- Iodide Ions can be used to precipitate lead (II) ions from 0.0030 M Pb(NO3)2 (

Question

- Iodide Ions can be used to precipitate lead (II) ions from 0.0030 M Pb(NO3)2 (aq). What minimum iodide ion concentration is required for the onset of PbI2 precipitation? What mass (in grams) of KI must be added for PBI2 to form?

- You decide to use sulfide precipiation to separate the copper (II) ions from the manganese(II) ions in a solution that is 0.20 M Cu^2+(aq) and 0.20 M Mn^2+(aq). Determine the minimum sulfide ion concentration that will result in the precipitation of one cation (identify the cation) but not the other.

- A lead electrode in 0.010M Pb(NO3)2 (aq) is connected to a hydrogen electrode in which the pressure of H2 is 1.0 bar. If the cell potential is 0.057 V at 25 degrees celsius, what is the pH of the electrolyte at the hydrogen electrode?

Explanation / Answer

A) PbI2 <=> Pb2+ + 2I-

Ksp = [Pb2+][I-]^2

Ksp = 1.4 x 10^-8

1.4 x 10^-8 = 0.003 [I-]^2

[I-]= 0.00216 M

With 1 L of solution

moles I- = 0.00216 x 1 L=0.00216 M

mass KI = 0.00216 M x 166 g/mol=0.359 g

B) Ksp of CuS = 6 x 10^-37

Ksp of MnS = 3 x 10^-14

So MnS is more soluble

Ksp = [Mn2+][S2-]

[S2-] = 3 x 10^-14/0.2 = 1.5 x 10^-13 M is the concentration of S2- when MnS is ready to precipitate. At this poin the less soluble CuS has already precipitated out of solution

C) Pb --> Pb+2 ==> + 0.13 volts

We have Nernst equation,

E = Eo - (0.0592 / n) (logQ)

0.057 = 0.13 - (0.0592 / 2) (log[ H2] [Pb+2] / [H+])

0.073 = (0.0296) (log[ 0.987 atm] [0.01] / [H+])

0.073 / (0.0296) = (log[ 0.987 atm] [0.01] / [H+])

2.466 = (log[ 0.987 atm] [0.01] / [H+])

292.561 = [ 0.987 atm] [0.01] / [H+])

H+ = [0.987 atm] [0.01] / 341.8

H+ = 3.374 x 10^-5 M

pH = -log[H+] = 4.47

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