==> H^+ + X^-
initial...0.1.....0......0
change...-x.......x......x
equil...0.1-x.....x.......x
Ka = (H^+)(X^-)/(HX)
Then dGo = -RTlnK
Consider a weak acid HX. If a 0.10 M solution of HX has a pH of 5.87 at 25°C, what is ΔG° for the acid's dissociation reaction at 25°C?
2 answers
t=35+273= 298k
HX ==> H^+ + X^-
I. 0.1. 0. 0
C -x. +x. +x
E. 0.1-x. +x. +x
[H^+]= 10^-ph= 10^-5.83 = 1.479x10^-6
x=[H^+]=[X^-]
[HX]=0.1-x= 0.099
k= 91.479x10^-6)^-2 / 0.099= 2.18x 10^-11
DG0= -RTln(k)
= -8.32x298xlm( 2.18x 10^-11)= 61kj
HX ==> H^+ + X^-
I. 0.1. 0. 0
C -x. +x. +x
E. 0.1-x. +x. +x
[H^+]= 10^-ph= 10^-5.83 = 1.479x10^-6
x=[H^+]=[X^-]
[HX]=0.1-x= 0.099
k= 91.479x10^-6)^-2 / 0.099= 2.18x 10^-11
DG0= -RTln(k)
= -8.32x298xlm( 2.18x 10^-11)= 61kj