0:00
Hello.
00:01
So in this question, when this n .a .2 .h .o .4, when it is di -basic, then it is 141 .95 gram per mole.
00:17
This is the molar mass.
00:19
So now, see, this 1m of n .a2 h .p .o4, it consists of one, in that case, 141 .95 gram per mole of, this.
00:36
Same molecule that is n -a2hp -o -4 it is present in 1 liter of solution.
00:46
So therefore when 0 .1m of n -a2 -h -o -4 then 141 .95 multiply by 0 .1 gram per mole of this same molecule it is present in 1 liter of solution.
01:05
So here as we can see that 1 liter solution contains this much of a molecule.
01:12
Molecules it is present here and while in 50 ml so this will end 0 .1m of n a 2 h of 4 it is present in 141 .95 multiply by 0 .1 in 50m it is present multiplied by 50 50 and divided by this much is the milliliter but in gram we will divide it by thousand so this much of gram of of n -a -2 -h -o -4, it is present in 50 ml of solution because 1 -l is equal to 1 ,000 ml.
01:57
So now we will, on solving this, we will have it, that is, 0 .709 gram of n -a -2 -h -p -o -4, when it is dibasic.
02:15
When this molecule it is monobacic.
02:19
So, 1m of n .a2hpo4, it is having 120 gram per mole or in 1 gram 120 gram it is there.
02:36
So simply we can say that 120 gram of n .a .h2 .p .o4 it is present.
02:51
So, in 1000 ml or 1 liter solution, so 0 .1m of n .h2, 1ah2, p .o4, it is present in 120 multiply by 0 .1 gram in 1 liter solution, and in, that is 120 multiply by 0 .5, multiply by 0 .5 ,000, multiplied by 1 ,000 gram.
03:27
Is present in 50 m l of solution so therefore on solving this we will have the value that is 0 .6 gram of n a h 2 po 4 now now from h heshehehehehehe equation we know that that is p h is equal to pka plus log of base divided by that is concentration of acid.
04:04
So ph of the buffer solution here this ka is the acid dissociation constant.
04:11
So now on putting the value here that is 7 .4 is equal to 7 .2 plus log of this base and acid it is there.
04:29
So on solving this the the ratio of this concentration of this base by acid, it is equal to 10 raise to the power.
04:41
It will become 10 raise to the power 0 .2.
04:44
That is equal to 1 .58.
04:46
Let us suppose this is 1.
04:49
Here, this sodium phosphate dibasic and sodium phosphate monobasic.
05:02
They both are acid -base conjugate pair.
05:09
Acid -based conjugate pair.
05:15
So therefore we know that, that is, 1m is equal to number of moles, number of moles divided by, or we can say that 1m, that means number of moles per liter...