00:01
Is happening we are looking at this reaction here n a h2 p o4 dot 2 h2 which is water of crystallization are reacting to give n a h p o 4 this is compound a let us assume and this compound b and along with this h3 o positive and h2 o we are getting and if we see the molecular weight of this asset it is we can say 156 and if we see molecular weight it is 142 gram approximately so what is happening here here we are having phosphate buffer and it is 1 liter with 0 .1 molarity so this is a phosphate buffer and ph it is 7 so what will be happening along with this ph we are having a pk value 6 .86 so we will be using henderson -hazelbeck equation anderson hazelbeck equation so it says ph it is equal to p -k plus log concentration divided by the acidic substances or we can say basic substances.
01:11
So from here, from here what we can say, 7 minus 6 .86, that will be equal to the log.
01:19
What is the salt here? n -a -2 -h -p -o -4 divided by the substance n -a -h -2 p -o4.
01:30
2h2.
01:31
So from here if we find the ratio of this n -a -h, n -a -2, h -p -o4 divided by the n -a -h2 -p -o -4.
01:45
That will be equal to 1 .38.
01:48
So this is the answer here in this case.
01:52
And also we see concentration of buffer.
01:56
The buffer solution which we are having, it is 0 .1 molar.
02:00
So what we can say here, here we can enforce these two things.
02:06
One base, conjugate base divided by asset with 1 .38 and conjugate base plus buffer it is 0 .1.
02:14
So we can substitute these values and from here 1 .38 a dad will be adding a and 0 .1...