00:01
To begin this problem, we should find the reaction of k -hp with n -a -o -h first.
00:07
So k -h plus n -a -o -h, and the right -hand side is k -nap plus h -2o.
00:15
So we need to determine the moles of k -hp.
00:18
So the moles of k -hp is equal to the molarity of k -hp times the volume, which is going to be 0 .015 moles because we do 0 .150 mole times 100 -millimeter.
00:30
And the number of n .a .o .h added is the molarity of naoh times the volume of naoh, and we get 0 .008 moles.
00:40
Now to calculate the total moles of knap, we add these two together, 0 .015 plus 0 .008, and we get 0 .023 moles.
00:53
And then to find the number of moles of k -hp reacted or that remain, then we subtract the values, 0 .015 moles minus 0 .008 moles, and we get 0 .007 moles.
01:11
So the poh must be determined before calculating the ph.
01:17
So the poh is equal to pkb plus log of k -hp divided by k -a, and ap.
01:25
And we can calculate kb or pkb by pkb is equal to negative log of kb, and kb is equal to kw divided by ka, or 1 .0 times 10 to the negative 14, divided by 3 .9 times 10 to the negative 6...