00:01
Hi, so for this question we're going to first look at in general what is going to happen when we reach the equivalence point for a weak acid.
00:10
So at the equivalence point for any weak acid that we're dealing with, we're going to generate a - plus water.
00:20
At the equivalence point we have added the same amount of concentration of each of them.
00:27
So that equilibrium, i'm going to omit the change because here we're going to assume that this goes to zero, this goes to zero, even though that is not completely true.
00:35
But we're generating an amount of c0 of a-.
00:40
So then we can use this quantity in order to find the ph as a - reactions with water are going to generate ha plus oh-.
00:59
And if we look at the ice table for this, what we're going to get is c0.
01:08
We're going to get that a certain amount minus x is going to dissociate in water.
01:15
So it's going to react with the water.
01:18
So plus x, plus x.
01:21
And at equilibrium we're going to get c0 minus x, x, and x.
01:25
If you look at the weak acids that you're dealing with, all of them are monoprotic weak acids.
01:31
So i can follow this structure for the equations.
01:35
This means that this reaction, first of all, corresponds to kb.
01:39
This is not the ka of the acid.
01:41
And this is equal to concentration of ha, concentration of oh - divided by concentration of a-.
01:50
So replacing the values at equilibrium for the concentration, this is x squared divided by c0 minus x.
01:57
So if we want to solve for x, this is going to be equal to kb times c0 minus x square root.
02:14
We can solve for x directly here by solving the quadratic expression that comes from this.
02:23
But in general, and in most cases, since the value for kb is going to be a small value, that means that the value for x is going to be in general much smaller than the value for c0.
02:35
So the approximation that we can make is that x is going to be much smaller than c0.
02:41
And this simplifies our equation because if that is true, c0 minus x is equal to c0.
02:48
So our equation over here becomes much simpler, right? and remember that this would be the concentration of oh-.
03:04
So after we have solved for this value, for all of the acids, what we can get is get the poh and then get ph as 14 minus poh.
03:19
So in all of the cases that we're going to deal with, we are starting with a 0 .1 molar solution that then it is going to react with 0 .08 molar of naoh.
03:40
But now we are going to have a difficulty because we are not given the volume of my initial solution, we're just given the concentration.
03:48
So in what other way can i calculate the ph for this solution? well, the other method that we can use if we're not given those quantities is we are still going to solve for x, but what we can use is henderson -hasselbalch equation because this x is also going to be equal to the concentration of ha.
04:10
So the other way to do it is finding the ph at the equivalence point as pka plus logarithm concentration of a minus divided by concentration of ha.
04:28
So this is going to become pka plus logarithm of x divided by c0.
04:35
So those are the two methods that we are going to have.
04:43
Unfortunately for both of them, in order to calculate the concentration that we are going to have at the equivalence point, we require the volume of the initial solution.
04:54
In any other way, there is no way for us to know.
04:57
So what i'm going to do is, in order to perform both of these methods, i'm going to assume that i'm going to have one liter of solution.
05:11
So this liter of solution, since we're going to have a one -to -one correspondence in between the weak acid and the strong base that we're going to employ, the amount of moles that we're going to get is going to be one liter of solution times my concentration of my weak acid.
05:34
It is going to be equal to the volume that i'm going to add of the strong acid, of oh minus, times the concentration of my strong base.
05:49
So the volume of oh minus that would be required is going to be 0 .1 divided by 0 .08.
05:59
So 1 .25 liters.
06:02
So under this assumption, the volume at equivalence point is going to be equal to 2 .25 liters...