00:01
For this problem, we're looking at buffers and we need to calculate the ph of the buffer when it's just the buffer, as well as after adding 0 .0100 moles of sodium hydroxide.
00:11
So our equation for buffers is as follows, ka times your acid info over your base info.
00:19
When we have just the weak acid and its conjugate base like we have here, so this is our weak acid, this is the conjugate base, the difference is one hydrogen ion, we need the concentrations that we're plugging in.
00:37
So here we know h plus equals our ka, which is 1 .8 times 10 to the negative fourth times 0 .200 over 0 .285.
00:49
So our initial hydrogen ion concentration is 1 .8 times 10 to the negative fourth times 0 .2 divided by 0 .285, which is 1 .26 times 10 to the negative fourth molar, and we take the negative log of that, and that gives us our ph.
01:14
So the ph initially would be at 3 .90.
01:19
Okay, after we add 0 .0100 moles of sodium hydroxide, then we would expect that this ph should go up, not a whole lot, but we're adding base to it, so that should increase the ph.
01:34
So same equation, h plus equals our ka, 1 .8 times 10 to the negative fourth.
01:40
This time we need our moles of our acid and base.
01:47
So we're going to take that 270 times 0 .2, which gives us 54 millimoles.
01:54
Okay, i'm going to change that.
01:55
0 .27 times 0 .2 is 0 .054, so let me change that.
02:02
0 .054 moles of our acid, and then 0 .285 times 0 .270 is 0 .0770.
02:20
Okay, and we're adding 0 .01 of the base, so put the plus on the bottom, minus on the top.
02:33
So 0 .054 minus 0 .01 divided by 0 .077 plus 0 .01 times 1 .8 times 10 to the negative fourth.
02:46
I hit something wrong there.
02:49
So 0 .054 times 1 .8 times 10 to the negative fourth gives us h plus equal to 9 .10 times 10 to the negative fifth molar.
03:10
Take the negative log of that, and your ph is now 4 .04.
03:16
So it did go up a little bit...
