In: Chemistry
Argentometric titrations can be used to determine the concentration of a number of halides (see Table 7-1 in the book). However, if the sample contains a mixture of halides they will precipitate in certain order. Considering that information, which of the following will preferentially form during a titration with AgNO3?
A .Silver acetate (Ksp = 2.0 x 10-3) |
B. Silver bromide (Ksp = 5.3 x 10-13) |
C. Silver chloride (Ksp = 1.8 x 10-10) |
D. AgF (very soluble) |
E. Silver carbonate (Ksp = 8.1 x 10-10) |
Given the Ksp values, we need to determine the solubility of the halides to answer the question.
a) Silver acetate, AgAc.
=
AgAc (s) <====> Ag+ (aq) + Ac- (aq)
Ksp = [Ag+][Ac-] = 2.0*10-3
===> (x).(x) = 2.0*10-3 (the solubilities of Ag+ and Ac- are the same due to 1:1 nature of the dissociation).
===> x2 = 2.0*10-3
===> x = 0.0447
The solubility of Ac- is 0.0447 M (ans).
b) Silver Bromide, AgBr.
AgBr (s) <=====> Ag+ (aq) + Br- (aq)
Ksp = [Ag+][Br-] = 5.3*10-13
===> (x).(x) = 2.0*10-3 (the solubilities of Ag+ and Br- are the same due to 1:1 nature of the dissociation).
===> x2 = 5.3*10-13
===> x = 7.280*10-7
The solubility of Br- is 7.280*10-7 M (ans).
c) Silver Chloride, AgCl.
AgCl (s) <=====> Ag+ (aq) + Cl- (aq)
Ksp = [Ag+][Cl-] = 1.8*10-10
===> (x).(x) = 1.8*10-10 (the solubilities of Ag+ and Cl- are the same due to 1:1 nature of the dissociation).
===> x2 = 1.8*10-10
===> x = 1.342*10-5
The solubility of Cl- is 1.342*10-5 M (ans).
d) Silver fluoride, AgF is soluble in water.
e) Silver Carbonate, Ag2CO3.
Ag2CO3 (s) <======> 2 Ag+ (aq) + CO32- (aq)
Ksp = [Ag+]2[CO32-] = 8.1*10-10
====> (2x)2.(x) = 8.1*10-10 (the solubilities of Ag+ and CO32- are in the ratio 2:1 due to the nature of dissociation).
====> 4x3 = 8.1*10-10
====> x = 1.423*10-5
The solubility of CO32- is 1.423*10-5 M (ans).
We note that Br- is the least soluble when a source of Ag+ is added and hence, AgBr will preferentially form when a solution of AgNO3 is added (ans).
When all the Br- is replenished, AgCl will start forming (ans).