Question

Question:The measurement of pH using a glass electrode obeys

the Nernst equation. The typical response of a pH meter

at 25.008C is given by the equation %meas 5 %ref 1 0.05916 pH

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 %ref 5 0.250 V and that %meas 5 0.480 V.

a. What is the uncertainty in the values of pH and [H1]

if the uncertainty in the measured potential is 61 mV

(60.001 V)?

b. To what accuracy must the potential be measured

for the uncertainty in pH to be 60.02 pH unit?

Short answer

Answer:

1. Uncertainty in pH and H⁺ are10^-4
2. Potential must be measured to the accuracy of

Step-by-step solution

calculating the uncertainty in values of pH

Determining the value of pH

E_meas= E_ref + 0.05916 pH

0.480 V = 0.250 V + 0.05916 pH

pH= 3.888

max + min Ph,

Hence, the uncertainty in values of pH will be +0.017

determining the uncertainty in values of H+

Determining the value of H⁺,

[H⁺] = 10^-pH

[H⁺] = 1.29 × 10^-4

Max+ min [H⁺],

[H⁺]_max = 1.24 × 10^-4m

[H⁺]_min = 1.35 × 10^-4m

[H⁺] uncertainty =10^-4p

Hence, the uncertainty in values of [H⁺] will be10^-4p

finding out the extent of accuracy necessary for uncertainty in pH to be 60.02 Ph

0.480 V = 0.250 V + 0.05916 pH

pH= 3.89 V

$$\begin{gathered} V=\pm \frac{0.02}{3.89} \\ V\pm 0.005 \end{gathered}$$

Hence potential must be measured to the accuracy of 0.005