00:01
Hi, so over here what we have is a titration of a strong acid with a strong base.
00:07
The strong acid in question is 31 .4 milliliters of a solution that is 0 .324 molar of hbr.
00:22
And the strong base that we're going to be titrating with has a concentration that is exactly the same as the one of my strong acid, that is 0 .324 molar.
00:36
So we want to know the ph at different points of my titration.
00:40
The first one is before addition.
00:47
Here the ph is going to be given directly by minus the logarithm of the concentration of h2o+.
00:55
Since this is a strong acid, this concentration is the same as the concentration of the acid, so minus logarithm of 0 .324 molar.
01:08
So the ph is, is equal to 0 .49 at the very beginning of my titration.
01:15
Next up, in the first step, we're going to add 15 .7 milliliters of naoh.
01:25
And now what we want to know is, first of all, what is going to be my final volume.
01:31
The final volume is going to be the volume that i added plus the volume that i had initially, that is 31 .4 milliliters.
01:40
So this final volume is 47 .1 milliliters or 47 .1 times 10 to the minus 3 liters.
01:56
Just to avoid any possible mistake when converting.
02:00
Next up we want to know the amount of mole of hbr that i'm going to have in my solution.
02:05
And the way to do it is remembering that molarity is equal to amount of mole divided by volume.
02:10
So amount of mole will be molarity times the volume that i have.
02:14
And the volume needs to be in liters because molarity is expressed in mole per liter.
02:20
So we have 0 .324 mole per liter times my volume that is of hbr, it is 31 .4 times 10 to the minus 3 liters.
02:42
And the result is 0 .01 mole of hbr.
02:52
Remember that this amount of mole of hbr is going to be the same as the amount of mole of hcl that i have added into my solution.
03:00
Now we can repeat the calculation with amount of mole of naoh that we have.
03:08
But you can notice quite easily that the volume added is exactly one half, right? and the concentration is the same in both cases.
03:18
So the amount of mole added will be directly one half of the amount of mole of hbr that i added.
03:28
Okay? so in this case the limiting reagent is naoh.
03:40
Recall that the reaction that we are studying is hbr plus naoh.
03:45
All of this in aqueous media are going to generate nabr plus water.
03:52
Properly these are going to be percent of ions but this is a simplified way of looking at the titration.
03:59
And all of this is in aqueous media except for water that is a liquid.
04:05
Okay? so since naoh is the limiting reagent the amount of hbr that is going to be left is amount initial minus amount that reacted.
04:17
This is 0 .01 initial minus the amount that reacted that is 0 .005.
04:24
Gives us a total amount of 0 .005 mole of hbr that is going to be left in solution.
04:33
With this we can calculate the concentration.
04:38
Recall again this is the same as the concentration as it of h2o plus.
04:43
And this is going to be 0 .005 mole divided by my total volume.
04:49
The total volume is 47 .1 times 10 to the minus 3.
05:02
So the result that we're going to get is 0 .106 molar...