00:01
First of all, let's have a look what all is given in the problem.
00:04
Given is, volume of naoh is given, volume of naoh it is given as 45 ml.
00:13
Similarly, molarity of naoh is given, molarity is given as 0 .100 molar.
00:21
Next is volume of ch3cooh is given, volume of ch3cooh is given, it is given as 50 ml and molarity of acetic acid that is ch3cooh is given, that is given as 0 .1 molar.
00:40
Ka is also given, that is 1 .8 multiplied with 10 to the power minus 5.
00:45
And what we are supposed to find out is, we need to find the ph of the solution.
00:51
We have to find the ph of the solution.
00:53
So, let's try to find out the ph of the solution.
00:58
See, millimoles of base we can calculate.
01:01
We can calculate millimoles of base that is our naoh, that is molarity into volume, that is 45 into 0 .1.
01:11
So, it will come around 4 .5 millimoles.
01:15
Similarly, we can calculate for acetic acid also.
01:19
Millimoles of acid, it will be like 50 multiplied with 0 .1.
01:24
So, it will be 5 .0 millimoles...