00:01
This question asked you to consider what happens when an aqueous solution of potassium iodide.
00:07
Potassium iodide is k .i., one of each of them, is mixed with an aqueous solution of lead nitrate.
00:19
I assume they mean lead to nitrate, as there's also a lead -4 nitrate.
00:25
If it's lead to nitrate, then we've got p .b.
00:29
Just one of them with two nitrates, because each nitrate only has.
00:33
A 1 minus charge.
00:37
When the cation switch places, the lead will go with the iodide, and according to solubility rules, lead iodide is insoluble, so that will be a precipitate that forms as a solid.
00:51
The potassium will then go with a nitrate, and according to solubility rules, this is soluble, so it stays as aqueous.
01:01
We'll only need one of each of these, as each has a plus one and minus one charge.
01:05
But to balance this with two iodides, we're going to need to put a two here.
01:10
That gave us two potassium, so we'll put a two here, giving us two nitrates, and we have two nitrates.
01:17
So it looks like the answer is the product, pbi2, will dissolve in water? no.
01:25
The net ionic reaction is that the spectator ions are, okay...