A solution prepared by mixing 55.0 mL of 0.260 M AgNO3 and 55.0 mL of 0.260 M Tl

Question

A solution prepared by mixing 55.0 mL of 0.260 M AgNO3 and 55.0 mL of 0.260 M TlNO3 was titrated with 0.520 M NaBr in a cell containing a silver indicator electrode and a reference electrode of constant potential 0.175 V. The reference electrode is attached to the positive terminal of the potentiometer, and the silver electrode is attached to the negative terminal. The solubility constant of TlBr is Ksp = 3.6 × 10–6 and the solubility constant of AgBr is Ksp = 5.0 × 10–13. (a) Which precipitate forms first?(b) Which of the following expressions shows how the cell potential, E, depends on [Ag ]?(c) Calculate the first and second equivalence points of the titration.What is the cell potential when the following volumes of 0.520 M NaBr have been added? d) 1.0 mL e) 17.7 mL f) 26.5 mL g) 27.4 mL h) 27.8 mL i) 45.2 mL j) 55.0 mL k) 62.8 mL

Explanation / Answer

Selective precipitation

a) Ksp of AgBr is lower than Ksp of TlBr. Therefore, AgBr is precipitated out of solution first.

b) Expression of cell potential dependence on [Ag]

E = [0.8 - 0.0592log(1/[Ag+])] - 0.175

c) First equivalence point

moles NaBr required = 0.26 M x 55 ml = 14.3 mmol

Volume NaBr needed = 14.3/0.520 = 27.5 ml

[Ag+] = sq.rt.(5 x 10^-13) = 7 x 10^-7 M

E = 0.8 + 0.0592log() - 0.175 = 0.99 V

Second equivalence point

moles NaBr required = 2 x 0.26 M x 55 ml = 28.6 mmol

Volume NaBr needed = 28.6/0.520 = 55 ml

[Ag+] = sq.rt.(3.6 x 10^-6) = 0.002 M

E = -0.34 + 0.0592log(0.002) - 0.175 = -0.015V

d) 1.0 ml

[Ag+] remained = (0.26 M x 55 ml - 0.52 M x 1.0 ml)/56 ml = 0.25 M

E = 0.8 + 0.0592(0.25) - 0.175 = 0.661 V

e) 17.7 ml

[Ag+] remained = (0.26 M x 55 ml - 0.52 M x 17.7 ml)/72.7 ml = 0.07 M

E = 0.8 + 0.0592(0.071) - 0.175 = 0.693 V

f) 26.5 ml

[Ag+] remained = (0.26 M x 55 ml - 0.52 M x 26.5 ml)/81.5 ml = 0.0064 M

E = 0.8 + 0.0592(0.0064) - 0.175 = 0.755 V

g) 27.8 ml

[Tl+] remained = (0.26 M x 55 ml - 0.52 M x 0.3 ml)/82.8 ml = 0.171 M

E = -0.34 + 0.0592(0.171) - 0.175 = -0.47 V

j) 55 ml

Second equivalence point

E = -0.34 + 0.0592log(0.002) - 0.175 = -0.015V

k) 62.8 ml

[Tl+] = Ksp/[Br-]

[Br-] excess = 0.52 M x 7.8 ml/117.8 ml = 0.034 M

[Tl+] = 3.6 x 10^-6/0.034 = 0.0001 M

E = -0.34 + 0.0592log(0.002) - 0.175 = -0.28 V

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