(2.303RT*ΔpH)/nF
R=8.314
T=25C=298K
T=37C=310K
n=1
F=26472.44
For 25 degrees:
(2.303*8.314*298*2.49)/(1*26472.44)=.1532V
=153.2mV
The pH difference across the membrane of a glass electrode is 2.49. How much voltage is generated by the pH gradient at:
25C?
37C?
3 answers
The person who responded is correct with their method except Faraday's constant is 96485.33.
(2.303RT*ΔpH)/nF
R= 8.314
T= 25 + 273.15 = 298.15 K
T= 37 + 273.15 =310.15 K
n= 1
F= 96485.33
For 25 degrees:
(2.303*8.314*298.15*2.49)/(1*96485.33) = 0.147 V
0.147 * 1000 = 147 V
(2.303RT*ΔpH)/nF
R= 8.314
T= 25 + 273.15 = 298.15 K
T= 37 + 273.15 =310.15 K
n= 1
F= 96485.33
For 25 degrees:
(2.303*8.314*298.15*2.49)/(1*96485.33) = 0.147 V
0.147 * 1000 = 147 V
Oops, 0.147 V * 1000 = 147 mV