Question

a solution prepared by mixing 60ml of .1m agno3 and 60ml 0f .1m tino3 was titrated with .2M nabr in a cell containing a silver indicator electrode and a reference electrode of constant potential .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 tibr is ksp=3.6x10^-6 and the solubility constant of agbr is ksp=5x10^-13.

AgBr precipitates first

the following expression shows how the cell potential E depends on {ag+]      E=.175-[.799-.05916log(1/[ag+])]

1. Calculate the first and second equivalence points of titration.

2. what is the cell potential when the following volumes of .2M nabr have been added: a)1ml, b)16.8ml, c)29ml, d)29.9ml, e)30.3ml, f)46.8ml, g)60ml, h)63.6ml

Answer 1

1) For first question the first equivalence point at 30 ml of NaBr an second equivalance point at 60 ml of NaBr. Mixture of solution contains 60 ml of 0.1 M AgNO3 an 60 ml of 0.1 M TiNO3 an this solution is titrated with 0.2 M NaBr. i.e, 120 ml of 0.05 M AgNO3 requires 30 ml of 0.2 M NaBr ( since stoichiometry is 1:1, 6mMol AgNO3 requires 6 mMol NaBr) simillarly for TiNO3.

2) Cell potential value for 1 ml of 0.2 M NaBr:

           AgNO3 + Ti NO3 + 2 NaBr -----------> AgBr + TiBr

intial   6mMol      6m Mol    0                        0       0

rxn     -0.2               0         0.2                     0.2       0

final    5.8    6    0.2                    0.2          0

[Ag+] = 5.8/ (120 +1) = 0.048 M

from given nernst equation, E = .175-(0.799-0.05916 log (1/[Ag+])) = -0.546 V

simillarly

for 16.8 ml,

iniital    6     6      0          0          0

rxn      3.36   0    3.36    3.36       0

final   2.64    0    3.36     3.36      0

[Ag+] = 0.019 M,   E = -0.522 V

for 29 ml

initial 6    6      0    0    0

rxn 5.8   0       5.8 5.8 0

final 0.2   0      5.8 5.8 0

[Ag+] = 0.0013M

E= -0.453 V

for 29.9 ml

[Ag+] = 0.000133 M , E = -0.395 V

for 30.3 ml

initila 6 6    0    0   0 0

rxn   6   0.6 6.06   6 0.6

final 0   5.4      6.06   6 0.6

[Ag+] = 0.0403 M

E = 0.175-(0.799-0.05916 log (1/0.0403))) = -0.54 V

simillary the remaining values are also rises...