00:01
Let's figure out what the concentration of our sodium ions or malarity is for an aqueous solution with different compounds.
00:10
So let's take our first compound, which is n -a -b -r.
00:17
Notice that we only have one atom of sodium and one atom of vr.
00:22
So that means that we have a one -to -one ratio.
00:25
So how do we determine then what our malarity is for only the sodium? well, we're going to take our 0 .025 molarity, or we can say that's moles per liter because that's the same thing.
00:44
And we take that and we say, okay, for every one mole of nabr, we have one mole of sodium ions.
00:58
Okay, we cancel these out and we end up getting 0 .024 .4.
01:04
0 .05 moles of n .a.
01:08
Per liter, which is the same as saying 0 .025 molarity of n .a plus.
01:19
Okay, let's do the second part of this problem.
01:24
B, our compound now changes from nabr to n .a .2 s .o4.
01:32
What is our ratio here? we have two sodium atoms to one sulfate atom.
01:41
2 to 1 ratio.
01:43
So let's take this same molarity here, which is moles per liter.
01:52
And let's use this compound.
01:54
So for every 1 mole of na2, so4, we have 2 moles of sodium ions...
