Question

Question 6

In an experiment 0.0500 M AgNO3 was used to titrate 20 mL of 0.12 M NaCl solution. When

silver electrode is used together with the pH electrode to measure potential at each point during

the titration, the potential measured at equivalence point is 0.055 V. Unfortunately, a student

doing the titration stopped at –0.055 V. Calculate the relative titration error (in %) of the

student’s result.

Answer 1

The potential of silver electrode measured is = 0.055V
E^o_cell = [0.0591 V / n ] * log K
Ag + e^- --------------------> Ag⁺¹ E^o_cell = 0.055 V
0.055V = [0.0591 V/ 1] * log [Ag⁺¹/Ag]
0.055V = [0.0591 V/1] * log [Ag⁺¹ / 0.0500]
0.930626 = log [Ag⁺¹] - log [0.0500]
[Ag⁺]_actual = 0.4261 M (concentration of product AgCl)
when the titration stopped at -0.055 V
-0.055 V =  [0.0591 V/1] * log [Ag⁺¹ / 0.0500]
[Ag⁺]_observed = 5.866025*10^-3M
Relative titration error = { [Ag⁺]_actual - [Ag⁺]_observed } / [Ag⁺]_actual
Relative titration error in % = { (0.4261 - 5.866*10^-3) / 0.4261 } *100 = 98.6233 %