00:01
Hello students in this question we are reacting an acid and base and we need to find the ph at the equivalent point and at the point after which 30 ml of base is added.
00:11
So volume of hno2 given is 40m and concentration of hno2 is given to us which is 0 .1 molar.
00:33
So from here we can easily get millimoles of hno2 which will be equals to the volume.
00:45
In ml multiplied by its concentration.
00:47
So 40 multiplied by 0 .1, which comes out to be 4 millimoles.
00:55
Now, concentration of base, which is k -o -h, concentration of k -oh is given to us as 0 .2 molar and ka for hno2 is given.
01:11
Ka for hno2 is given as 4 .6 multiplied by 10 power minus 4 .4.
01:22
If we write the, if we first solve the ph at the equivalence point, ph at equivalence.
01:36
At equivalence point, the millimoles of k -o -h will be equal to millimoles of hno2.
01:44
So, millimoles of k -o -h will be equal to millimoles of hno2.
01:58
From here, we can get the volume of h -n -o -h.
02:03
This implies, millimoles of k -o -h will be the volume of k -h multiplied by its concentration, which is 0 .2, which is equal to millimoles of hno2 as we calculated before it is 4.
02:15
From here we get volume of k -o -h is equal to 20 ml.
02:23
Therefore, total volume of the salt formed will be total volume will be equals to 20 of k -o -h plus 40 of hno2 in ml and this is 60 eml.
02:41
Now we have volume with us.
02:43
We have millimoles of final salt form which is 4.
02:46
Therefore concentration of salt will be equals to the millimoles which is 4 divided by the volume which is 60 and if we calculated it will come out to be 0 .067 molar.
03:11
0 .067 molar.
03:15
Now we have the concentration of salt with us so, we can easily calculate the ph, but before that we need to calculate the value of pca.
03:27
Pca is equal to minus logarithm value of ka, and the value of ka is given to us in question as 4 .6 multiplied by 10 power minus 4.
03:37
So if we calculate it, it will come out to be 3 .337.
03:44
So for finding the ph, we will apply this formula, ph, ph, ph, is equal to.
03:55
To half of pkw, half of pkw plus half of pka plus logarithm of the concentration.
04:21
So this will be the working formula for part a of the equation...