Question

An important source of Ag is recovery as a by-product in the metallurgy of lead. The percentage of Ag in lead was determined as follows. A 1.030 −g sample was dissolved in nitric acid to produce Pb2+(aq) and Ag+(aq). The solution was then diluted to 500.0 mL with water, a Ag electrode was immersed in the solution, and the potential difference between this electrode and a SHE was found to be 0.490 V .

What was the percent Ag by mass in the lead metal?

It is not 0.12%

Answer 1

The amount of sample = 1.03 g

it gives Pb+2 and Ag+

The volume = 500mL

Reaction will be

Ag⁺(aq) + 1/2H₂(g) --> Ag(s) + H⁺(aq)

The potential difference = 0.49 = Ecell

E0cell = 0.799V

Ecell = E0cell - 0.0592 / 1 log [H+] / [Ag+]

0.49 = 0.799 - 0.0592 log (1/[Ag+])

-0.309 = -0.0592 log (1/[Ag+])

5.21 = log (1/[Ag+])

taking antilog

162181 = 1/[Ag+]

[Ag+] = 6.17 X 10^-6

So moles of Ag+ = concentration X volume in litres = 6.17 X 10^-6 X 0.5 = 3.08 X 10^-6

Mass of Ag = Moles X atomic weight = 3.08 X 10^-6 X 108 = 3.329 X 10^-4

% of Ag+ = Mass of Ag+ X 100 / Total mass = 3.329 X 10^-4 X 100 / 1.030 = 0.032 %