00:01
Long problem coming up.
00:03
We're given the following information.
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We have a 100 mil solution that contains 0 .120 molar chromium 2 nitrate and 0 .50 molar nitric acid.
00:31
We add to this 20 milliliters of 0 .250 molar potassium dichromate.
00:51
The chromate and dichromate and chromium 3, excuse me, the dichromate and the chromium 2 ions, all react to produce chromium 3.
01:05
For part a, we're writing the net ionic equation.
01:14
And let's go ahead and get started with that.
01:16
The net ionic equation is as follows.
01:58
There's a.
01:59
For b, we're asked to calculate the concentrations of all the ions after the reaction comes to completion.
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So what i'm going to do is i'm going to figure out the initial concentration of every ion that is pertinent here.
02:22
And when we look at our chemical equation, we can see that our nitrate ions and our potassium ions are spectators.
02:32
They're not in.
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So those ions are going to be the same before and after.
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So let's go ahead and start with previous page.
02:48
Oops, see, i just dropped my pen.
02:55
Okay.
02:58
Concentration of all ions.
02:59
Shall we start with cr2 plus? i started with 100 milliliters, and i added 20 milliliters.
03:16
So i'm going from 100 to 120 milliliters, and my initial concentration was 0 .120 moles per liter.
03:30
So my final concentration will be, or my initial concentration will be 0 .10 molar cr2 plus.
03:40
That's my first concentration of that substance.
03:45
Next, let's do our cr207.
03:51
2 -minus.
03:56
That had cr207.
04:03
We started out with 20 millimeters.
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I should go 20 .0 millimeters.
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And diluted that also to 120 millimeters.
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And that initial concentration was, i believe, 0 .250 0 .250.
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And that would equal 0 .0467 for my chromium or excuse me my dichromate.
04:38
So here's my first two initial concentrations.
04:44
Next let's do where's my h's.
04:49
For my h pluses, that i also started out with 100 milliliters and diluted it to 120.
04:57
And my initial concentration was given as 0 .5 .00 molarity.
05:05
And that equals 0 .4167 molar h -1 .6 -ion.
05:18
What else here? let me look at my ions again.
05:24
We did our h -plus, we did our cr, we did our dichromate.
05:29
We've got k and no3.
05:38
For potassium, that was also an initial volume of 20 milliliters that we diluted to 1 .20 milliliters.
05:52
But this is going to have 2 times 0 .250 moles per liter because it's k2.
06:02
See, i shouldn't have done that.
06:04
That would be there.
06:06
So my potassium, i believe, was 0 .08 something.
06:19
And two of my substances has nitrates in it.
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Let's go back to this page.
06:23
You'll see that this, this both had nitrate.
06:28
My h plus concentration has an equivalent from my hn03, i will have an n03 concentration of 0 .4167 molar.
06:47
From my chromium to nitrate, my nl3 concentration will be this value times 2 because we have a 2 to 1 mole ratio for my nitrate ions to my chromium nitrate.
07:23
So that will equal my total will equal 0 .4167 plus 2 times 0 .10 equals 0 .617.
07:42
Molar n -o -3...