Answer to Question #131721 in General Chemistry for Anon

Question

AgNO3 is slowly added to a solution containing 0.010 M Na2CrO4 and 0.050 M KBr.
Given: Ksp(Ag2CrO4) = 1.1 × 1O−12 ; Ksp(AgBr) = 5.0 × 1O−13 .
3.1 Show all your calculations and reasoning to prove that AgBr will precipitate first.
[6]
3.2 What is the percentage of Br− that remains in solution at the instant when the
precipitation of Ag2CrO4 starts?

Answer 1

3.1 First we will need to calculate [Ag⁺] at which each compound will precipitate. (The one which will precipitate at a lower [Ag⁺] will be the one to be precipitate first)

Thus, for Ag2CrO4 : $[Ag+] = Ksp/[CrO42-]$

$[Ag+] = (1.1x10-2)/0.01M$

$[Ag+] = 1.1M$

For AgBr: $[Ag+] = Ksp/[Br-]$

$[Ag+] = (5.0x10-13)/0.05M$

$[Ag+] = 1.0x10-11$

From the calculation it is evident that [Ag⁺] for AgBr is less than [Ag⁺] for Ag₂CrO₄

Therefore, it is true that AgBr will precipitate first.

3.2 From (3.1) above, Ag₂CrO₄ begins to precipitate when [Ag⁺] = 1.1M

So, [Br^-] remaining when [Ag⁺] = 1.1M is given by;

[Br^-]$=Ksp/[Ag+]$

[Br^-] $=(5.0x10-13)/1.1M$

[Br^-] $=4.54x10-13M$

To find the percentage of Br- reamaining we need to find the fraction of Br^-remaining, then multiply by 100 to change it to percentage

$Percentage of [Br] remaining=([Br-] remaining/[Br-] initial) x 100$

$=((4.5x10-13M)/(0.05M)) x100$

$=9.09x10-10 percent$

=9.09x10^-10%