Reading (in volts) = 0.453 - (2.303) RT pH / nF
where R = 8.314 volt coulomb/K/mole
T = temperature of the solution in degrees Kelvin
pH = pH of the solution
n = number of electrons involved = 1
F = 96,487 coulombs
What reading would I expect from this pH 4.0 solution if it was
a) 19oC
b) 26oC
2007-01-12 08:58:21 · 3 answers · asked by elizangela r 2 in Science & Mathematics ➔ Mathematics
T = 273 + oC, so
a) V = 0.453-2.302. 8.314. (273+19).4 /96,487 = 0.221219 volts
b) V = 0.453-2.302. 8.314. (273+26).4 /96,487 = 0.215663 volts
2007-01-12 09:12:36 · answer #1 · answered by Jano 5 · 0⤊ 0⤋
Reading (in volts) = 0.453 - (2.303) RT pH / nF
= (.453-2.303*8.314*T*pH)/1*96487
= -1.9375*T*pH
So when t = 19C = 292.15K, we get
Reading = -1.9375*4*292.15 = -2264.163 volts
when T = 26C=299.15K we get
Reading = -1.9375*4*299.15 = -2318.413 volts
I was not sure about units. I could not get mole to cancel out.
My answer seems to be in volt moles times whatever the units of pH is (moles/liter?.)
2007-01-12 17:43:21 · answer #2 · answered by ironduke8159 7 · 0⤊ 0⤋
Hey, that answer is good enough for me.
2007-01-12 17:15:12 · answer #3 · answered by ragnar1002000 2 · 0⤊ 0⤋