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.

To find the uncertainty in the values of pH and [H+], we need to calculate the uncertainties using the given equation and information.

1. Uncertainty in pH:
The equation relating the measured potential to the pH is given as follows:
E(measured) = E(ref) + 0.591pH

Let ΔE_meas be the uncertainty in the measured potential, which is 0.001 V. As we want to find the uncertainty in pH, we need to differentiate the equation with respect to pH and multiply it by the uncertainty in the measured potential:

ΔpH = (ΔE_meas / 0.591)

Substituting the given ΔE_meas = 0.001 V into the equation:
ΔpH = (0.001 V / 0.591) ≈ 0.0017 pH units

Therefore, the uncertainty in pH is approximately 0.0017 pH units.

2. Uncertainty in [H+]:
[H+] is related to pH using the equation:
[H+] = 10^(-pH)

To find the uncertainty in [H+], we multiply the uncertainty in pH by the derivative of the equation with respect to pH:

Δ[H+] = ln(10) * [H+] * ΔpH

Substituting the given ΔpH = 0.0017 pH units:
Δ[H+] = ln(10) * 10^(-pH) * 0.0017

Since we don't know the specific value of pH, we cannot calculate the exact uncertainty in [H+] without knowing the pH value.

3. Accuracy in Potential:
To determine the accuracy required in the measured potential for the uncertainty in pH to be 0.02 pH units, we need to rearrange the equation for ΔpH:

ΔpH = (ΔE_meas / 0.591)

Substituting ΔpH = 0.02 pH units:
0.02 = (ΔE_meas / 0.591)

Solving for ΔE_meas:
ΔE_meas = 0.02 * 0.591 ≈ 0.01182 V

Therefore, the potential needs to be measured with an accuracy of approximately 0.01182 V for the uncertainty in pH to be 0.02 pH units.