00:01
First we need to calculate the phosphate concentration required to initiate precipitation of each of these.
00:10
Aluminum phosphate as a solid is involved in an equilibrium producing aluminum ion and phosphate ion according to its ksp, where ksp will be equal to the aluminum concentration multiplied by the phosphate concentration.
00:37
And they didn't give us that value, so we'll have to look it up in the table i have.
00:42
It's 6 .3 times 10 to the negative, i just lost it, 19.
00:57
So the phosphate concentration that will initiate precipitation will be equal to the phosphate concentration when multiplied by aluminum, gets us this value.
01:08
Rearranging the equation, phosphate is equal to ksp, 6 .3 times 10 to the negative 19, divided by the aluminum concentration which they gave to us in this solution.
01:26
We have an aluminum concentration coming from aluminum nitrate at 0 .017 molar and we'll need a phosphate concentration of 3 .71 times 10 to the negative 17 molar to initiate precipitation of the aluminum as aluminum phosphate.
01:50
Then the second potential precipitate is calcium phosphate ca3po4 2.
01:58
This dissociates into three calcium ions and two phosphate ions.
02:08
Ksp then is going to be equal to the calcium concentration raised to the third and the phosphate concentration raised to the two and the ksp i have is 2 .0 times 10 to the negative 29.
02:31
So to calculate the phosphate concentration that initiates precipitation of the calcium, we rearrange this equation and phosphate will be equal to the square root of ksp 2 .0 times 10 to the negative 29 divided by the calcium concentration cubed...