00:01
In answering this question, let's determine the concentrations, that is, the concentration of hno2, which is equal to 1, multiplied by 1 ,000 milliliters, divided by 1 ,000 milliliters plus 50 milliliters, which is equal to 0 .952m.
00:20
And as for the concentration of n -a, n02, this is going to be equal to 1, 1 per liter, multiplied by 1, 1 ,000.
00:31
Divided by a thousand plus 50 which is also equal to 0 .952m.
00:40
So to determine the concentration of the h plus ions, this is going to be equal to 0 .65 multiplied by 50 milliliters divided by 1000 plus 50 milliliters which is equal to 0 .0 31m.
00:58
So looking at this, what we want to do here is to have a reaction where we have h plus reacting with n02 minus to produce the hno2 and h2o.
01:15
So looking at this, we don't need this part.
01:18
So the initial concentrations here we've got 0 .031 change.
01:26
Then here we've got 0 .952.
01:28
This we have 0 .952.
01:31
So looking at the change, this decreases by 0 .031.
01:35
This also decreases by 0 .0 .31 and this increases by 0 .031.
01:41
So what this means is at equilibrium, this we have 0 and we have 0 .921 and here we've got 0 .983...