00:01
We are asked to find the moles of sodium hydroxide we have to add to 250 milliliters of 0 .367 molar nitrous acid to prepare a buffer with 3 .55 ph.
00:13
Okay, so let's first look up a ka for hno2 and make sure this matches to make sure that one matches and i'm going to use 4 .0 times 10 to the minus 4.
00:30
Okay, then we're going to write down and this will produce my water and my no2.
00:47
Okay, so we know that this must be limiting.
00:54
If not, it would be a strong base, so that has to be limiting.
01:20
Okay, so the final mixture will therefore have to be hno2 and no2 minus.
01:36
So we're going to be doing this.
01:40
My ph will be equal to my pka plus the log of my base over my acid.
01:50
I'm going to pause.
01:51
I'll be back.
01:52
We are asked to find the moles of sodium hydroxide we have to add to 250 milliliters of 0 .367 molar nitrous acid to prepare a buffer with 3 .55 ph.
02:06
Okay, so let's first look up a ka for hno2 and make sure this matches to make sure that one matches and i'm going to use 4 .0 times 10 to the minus 4.
02:22
Okay, then we're going to write down.
02:30
Okay, let's see here.
02:32
I stepped away for a little bit, so i got to figure out where i am.
02:38
We're told that we want a buffer of this and i've got this, so let's use those two values to set up 3 .550 will equal the negative log of 4 .0 times 10 to the minus 4 plus the log of my base over my acid and that will be, let me do this now, negative log 4 times 10 to the minus 4, enter, whoops, negative log 4 times 10 to the minus 4, enter, is 3 .9, 3 .398 plus the log of base over acid and 3 .55 minus second answer, enter, equals 0 .152 equals the log of my base over my acid.
04:12
Okay, let's see if i can figure this out.
04:14
So base over acid will equal second log 0 .152 is 1 .41.
04:50
Okay, now i know that my acid, what was my concentration of my acid here? i don't know.
05:01
Let me go back up here.
05:02
Concentration of my acid, 0 .367 molar.
05:07
So my acid, these are concentrations and we're going to assume naoh results in no volume change.
05:21
We're given that, i think.
05:28
Okay, let me think here.
05:35
So this will be x over 0 .367 minus x equals 1 .419.
05:54
Let me see how that works.
06:02
This equals negative 1 .419 x plus 0 .520773 equals x.
06:17
So i get 2 .419 x equals 0 .5207 x equals that number divided by 2 .419.
06:37
And i get 0 .215 molar moles of x, 2 .1528.
06:54
Okay, now let's get the volume.
06:59
No, that's how many moles i need.
07:20
Let me double check.
07:21
I just want to make sure i've got this.
07:26
I'm going to pause again.
07:29
We were asked to find the moles of sodium hydroxide.
07:32
We have to add 250 milliliters of 0 .367 molar nitrous acid to prepare a buffer with 3 .55 ph.
07:43
Okay, so let's first look up ka for hno2 and make sure this matches to make sure that one matches...