Question

A pH electrode is attached to a voltmeter that reads 0.253V when the electrode is immersed in a buffer solution with a pH of 3.25 at 25 degree C. After the pH electrode is moved to an unknown buffer solution at 25 degree C, the reading of the voltmeter is lowered by 43%. From the Nernst potential equation, come up with an equation relating the measured voltage (delta psi = psi_sensor - psi_solution) to the pH of the solution. Calculate the pH of the unknown buffer.

Answer 1

a) we know that

Nernst's Equation is

Ecell = E^ocell - (RT/nF) lnQ

Q = reaction quotient

R= gas constant

T = temperature

n = number of electrons

F= Faraday's constant = 96485 C
The half reaction in the pH electrode is ionisation of hydrogen, so
ln Q =2.303 log₁₀ Q = 2.303 log₁₀ ( [1] / [H⁺] )

2.303 log₁₀ Q = - 2.303 log₁₀[H⁺] = 2.3026 P^H

as   P^H= - log₁₀[H⁺]

Now , E = ψ_sensor ; E^o = ψ_solution
   ψ_sensor = ψ_solution - 2.3026 (RT/ZF) P^H
   ψ_sensor - ψ_solution = Δψ
n =1 ; F = 96485 C mol^-1 , R = 8.314
     Δψ = - 2.303 (8314 X T / 1 X 96485) P^H = 1.9844 X 10^-4 T P^H
   b) For second part it is given that ψ_sensor = 0.253 V

Putting value in the above equation
ψ_solution = ψ_sensor+ 1.9844 X 10^-4 T P^H

T = 25⁰C = 25+ 283.15 = 298.15 K ; pH = 3.25
ψ_solution = 0.253 + 1.9844 X 10^-4 X 298.15 X 3.25

ψ_solution = 0.253 + 0.1923 = 0.4453 V

We have to consider the fact that the voltameter reading gets lowered by 43%
so the solution reading will be 100-43 % of 0.253 = 0.57 X 0.253 = 0.1442 V

ψ_solution = ψ_sensor+ 1.9844 X 10^-4 T P^H

ψ_solution - ψ_sensor = 1.9844 X 10^-4 T P^H
0.1442 - 0.4452 = -1.9844 X 10^-4 T P^H

on calculating

pH = 5.09