00:01
For this question we will require two formula.
00:03
One is to calculate the ph which can be calculated as minus log of concentration of h positive.
00:11
And then we need to calculate the poh which would be equal to 14 minus ph.
00:18
So let's go on to part a of the question.
00:22
Part a of the question they've given us 0 .03 molar hcl which can be written as 3 .3.
00:26
H .c .l which can be written as 3.
00:30
10 raise to power minus 2 molar hcl.
00:33
As hcl is a strong acid, therefore concentration of h positive over here would be 3 into 10 raise to power minus 2.
00:43
Let's calculate the ph which would be minus log of concentration of h positive which is 3 into 10 raise to power minus 2.
00:54
Finding the log of this would give us 1 .523.
01:00
And the value of poh would be equal to 14 minus 1 .523 which becomes equal to 12 .477.
01:18
This is the answer to part 1.
01:21
Let's go on to part 2 of the question...