Solubility Equilibria Assignment Question [2005/RI/III/3]
The chromate(VI) and chloride salts of silver are only slightly soluble. Silver chromate(VI), Ag 2CrO4, is
a red compound with Ksp = 1.3 x 1012 mol3 dm9 while silver chloride is a white compound with Ksp =
1.8 x 1010 mol2 dm6.
An aqueous solution, F1, contains chromate(VI) and chloride ions, both of concentration
1.00 x 103 mol dm3.
(a)
Calculate the concentration of silver ions needed to precipitate out
(i) silver chromate(VI)
(ii) silver chloride
in F1 and hence prove that silver chloride precipitates out before silver chromate(VI).
[3]
(i)
Ksp of Ag2CrO4 = [Ag+(aq)]2[CrO42(aq)] = 1.3 x 1012 mol3 dm9
To precipitate out Ag2CrO4, [Ag+(aq)]2[CrO42(aq)] > 1.3 x 1012
[Ag+(aq)]2> 1.3 x 1012 /[CrO42(aq)] = (1.3 x 1012) /(1.00 x 103) = 1.3 x 109
[Ag+(aq)] >
(ii)
1.3 x 10-9 = 3.61 x 105 mol dm3
[1]
Ksp of AgCl = [Ag+(aq)][Cl(aq)] = 1.8 x 1010 mol2 dm6
To precipitate out AgCl, [Ag+(aq)][Cl(aq)] > 1.8 x 1010
[Ag+(aq)]> 1.8 x 1010 /[Cl (aq)] = (1.8 x 1010) /(1.00 x 103)
= 1.80 x 107 mol dm3
[1]
Since [Ag+(aq)] needed to ppt out AgCl is lower than that needed to ppt out Ag2CrO4, AgCl
ppt out before Ag2CrO4.
[1]
(b) Calculate the concentration of chloride ions in F1 when silver chromate(VI) begins to precipitate
out. Comment on whether potassium chromate(VI) is a suitable indicator in determining the
concentration of chloride ions by titration against standard silver nitrate solution.
[2]
By the time Ag2CrO4 starts to ppt out, [Ag+ (aq)] = 3.61 x 105 mol dm3
At this time, [Cl (aq)] = (Ksp of AgCl)/[Ag+ (aq)] = (1.8 x 1010) / (3.61 x 105)
= 4.99 x 106 mol dm3
[1]
Potassium chromate(VI) is a suitable indicator since red Ag2CrO4 only appears when
negligible amount of Cl is left in solution so that the endpoint at which the colour change
is observed is very close to the equivalence point.
[1]
(c) Chromate(VI) ions can be converted to dichromate(VI) ions by reaction with hydrogen ions.
(i) Write an equation for the equilibrium that exists between chromate(VI) ions and
dichromate(VI) ions in acidified solution.
[1]
2CrO42 (aq) + 2H+ (aq) ⇌ Cr2O72 (aq) + H2O (l)
[1]
(ii) 50.0 cm3 of F1 is buffered at a pH of 3. It is found that 90% of the chromate(VI) present is
converted to dichromate(VI). Determine the equilibrium constant for the reaction.
[4]
[Total: 10]
After being buffered at a pH of 3,
[CrO42(aq)] at equilibrium = 0.10 x 1.00 x 103 = 1.00 x 104 mol dm3
[CrO42(aq)] reacted = 0.90 x 1.00 x 103 = 9.00 x 104 mol dm3
[Cr2O72(aq)] formed at equilibrium = ½ x 9.00 x 104 = 4.50 x 104 mol dm3
Kc =
[Cr2O7 2- (aq)]
(4.50 x 10-4 )
= 4.50 x 1010 mol3 dm9
=
22
+
2
-4 2
-3 2
[CrO4 (aq)] [H (aq)]
(1.00 x 10 ) (1.00 x 10 )
Common mistake:
 Substitute [H+] as 1.0 x 10-4, forgetting that pH of solution is maintained at 3.
[2]
[2]