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The measurement of pH using a glass electrode obeys the Nernst equation. The typical response of a pH meter at 25.00 degrees C is given by the equation:
E(measured)= E(ref) +.591pH

where ref contains the potential of the reference electrode and all other potentials that arise in the cell that are not related to the hydrogen ion concentration. Assume that E ref = 0.250 V and that E meas = 0.480V

What is the uncertainty in the values of pH and [H+] if the uncertainty in the measured potential is 1 mV (0.001 V)? pH=+/- ___ H+ = +/-_____

To what accuracy must the potential be measured for the uncertainty in pH to be 0.02 pH units?

___Potential

Please! help me its so confusing i m not sure how to do this at all. i tried the delta G equations but it doesnt make sense

I would use the 0.480 and 0.250 given in the problem and solve for pH. See what that value is. Then change 0.480 by +0.001 and recalculate the pH. Do the same when changing by -0.001. That will give you the change in pH when changing by +/- 0.001 v.

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