Enter the pH into the calculator to estimate the ideal electrode voltage (relative to 0 V at pH 7) at 25 °C. Actual pH-probe voltage depends on calibration, temperature, and the specific sensor/reference system.
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Voltage from pH Formula
The voltage output of an ideal pH electrode is described by the Nernst equation. The voltage depends on temperature and on how the electrode is referenced/calibrated. A common calibrated form assumes the meter is set to 0 V (0 mV) at pH 7.
E = E_7 – (2.303·R·T / (n·F)) · (pH – 7)
- E is the electrode potential (voltage)
- E7 is the electrode potential at pH 7 (often taken as 0 V after calibration)
- R is the gas constant
- T is the absolute temperature (K)
- n is the number of electrons transferred (for the H+ relationship, n = 1)
- F is the Faraday constant
At 25 °C (298.15 K) with n = 1, the ideal Nernst slope magnitude is about 0.05916 V per pH unit (59.16 mV/pH). With E7 = 0 V, the simplified relationship is: E(V) ≈ 0.05916 × (7 − pH).
| pH | Voltage (V) |
|---|---|
| 0 | 0.414 |
| 1 | 0.355 |
| 2 | 0.296 |
| 2.5 | 0.266 |
| 3 | 0.237 |
| 4 | 0.177 |
| 4.5 | 0.148 |
| 5 | 0.118 |
| 5.5 | 0.089 |
| 6 | 0.059 |
| 6.5 | 0.030 |
| 7 | 0.000 |
| 7.5 | -0.030 |
| 8 | -0.059 |
| 9 | -0.118 |
| 10 | -0.177 |
| 11 | -0.237 |
| 12 | -0.296 |
| 13 | -0.355 |
| 14 | -0.414 |
| *Calculator/table assumption: ideal Nernst slope at 25°C with E7 = 0 V, so E(V) ≈ 0.05916 × (7 − pH). Therefore, 1 pH unit corresponds to about 59.16 mV in magnitude (sign depends on whether pH is above or below 7). | |
How to Calculate Voltage from pH?
The following two example problems outline how to calculate the Voltage from pH (using the ideal 25 °C slope and assuming 0 V at pH 7).
Example Problem #1:
- First, determine the pH. In this example, the pH is measured to be 4.
- Finally, calculate the voltage using: E(V) = 0.05916 × (7 − pH).
E = 0.05916 × (7 − pH)
Inserting the values from above and solving the equation with the input values gives:
E = 0.05916 × (7 − 4) = 0.17748 (volts)
Example Problem #2:
Using the same process as example problem 1, we first define the variables outlined by the formula. In this case, the values are:
pH = 10
Entering these values into the formula above gives :
E = 0.05916 × (7 − 10) = −0.17748 (volts)
