Question

Near the endpoint, the red color of the silver chromate forms, but then disappears, as you swirl. This is because the Ksp of silver chromate may be temporarily exceeded locally (where the silver nitrate is added), but as you swirl, the chloride ions still in solution convert the silver chromate into silver chloride. Write an equation for this process, (the process of converting the silver chromate into white silver chloride) and calculate the equilibrium constant of the reaction in terms of the solubility products of silver chloride and silver chromate.

Ksp = 1.6 x 10-10 for AgCl Ksp = 1.1 x 10-12 for Ag2CrO4

b.   Assume the chromate concentration of indicator in the flask is 0.0050 M. What is the concentration of Cl- remaining in the flask when silver chromate first precipitates (that is, at the endpoint of the titration).

c.   What percentage of the Cl- in the original unknown remains when the endpoint is reached? You should use your answer to question 2, and estimate the volume of solution in the flask.

Answer 1

For the given precipitation reaction,

a. Equations,

Ag+(aq) + Cl-(aq) <==> AgCl(s) ---(1)

Ag2CrO4(s) <==> 2Ag+(aq) + CrO4^2-(aq) ---(2)

multiply (1) with 2 and add both equations,

Ag2CrO4(s) + 2Cl-(aq) <==> 2AgCl(s) + CrO4^2-

Keq = (Ksp of AgCl)^2/Ksp of K2CrO4

       = (1.6 x 10^-10)^2/1.1 x 10^-12

       = 2.33 x 10^-8

b. when [CrO4^2-] = 0.005 M

concentration of Ag+ when Ag2CrO4^2- precipitates out,

[Ag+] = sq.rt.(Ksp/[CrO4^2-])

         = sq.rt.(1.1 x 10^-12/0.005)

         = 1.48 x 10^-5 M

Now,

[Cl-] precipitated by this [Ag+]

[Cl-] = Ksp/[Ag+]

        = 1.6 x 10^-10/1.48 x 10^-5

        = 1.08 x 10^-5 M

[Cl-] initially present = sq.rt.(Ksp AgCl)

                                = sq.rt.(1.6 x 10^-5)

                                = 1.26 x 10^-5 M

So,

[Cl-] remaining in the flask = 1.26 x 10^-5 - 1.08 x 10^-5

                                           = 1.80 x 10^-6 M

c. percentage of [Cl-] remaining in the flask = 1.80 x 10^-6 x 100/1.08 x 10^-5

                                                                     = 16.67%