00:01
In this problem, we are asked to calculate the ph after each addition of acid.
00:05
And i want to make a couple of notes here.
00:07
This is a strong base and this is a strong acid.
00:17
And please notice that our concentrations for each are the same.
00:23
So first is zero milliliters.
00:25
And since we have a strong base, i will have 0 .1000 molar koh will yield the same concentration of oh-, which means my ph will be equal to, since this is oh, 14 minus the negative log of 0 .1000 molar.
01:04
And this, i believe, will equal 13.
01:08
Let me do this.
01:09
14 minus negative log 0 .1 is 13, and that'll be 0 .000, will be my ph.
01:29
And i did three decimal places because i have, i should probably do 0 .000.
01:33
Okay, for my second one, my second addition, my second, my first addition will be 15 milliliters of acid.
01:59
Okay, 15 milliliters of acid.
02:04
Now, since i have 30 and 15 and my concentrations are the same, i will have 15 milliliters.
02:22
Let's do this first.
02:24
My volume total will now equal 15 .00 milliliters plus 30 .00 milliliters will equal 45 .00 milliliters.
02:48
My moles of excess hydroxide can be found by taking 30 .00, and i'm going to go with millimoles, 30 .00 ml of oh - times 0 .1000 mmol of oh - per ml of oh - equals 3 .0 mmol of oh-.
03:47
And 15 .00 milliliters of h plus at 0 .1000 millimoles per milliliter will equal 1 .500 millimoles.
04:11
We're going to subtract these two values and i'll 0 .5 1 .500 millimoles of excess oh-.
04:30
So my concentration will equal 1 .500 millimoles divided by 45 .00 milliliters will equal 1 .5 divided by 45 is 0 .0333 molar.
05:21
And my ph will equal, again, 14 minus the negative log of 0 .0333.
05:37
So my ph will equal 14 minus negative log, second answer, enter.
05:46
Whoops.
05:48
14 minus negative log, second answer, enter.
05:52
Now i don't have my answer anymore.
05:58
0 .0333.
06:02
And i get 12 .5224.
06:16
There's two.
06:19
There's a lot of these to do, and i thought this would be sort of fun, but i might be wrong.
06:25
C is 29 .00 milliliters.
06:33
So i still have three millimoles base.
06:40
And now i'm going to have 29 .00 milliliters times 0 .1000 millimoles per milliliter equals 2 .900 millimoles.
07:04
So subtract and i get 0 .100 millimoles and my molarity will equal 0 .100 millimoles divided by my new volume total, which will be 29 plus 30 milliliters.
07:39
0 .1 divided by 59 is 0 .001695 molar oh minus.
07:56
So my ph will equal 14 minus the negative log of 0 .001695.
08:12
So my ph will be 14 minus negative log, second answer.
08:18
And here i get 11 .229...
