A solution prepared by mixing 60.0 mL of 0.220 M AgNO3 and 60.0 mL of 0.220 M Tl

Question

A solution prepared by mixing 60.0 mL of 0.220 M AgNO3 and 60.0 mL of 0.220 M TlNO3 was titrated with 0.440 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.

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

(d) What is the cell potential when the following volumes of 0.440 M NaBr have been added? a. 1.0 ml b. 16.0 ml c. 29.0 ml d. 29.9 ml e. 30.3 ml f. 46.0 ml g. 60.0 ml h. 62.0 ml

Explanation / Answer

Titration

(a) Equivalence points

First equivalence point : all of AgBr gets precipitated

E = [(1.066 + 0.80)/2] - 0.175 = 0.758 V

Second equivalence point : all of TlBr gets precipitated

E = [(1.066 + 0.34)/2] - 0.175 = 0.528 V

(d) Cell potential when,

(a) 1 ml NaBr added

moles of AgNO3 = 0.22 M x 60 ml = 13.2 mmol

moles of NaBr = 0.44 M x 1 ml = 0.44 mmol

Excess [Ag+] remain = 12.76 mmol/61 ml = 0.21 M

E = [Eo - 0.0592 log(Ag+)] - 0.175

    = [0.8 - 0.0592 log(0.21)] - 0.175 = 0.665 V

(b) 16 ml NaBr added

moles of AgNO3 = 0.22 M x 60 ml = 13.2 mmol

moles of NaBr = 0.44 M x 16 ml = 7.04 mmol

Excess [Ag+] remain = 6.16 mmol/76 ml = 0.81 M

E = [Eo - 0.0592 log(Ag+)] - 0.175

    = [0.8 - 0.0592 log(0.81)] - 0.175 = 0.690 V

(c) 29 ml NaBr added

moles of AgNO3 = 0.22 M x 60 ml = 13.2 mmol

moles of NaBr = 0.44 M x 29 ml = 12.76 mmol

Excess [Ag+] remain = 0.44 mmol/89 ml = 0.005 M

E = [Eo - 0.0592 log(Ag+)] - 0.175

    = [0.8 - 0.0592 log(0.005)] - 0.175 = 0.761 V

(d) 29.9 ml NaBr added

moles of AgNO3 = 0.22 M x 60 ml = 13.2 mmol

moles of NaBr = 0.44 M x 29.9 ml = 13.156 mmol

Excess [Ag+] remain = 0.044 mmol/89.9 ml = 0.0005 M

E = [Eo - 0.0592 log(Ag+)] - 0.175

    = [0.8 - 0.0592 log(0.0005)] - 0.175 = 0.821 V

(e) 30.3 ml NaBr added

moles of AgNO3 = 0.22 M x 60 ml = 13.2 mmol

moles of NaBr = 0.44 M x 30.3 ml = 13.332 mmol

Excess [Br-] remain = 0.132 mmol

moles of TlNO3 = 0.22 M x 60 ml = 13.2 mmol

Excess [Tl+] remain = 13.068 mmol/90.3 ml = 0.145 M

E = [Eo - 0.0592 log(Tl+)] - 0.175

    = [-0.34 - 0.0592 log(0.145)] - 0.175 = -0.465 V

(f) 46 ml NaBr added

moles of AgNO3 = 0.22 M x 60 ml = 13.2 mmol

moles of NaBr = 0.44 M x 46 ml = 20.24 mmol

Excess [Br-] remain = 7.04 mmol

moles of TlNO3 = 0.22 M x 60 ml = 13.2 mmol

Excess [Tl+] remain = 6.16 mmol/106 mmol = 0.06 M

E = [Eo - 0.0592 log(Tl+)] - 0.175

    = [-0.34 - 0.0592 log(0.06)] - 0.175 = -0.442 V

(g) 60 ml NaBr added

moles of AgNO3 = 0.22 M x 60 ml = 13.2 mmol

moles of NaBr = 0.44 M x 60 ml = 26.4 mmol

Excess [Br-] remain = 13.2 mmol

moles of TlNO3 = 0.22 M x 60 ml = 13.2 mmol

Equivalence point

E = (-0.34 + 1.066)/2 - 0.175 = 0.188 V

(h) 62 ml NaBr added

moles of AgNO3 = 0.22 M x 60 ml = 13.2 mmol

moles of NaBr = 0.44 M x 62.0 ml = 27.28 mmol

Excess [Br-] remain = 14.08 mmol

moles of TlNO3 = 0.22 M x 60 ml = 13.2 mmol

excess [Br-] remain = 0.88 mmol122 ml = 0.0072 M

E = [-0.34 - 0.0592 log(3.6 x 10^-6/0.0072)] - 0.175 = 0.020 V

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