Question

The pH electrode responds according to the following Nernst equation à E (V) = 0.0592 pH + Q. If the concentration of [H⁺] increases 10 000-fold, how much will we expect the measured voltage to change?

Answer 1

Nernst equation

E (V) = 0.0592 pH + Q

Consider

[H+] = 100

pH = - log[H+] = - log (100) = - 2

E1 = 0.0592 (-2) + Q = - 0.1184 + Q

If

[H+] = 100 x 10^4 = 10^6

pH = - log[10^6] = - 6

[H+] increases then pH decreases

E2 = 0.0592 (-6) + Q = - 0.3552 + Q

[H+] increases then pH decreases, the voltage also decreases from E1 to E2.

For the Constant value of Q

Change in voltage = (E2 - E1)*100/E1

= [ ( - 0.3552) - (- 0.1184)] *100/ (- 0.1184)

= (-0.3552 + 0.1184)*100/(-0.1184)

= 200%

If the concentration of [H+] increases 10 000-fold, then 200% voltage drop we can expect.