00:01
How much, what is stuck on me, got like a lego stuck on me, 5 .6 molar sodium hydroxide has to be added to 650 milliliters of buffer that is 0 .0205 molar acetic acid with a pka of 4 .76 and 0 .0270 molar sodium acetate.
00:20
If we want to raise the ph to 5 .75, let's begin.
00:25
Get.
00:26
Okay, so i'm going to use henderson -hasselbalch here of our conjugate base, our acid, and this will be 5 .75, right? 5 .75 equals 4 .76 plus the log of my base is 0 .0270 molar and 0 .020.
01:10
Five molar what hang on this isn't right so i don't know what my ratio is going to be let me figure this out so i'm going to have 5 .75 minus 4 .76 this will be 0 .99 will equal my log of my conjugate base over my acid and that'll be and this will be did i do that right 9 .77 so my ratio, i'm getting a big ratio here.
03:10
I'm getting a ratio of my conjugate base over my acid equal 10 to the 0 .99 equals 9 .77.
03:26
That seems awfully big, but let's keep going.
03:32
So let's figure out the moles of acetic acid that we have and that will equal 0 .0205 molar times 0 .650 liters.
04:01
And this will give me 0 .0205 times 0 .65 is 0 .013325 moles of acetic acid.
04:51
So my conjugate base, i will need x over 0 .013325 equals 9 .77, and that will equal 0 .130185 moles of my conjugate base...