00:01
In this setup, we are titrating 30 milliliters of 0 .120 molarity, thioacetic acid, with 0 .06 molarity sodium hydroxide.
00:11
If you needed to know what volume of base it takes to get to the equivalence point, you could use this equation m -a -v -a equals m -b -b -b.
00:31
M .a stands for the malarity of the acid.
00:35
That subscript a stands for the acid.
00:37
So that would be 0 .120 malarity.
00:41
Va stands for the volume of the acid, and that should be in liters.
00:47
So that would be the 30 .0 milliliters divided by a thousand milliliters.
00:53
So that would be 0 .030 liters.
00:57
M .b on the other side is the molarity of the base.
01:03
That would be our 0 .060.
01:05
And what we would be solving for in this problem is the volume of the base that's symbolized by v -subscript b.
01:13
That would be the volume of base in liters to reach the equivalence point.
01:19
So if we fill in this equation and solve for vb, it would look like this.
01:24
M .a, we would put in 0 .120 molarity.
01:31
For va, again, that's our 30 milliliters, but we want that in liters.
01:36
So we want to divide that by 1 ,000.
01:39
That would be 0 .03 -0 -0 liters.
01:44
That would be equal to mb, the molarity of our base, 0 .060 -mularity, times vb, which, is what we're solving for.
01:57
Simplifying this expression, solving for vb, we find vb is equal to 0 .060.
02:06
That would be liters, or if you need it in milliliters, it would be 60 milliliters.
02:14
So either of those answers would be correct, depending on which unit you wanted in.
02:20
Now, if you needed to know what volume it would take to get to the halfway point, or halfway to the equivalence point, you would just take either of these values, depending if you want it in liters or milliliters, and multiply it by one or divide it by two.
02:38
So if it takes 0 .060 liters to get to the equivalence point, it would take 0 .060 liters.
02:49
It would take half of that volume to get to the halfway point to the equivalence.
02:55
That would be 0 .030.
02:59
Or 30 milliliters to get to the halfway point.
03:06
Now if you're halfway to the equivalence point, there's an interesting expression that you should know.
03:14
Halfway to the equivalence point, i'll write that out, halfway to the equivalence point, the ph equals the pca.
03:28
So the p .h equals the p .k .a.
03:32
And p -k -a is the negative log of k -a.
03:38
So if you know the k -a value for this acid, you could take the negative log of it, and that would be the ph.
03:46
But it's the ph only at that one specific point during the titration when you are halfway to the equivalence point.
03:53
And this problem would be when we've added 30 milliliters.
03:56
Now the k .a.
03:59
Of thioacetic acid is known.
04:02
So we can solve this.
04:04
The k .a.
04:06
Value for thioacetic acid is 4 .68 times 10 to the negative 4.
04:15
So if we take the negative logarithm of that, we'll get the ph at the halfway point.
04:21
And solving that gives us a value for the ph equal to 3.
04:29
3 .3...