A solution prepared by mixing 40.0 mL of 0.300 M AgNO3 and 40.0 mL of 0.300 M Tl

Question

A solution prepared by mixing 40.0 mL of 0.300 M AgNO3 and 40.0 mL of 0.300 M TlNO3 was titrated with 0.600 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?

(c) Calculate the first and second equivalence points of the titration.

(d) What is the cell potential when the following volumes of 0.600 M NaBr have been added? at

1mL

20.3 mL

10.7 mL

30.7 mL

19 mL

40 mL

19.9 mL

41 mL

Explanation / Answer

Titration

(a) Ksp of AgBr is much lower than the Ksp of TlBr, therefore,

the precipitate to form first is,

AgBr

(b) first equivalence point

all of AgBr precipitated

[Ag+] = [Br-]

[Ag+] = sq.rt.(Ksp) = sq.rt.(5 x 10^-13) = 7.1 x 10^-7 M

Ecell = 0.175 - ([0.799 + 0.05916 log[Ag+])

         = 0.175 - ([0.799 + 0.05916 log(7.1 x 10^-7))

         = -0.260 V

Second equivalence point

all of TlBr precipitates

[Br-] = ksp(3.6 x 10^-6) = 1.9 x 10^-3 M

[Ag+] = 5 x 10^-13/1.9 x 10^-3 = 2.63 x 10^-10 M

Ecell = 0.175 - ([0.799 + 0.05916 log[Ag+])

         = 0.175 - ([0.799 + 0.05916 log(2.63 x 10^-10))

         = -0.057 V

(d) after volume of NaBr added

1.0 ml

[Ag+] remained = (0.3 M x 40 ml - 0.6 M x 1 ml)/81 ml = 0.141 M

Ecell = 0.175 - ([0.799 + 0.05916 log[Ag+])

         = 0.175 - ([0.799 + 0.05916 log(0.141))

         = -0.574 V

20.3 ml

this is past first equivalence point

[Tl+] remained = (0.3 M x 40 ml - 0.6 M x 0.3 ml)/100.3 ml = 0.12 M

[Ag+] = 5 x 10^-13/0.12 = 4.24 x 10^-12 M

Ecell = 0.175 - ([0.799 + 0.05916 log[Ag+])

         = 0.175 - ([0.799 + 0.05916 log(4.24 x 10^-12))

         = 0.049 V

30.7 ml

this is past first equivalence point

[Tl+] remained = (0.3 M x 40 ml - 0.6 M x 10.7 ml)/110.7 ml = 0.05 M

[Ag+] = 5 x 10^-13/0.05 = 9.92 x 10^-12 M

Ecell = 0.175 - ([0.799 + 0.05916 log[Ag+])

         = 0.175 - ([0.799 + 0.05916 log(9.92 x 10^-12))

         = 0.027 V

19 ml

[Ag+] remained = (0.3 M x 40 ml - 0.6 M x 19 ml)/99 ml = 0.0061 M

Ecell = 0.175 - ([0.799 + 0.05916 log[Ag+])

         = 0.175 - ([0.799 + 0.05916 log(0.0061))

         = -0.493 V

40 ml

this is second equivalence point

Ecell = -0.057 V

19.9 ml

[Ag+] remained = (0.3 M x 40 ml - 0.6 M x 19.9 ml)/99.9 ml = 0.0006 M

Ecell = 0.175 - ([0.799 + 0.05916 log[Ag+])

         = 0.175 - ([0.799 + 0.05916 log(0.0006))

         = -0.433 V

41 ml added

excess [Br-] = 0.6 M x 1 ml/121 ml = 0.005 M

[Ag+] = 5 x 10^-13/0.005 = 1.01 x 10^-10 M

Ecell = 0.175 - ([0.799 + 0.05916 log[Ag+])

         = 0.175 - ([0.799 + 0.05916 log(1 x 10^-10))

         = -0.033 V

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