00:01
Everyone, so here in this question we're given that calculate the h -positive concentration and the ph for the following solutions.
00:10
Okay, so we're given 0 .0 .08 molar, h -no3.
00:17
Right.
00:18
So here, h -positive concentration is 0 .018.
00:25
1 -8.
00:26
And ph, right, it is equal to minus of log.
00:33
Of each positive that is minus of log of 0 .018 which will give us 1 .7 so okay then number b we have 0 .037 molar k -oh so here oh it's negative concentration is 0 .037 7.
01:04
Each positive concentration will be 10 ways to 2 .15 divided by o .h.
01:11
Negative concentration because we know h.
01:13
Positive plus o .h negative is equal to 10 based 3 % minus 14.
01:17
Right? so 10 ways to dipur minus 14 divided by 0 .037.
01:23
So this will give us 2 .70 multiplied by 10 way to the power minus 13.
01:30
And p .h will be equal to minus or log is positive concentration.
01:38
So this will give us minus of log 2 .70 multiplied by 10 raised to the power minus 13, which will be equal to 12 .57.
01:55
Or we can write to this approximately equal to 12 .6...