00:01
To determine which one precipitates first, we need to calculate the silver concentration that results in precipitation of both of them.
00:12
In the case of silver chloride, as a solid, it establishes an equilibrium with silver ion and chloride, where ksp is equal to the product of the concentration of silver multiplied by the concentration of chloride, and that value they gave us as 1 .6 times 10 to the negative 10.
00:40
So to calculate the concentration of silver that initiates precipitation, we simply rearrange this equation and plug in the chloride concentration that we have.
00:56
Silver equals ksp divided by the chloride concentration, 1 .0 times 10 to the negative 3, coming from potassium chloride, giving us 1 .6 times 10 to the negative 7 molar.
01:14
Then we carry out a similar calculation with silver bromide.
01:21
It establishes an equilibrium with silver ion and bromide ion according to its ksp, which will be equal to the concentration of silver plus the concentration of bromide, and that value is even smaller at 7 .7 times 10 to the negative 13.
01:45
So the silver concentration that initiates precipitation of silver bromide will be equal to 7 .7 times 10 to the negative 13, divided by the bromide concentration, 1 .0 times 10 to the negative 6, giving us 7 .7 times 10 to the negative 7.
02:14
So the one that's going to precipitate first will be the one that requires the smaller concentration...