00:01
The equilibrium reaction that we're considering goes essentially to completion because we have such a large k value m2 plus plus two oxalates becomes the complex ion mn, c204, 2, 2 minus.
00:28
So the k expression, or the kf expression, is going to be the concentration, of mn c204 2 minus divided by the concentration of mn 2 plus and the concentration of c204 2 minus squared.
01:02
So we can assume that the reaction goes to completion first and manganese is the limiting reactant.
01:12
We have the smallest volume and the smallest concentration of manganese.
01:17
So we can assume all of the manganese reacts, creating the maximum amount of mn2 -0 -4 -2 -2 -2 -minus.
01:29
So this concentration can be calculated mnc204 -2 -2 -2 -1, as being the moles of manganese that we start with, which is 550 -millimeter, or 050 -550 liters.
01:53
At a concentration of 0 .015 moles per liter, and then essentially one mole of it becomes one mole of our complex ion.
02:10
And we'll divide that by the new volume.
02:13
The new volume is going to be the 550 milliliters plus the one liter.
02:21
So that'll be 1 .550 liters liters.
02:24
So we get a complex ion concentration of 5 .32 times 10 to the negative 3 molar.
02:39
And what about the oxalate concentration? well, the oxalate concentration will be equal to what we started with, the moles that we start with.
02:58
So we're going to start with one liter at one molar of oxalate.
03:04
And then when we form this, we will consume two moles of oxalate for every one mole of this that we formed...