pH=7 - millivolt current=0

For growing ph, remove roughly 57.14 mV per unit of pH (ph=8, millivolt current=-57.14)

For curbing ph, add roughly 57.14 mV per unit of pH

If you are lazy, use (for example):

http://www.traditionaloven.com/conversions_of_measures/ph-voltage.html

Thanks Michel ,

I have interpreted like this pH = 7 - mV/ 57.14 .

I am afraid there might be a misunderstanding... I give you some examples:

pH5 --> mV: 114.29

pH6 --> mV: 57.14

pH7 --> mV: 0

pH8 --> mV: -57.14

ph9 --> mV: -114.29

And so on...

No Michel there is no misunderstanding ,I understood what you meant .

In my formula ,I can also able to calculate fractional values .

Formula --- pH = 7 - mV/ 57.14 .

For example when mV value = 57.14 i will get pH value of 6 by substituting mV value in the formula i .e pH = 7 - (57.14/57.14) = 7-1 = 6 .

lly for -57.14 pH = 7 -(-57.14/57.14) = 8

for -114.29 pH = 7 -(-114.29/57.14) = 9

This can also help to get fractional values of pH.

for 64.2 mV the pH value will be pH = 7 -(64.2/57.14) = 5.88

Kindly let me know if i am wrong .

1. Choose two buffers, let say pH1 = 7.00 and pH2 = 4.00 and measure the corresponding E1 and E2 (in mV).

2. For a sample with pHx, you measure Ex.

3. Calculate pHx = pH1 + (Ex – E1)(pH2 – pH1)/(E2-E1)

Important notes:

- all buffers and samples must be measured at the same temperature

- use a stabilisation time of about 1 minute for the electrode response

- the instrument must have a very high impedance (500 Gohm, not used in common voltmeters; use a pH/mV-meter in "mV" mode to test the equation)

- use a 3rd known buffer as sample (e.g., pH= 9.00 or 10.00) to verify if the sensor response is linear).

- at pH=7.00 , the common measured E is not zero; in the given equation, (Ex – E1) is the correction for this asymmetry

- in the given equation, (pH2 – pH1)/(E2-E1) is the electrode slope.

Thank you Mr. Cornel and Mr.Michel