Solution
(a) In Example 20-8 we found that AgBr begins to precipitate when [Ag]3.3 10 ^10 M.
This value for [Ag] can be substituted into the Kspexpression for AgI to determine [I]
remaining unprecipitatedwhen AgBr begins to precipitate.[Ag][I]1.5 10 ^16[I]unppt’d
1.5
[
Ag1
0
] 16
1
3.
.5
3
1
10
0
1
16
0 4.5 10 (^7) M
The percentage of Iunprecipitated is
% Iunppt’d100%
4
1
.
.
5
0
1
1
0
0
7
3
M
M
100%
0.045% Iunprecipitated
Therefore, 99.955% of the Iprecipitates beforeAgBr begins to precipitate.
(b) Similar calculations show that just beforeAgCl begins to precipitate, [Ag]1.8 10 ^7 M,
and the [I] unprecipitated is calculated as in part (a).
[Ag][I]1.5 10 ^16
[I]unppt’d
1.5
[
Ag
1
0
]
16
1
1
.
.
5
8
1
1
0
0
1
7
6
8.3 10 ^10 M
The percentage of Iunprecipitated just before AgCl precipitates is
% Iunppt’d100%
8
1
.
.
3
0
1
1
0
0
1
3
0
M
M
100%
0.000083% Iunprecipitated
Therefore, 99.999917% of the Iprecipitates before AgCl begins to precipitate.
A similar calculation for the amount of Brprecipitated just before AgCl begins to precip-
itate gives
[Ag][Br]3.3 10 ^13
[Br]unppt’d
3.3
[
Ag
1
0
]
13
3
1
.
.
3
8
1
1
0
0
1
7
3
1.8 10 ^6 M
% Brunppt’d100%
1
1
.
.
8
0
1
1
0
0
6
3
M
M
100%
0.18% Brunprecipitated
Thus, 99.82% of the Brprecipitates before AgCl begins to precipitate.
You should now work Exercises 36 and 38.
We have described the series of reactions that occurs when solid AgNO 3 is added slowly
to a solution that is 0.0010 Min Cl, Br, and I. Silver iodide begins to precipitate first;
99.955% of the Iprecipitates before any solid AgBr is formed. Silver bromide begins to
[Br]unppt’d
[Br]orig
[I]unppt’d
[I]orig
[I]unppt’d
[I]orig
In Example 20-7 we did a similar
calculation, but we did not express the
result in terms of the percentage of an
ion precipitated.
836 CHAPTER 20: Ionic Equilibria III: The Solubility Product Principle
We have subtracted 0.045% from
exactly100%. We have therefore not
violated the rules for significant
figures.