00:01
A solution is .015 molar in both bromine and sulfate.
00:06
So a .204 molar solution of lead to nitrate is slowly added to it with a burette.
00:10
The blank anion will precipitate from the solution first.
00:15
So what you need to do is calculate q for both of those situations.
00:21
For the pbbr2, you'll form lead two ions and two bromide ions.
00:30
So the ksp expression would be the lead to ions times the bromide ions squared.
00:37
So when you're calculating q, you would plug in the lead to concentration, which they gave us was 0 .204, and then the bromide ion concentration squared.
00:50
So then you'll solve for the value of q, which is 0 .204 times 0 .015 squared.
00:55
So q is 4 .59 times 10 to the negative 5.
01:02
Then you'll do the same thing for the lead to sulfate, pbso4.
01:08
When it dissolves, it forms the lead two ions and the sulfate ions.
01:13
And so its ksp expression is the lead two ions times the sulfate ions.
01:21
And so its value for q, you're going to plug in the lead concentration, which is the same, and the sulfate ion concentration, which is .015.
01:34
And when you calculate that, you get that the q value for the lead to sulfate is 0 .00306.
01:48
So based on this, we need to compare q to k.
01:52
So the ksp for pbr2 is 6 .60 times 10 to the negative 6 .6.
01:59
Whereas the ksp value for the pvso4 is 2 .53 times 10 to the negative 8.
02:08
For both of these, the q is smaller, i'm sorry, it's larger than k.
02:13
Q is larger than k on both of them.
02:16
So they will both eventually precipitate.
02:18
It's a question of which one will precipitate from the solution first.
02:21
That one is going to be the lead -2 sulfate will precipitate first because it has a smaller value of k...