Question

The solubility of AgCrO4 is several times greater than that of AgCl, so the red precipitate cannot form until all the chloride ions have effectively been removed from the solution. If a student uses 1mg of K2Cro4 as an indicator of 100 ml sample solution, what is the concentration (M) of chloride ion in the solution when the red precipitate (AgCro4) starts to form?

Answer 1

Assume:

2Ag+(aq) + CrO4-(aq) --> Ag2CrO4(s)

if we use

1 mg of K2CrO4 --> change to mol

mass = 10^-3 g o K2CrO4

MW of K2CrO4 = 194.1903 g/mol

mol = mass/MW = (10^-3) / 194.1903 = 0.00000514958 mol of K2CrO4 are added

V = 100 mL = 0.1 L

[K2CrO4] = mol/V = 0.00000514958 = 0.1 = 0.0000514958 M

[CrO4-2]

Ksp = [Ag+]^2[CrO4-]

From tables

Ksp = 1.1*10^-12

substitute data:

Ksp = [Ag+]^2[CrO4-]

1.1*10^-12 = (M)^2 * (0.0000514958 )

M = sqrt((1.1*10^-12) / 0.0000514958)

M = 0.0001461 of Ag+

since ratio between Ag+´and Cl- is 1:1

[Cl-] = 0.0001461 M