00:01
In this question it is saying that there is a solution which is prepared by dissolving sodium succinate and disodium succinate by dissolving these two in one liter to produce a solution with ph value 6 .0 and the total concentration of the solution is 50 millimolar and the p .k value of 6 .64.
00:29
You need to calculate the amount of our mass of both compound, means mass of sodium suction and mass of disodium suction in solution.
00:47
Okay, so this is our task.
00:52
Here it is given at the total concentration.
00:57
First of all, we need to consider sodium succinate as a we are taking it as a pic acid and disodium succinate as conjugate pair as a minus so here it is a given as ha haeus a means the concentration of the solution is total concentration means h a plus a minus it is given as 50 millimol and we will directly we will convert it into molar so we will get it like 0 .05 molar okay here it is a millimolar so this is in molar concentration and from this we can also say concentration of a minus means di -sodium succinate will be 0 .05 minus concentration of sodium succinate.
02:25
Now we will apply the henderson -hassel -valk equation.
02:32
Now we know we will apply henderson -henderson.
02:43
Hesel ball equation and this equation is given as p h is equal to pk a plus log a minus divided by h a okay so here ph is 6 .0 pka is given as 5 .64 p h is 6 .0 pkh 5 .64 plus log.
03:27
Now, a minus or h .a minus is not given.
03:35
So, when we solve this, we will get the value of a minus divided by h .a.
03:45
And this will come nearly 2 .29...