Question

When you add 2.0ml of 0.50 M S2O32- to the 2.0 ml of 0.10M silver nitrate, formation of Ag(S2O3)23- occured by equation Ag+ +2S2O32- <---->Ag(S2O3)23- Ecell increased even though formation of a solid was not observed.

a) Using the value Ecell = 0.70V, calculate the [Ag+] in the test solution after equilibrium is reached.

b). Setup a reaction table for equation above and calculate the formation constant for Ag(S2O3)23- .

Show your calculations clearly.

Answer 1

a) Ecell = -RT/nF lnK

K = [Ag(S2O3)2^3-]/[Ag+][S2O3^2-]^2

From the reaction,

[Ag(S2O3)2^3-] = 0.10 M x 2 ml/4 ml = 0.05 M

remaining [S2O3^2-] = (0.5 x 2 - 0.1 x 2 x 2)/4 = 0.15 M

Feed the values,

0.70 = -8.314 x 298/1 x 96485 (ln(0.05/[Ag+](0.15)^2))

[Ag+] = 1.54 x 10^12 M

b) Ag(S2O3)2^3- = 0.05 M

                   Ag+    +      S2O3^2-      -----> Ag(S2O3)2^3-

initial           0.2               1.0                            -

change       0.2                0.4                          0.2

Eq                -                  0.6                          0.2

Kf = (0.2)/(0.6)^2 = 5.55 x 10^-1