A solution prepared by mixing 43.2 mL of 0.140 M AgNO3 and 43.2 mL of 0.140 M Tl

Question

A solution prepared by mixing 43.2 mL of 0.140 M AgNO3 and 43.2 mL of 0.140 M TlNO3 was titrated with 0.280 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) Write an expression that shows how how cell voltage depends on [Ag+]

b) what is the cell voltage when the following volumes of 0.280M NaBr have been added? 1.0mL, 15.6mL, 20.6mL, 21.5mL, 21.9mL, 37.2mL, 43.2mL, 52.8mL

Explanation / Answer

ANSWER:

(a) Ecell = E ca - Ean

E ca = Electrode potential of reduction half cell and Ean = Electrode potential of oxidation half cell

Or Ecell = EAg - Eref --------------- (1)

E Ag = EAgo + 0.0591log [Ag+] --------------- (2)

substituting (2) in (1)

Ecell = EAgo + 0.0591log [Ag+] - Eref ((Required expresion.)

(b) Step (I) Calcultion of molarity:

Moles of AgNO3 = Volume (in L) X Molarity = 0.043.2 X 0.140 = 0.006 Moles   (43.2mL = 0.043L)

Moles of TlNO3 = 0.043.2 X 0.140 = 0.006 Moles

Total volume = 0.0432 + 0.0432 = 0.0864L

Molarity of AgNO3 = No. Of moles /Volume = 0.0.006/0.086 = 0.07M

Molarity of TlNO3 = No. Of moles /Volume = 0.006/0.086 = 0.07M

Concentration of different ions in solution:

Ag+ = 0.07 moles (From AgNO3)

Tl+ = 0.07moles (From AgNO3)

NO3- = 2 X 0.07 = 0.14 moles (FromAgNO3 + TlNO3)

Step (ii) Write balance rections

AgNO3 and TlNO3 react with NaBr

AgNO3 + NaBr --------------> AgBr + Na NO3

TlNO3 + NaBr --------------> TlBr + NaNO3

Step (iii) Calculation of Ag+ concentration on addition of different volumes of NaBr

(a) when 1mL of 0.28M NaBr is added.

moles of Br- added = 0.28 X 0.001 = 0.0.00028

therefore Concentration of Ag+ = 0.07 - 0.00028 = 0.069 Moles

Correction of Ag+ Concentration: There will be some addition of Ag+ concentration from AgBr and TlBr Given as

Ksp (AgBr) = Ag + + Br-

(NOTE: TlBr is also sparingly soluble hence will not affect concentration of Ag+ by commonion effect.)

5.0 × 10–13 = x X x

Or x = 7.0X 10-7

total Ag+ concentration = 0.069 + 7.0X 10-7 = 7.069 X10-7

(iv) Calculate Ecell

Ecell = EAgo + 0.0591log [Ag+] - Eref

Ecell = 0.799 + 0.0591log [7.069 X10-7 ] - Eref ---------------(3)

(Note reference electrode has not been mentioned by student. lets take as an example, calomel as reference electrode. Then Eref = 0.2422)

Simillarly Ecell can be measured for other volumes of NaBr as well folloeing the same procedure. There must be molarity corrections with increased volume as shown below. Rest is same as above.  

(b) when 52.8 mL of NaBr was added.

moles of NaBr = 0.052 X 0.28 = 0.014

total volume = 0.086 + 0.0528 = 0.139L

molarity (NaBr) = 0.014/0.139 = 0.1M

AgNO3 = 0.006/0.139 = 0.04M

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.