00:01
Hello students, let's do a question from ionic equilibrium.
00:06
It has been given that carbonic acid, h2, co3, was titrated against k -o -h, and we have to calculate the ph at different intervals.
00:21
Carbonic acid that was taken was 0 .02 molar and its volume was fixed every titration, that it was 25 milliliteration, and at the very first time, k -o -h was added and k -h was added, the volume of the k -o -h that was added was 30 -millimeter and its molarity was 0 .025 molar.
00:52
Besides, ka1 for the carbonic acid is mentioned as 4 .3 multiplied by 10 raised to the power minus 7 and since it is a diaprotic acid.
01:06
So, k -a -2 is equal to 5 .6 multiplied by 10 -drance to the power minus 11.
01:14
And we have to calculate when this amount of the k -o -h was added, at that instant, what was the ph of the solution.
01:25
Before asserting the ph, let's try to see the chemistry behind this reaction.
01:30
When h2 co3 was made to titrate with k -o -h.
01:37
First of all, all the h -2 -co -3 will be converted into k -h -co -3 and water will be formed.
01:49
And at the next step, k -h -c -o -3 will once again react with k -o -h and k -2 -c -o -3 plus water will be formed.
02:11
In this particular reaction, this is the molar ratio 1 is to 1 and once again the molar ratio is 1.
02:19
So, at the first experiment, let's name it as a, h2 -c -o -3 will first of all react with k -o -h to convert itself into kh -c -o -3 plus h2.
02:39
Number of millimoles of h2 -c -3 can be calculated by the number of mimilie -moles of h -2 -c -3 can be calculated by its molarity multiplied by its volume and this will be 0 .02 multiplied by volume was 25 it means we started with 0 .5 millimoles of h2 co3 number of millimoles of k -oh what we added was 0 .025 molarity multiplied by its volume that is 30 milliliter so this value is coming out to be 0 .75 millingoles it means we want to calculate the ph when 0 .5 millimoles of h2 co3 reacted with 0 .75 millimoles of k -oh so according to the molar ratio when the t was equal to 0 its number of millimoles were this at the time of observation that is called as the infinitive it becomes zero it remains pointative to five and it is formed point five so excess k -o -h point two five millimeters are left so the second reaction will start and kh co3 will react with k -o -h and hence k2 co3 plus 2 will be formed.
04:13
So initially this was 0 .5 this was 0 .25.
04:19
At the end of the reaction 0 .25 of this will be formed and will be remaining and k -oh will get exhausted and this will be formed 0 .25.
04:31
It means in the solution we are having acid and it's salt.
04:37
So this is a buffer solution and for the buffer solution we will apply the handle henderson's equation and henderson's equation formula is ph is equal to p -k -a plus log -off concentration of the conjugate base divided by concentration of the base acid i'm sorry so ph is equal to p -k -a and p -k -a we are given us two values but here the p -k -a we will take will be p .k .a .2.
05:24
So this will be minus log of 5 .6 multiplied by 10 raise to the power minus 11 plus log of.
05:33
Concentration of the conjugate base.
05:35
This is the conjugate base and this is 0 .25 divided by concentration of the asset.
05:43
Concentration of the asset is also 0 .25.
05:46
And at the very first instance the ph will be equal to and the value for this is coming as 10 .44 10 .44 plus log 1.
06:01
Log 1 is equal to 0 so ph will be equal to 10 .44 this was the answer for the first titration let's move to the second titration.
06:14
In the second titration that we are tagging as b, k -o -h is reacting with h2 -co -3 to form k -h -co -3.
06:27
Number of millimoles of k -o -h we have taken is 0 .025 molar and the volume is 40 -millimeter, so the number of millimole 1 .00.
06:40
For h -2 -co -3 it was constant 0 .0 -2 molar and it is 25 milliliter that is was equal to 0 .5.
06:54
So at the time is equal to 0, this is the number of millimoles of the acid in the base, and at the time of observation, this is excess, so 0 .5 will be remaining, this will get exhaust, and khco3 that will be formed is equal to 0 .5...