00:01
Ionization of a 0 .2 molar solution of monoprotic acetylacetic acid.
00:05
And we're given the acid dissociation constant of 3 .0 times 10 to the minus 4th.
00:11
What am i asked to do? okay, percentionization.
00:14
So let's go ahead and write down our dissociation.
00:25
That'll be h plus and c9h704 minus.
00:34
We're going to do an ice table for this.
00:37
And that will be, i believe, it was 0 .20 minus x plus x and plus x 0 .20 minus x, x, x and x.
00:52
So our ka will be equal to our hydrogen ion or hydronium, if you prefer to write it that way, concentration times our anion, and i think it's called acetylacillate, acetylacillate, and then my acid, i hope i can, yes, i'm going to assume that my 5 % rule applies here.
01:29
So my ka is 3 .0 times 10 to the minus 4th equals x squared, and then i'm just going to do 0 .20, and x will equal 7 .746 times 10 to the minus 3rd more.
01:46
So my percent ionization will be equal to 7 .746 times 10 to the minus third divided by 0 .20.
02:04
Both these are molar times 100.
02:08
And this will equal 3 .87%.
02:14
And it looks like, let me get down to my actual problem again.
02:19
I bet i have to go to two sake bakes.
02:21
Yes.
02:22
So i'm going to go to 3 .9.
02:27
So i satisfy my significant figures rules.
02:31
And this is my percent ionization.
02:39
That's my answer for a...