Water Module: Session 8, Potentiometric Titrations

How does the [Cl^-] change and thus the potential in the titration of Cl^- with AgNO₃?

K_sp = 1.8 x 10^-10

E = E˚ - 0.059/n{log (prod. prod/prod react)}

2Ag⁺ + Cu à 2Ag + Cu⁺²

E = .46 – 0.059/2{log([Cu⁺²]/[Ag⁺]²)}

E = .46 – 0.059/2{log(0.1M/[Ag⁺]²)}

E = .46 + 0.059/2{log([Ag⁺]²/0.1M)}

E = .46 + 0.059/2{log([Ag⁺]²) – log(0.1M)}

E = .46 + 0.059/2{log[Ag⁺]²} – 0.059/2{log(0.1M)}

E = .46 + 0.059/2{log[Ag⁺]²} + 0.0295

E = .46 + 0.059(log[Ag⁺]) + 0.0295

E = .49 + 0.059(log[Ag⁺])

K_sp = [Ag⁺] [Cl^-] so [Ag⁺] = K_sp/[Cl^-]

E = .49 + 0.059{log(K_sp/[Cl^-])}

E = .49 + 0.059(log K_sp) – 0.059 (log[Cl^-])

E = 0.49 - 0.57 + 0.059pCl

E = -0.08 + 0.059pCl

Titrate 100.0 ml of 0.010M Cl^- with 0.10M AgNO₃

Before titration:

[Cl^-] = 0.010M so pCl = 2.00 and E = 38 mV

After the addition of 5.00 ml of AgNO₃

Ag⁺ + Cl^- à AgCl

.5 1.0

-.5 -.5 +.5

0 .5 .5 pCl = -log (.5/105) = 2.32 so E = 57 mV

After the addition of 10.00 ml of AgNO₃

Ag⁺ + Cl^- à AgCl

1.0 1.0

-1.0 -1.0 +1.0

0 0 1.0 Ksp = 1.8 x 10^-10 = [Ag⁺] [Cl^-] = x²

x = 1.34 x 10^{-5 so} pCl = 4.87 and E = 207 mV

After the addition of 15.00 ml AgNO₃

Ag⁺ + Cl^- à AgCl

1.5 1.0

-1.0 -1.0 +1.0

.5 0 1.0 Ksp = 1.8 x 10^-10 = [Ag⁺] [Cl^-] = .5/115 [Cl^-]

[Cl^-] = 4.14 x 10^-8

so pCl = 7. 38 and E = 356 mV

Or use E = .49 + 0.059(log[Ag⁺])

Potentiometric Titration of CO₃^-2 and HCO₃^-

Very similar to unknown acid titration

Caution!!

In the titration, CO₃^-2 is converted to HCO₃^- so you need to subtract the volume to neutralize CO₃^-2 from the volume needed to titrate HCO₃^- to calculate the amount of HCO₃^- in the sample.