00:01
In this question to calculate the initial ph, according to this question a, initial ph as plus is equal to 0 .175 molarity.
00:16
So the ph is minus low 0 .175 is equal to 0 .76.
00:25
This is the value of ph.
00:29
Next, articulence point number, moles, of acid equal to number of, number of moles of, moles, space.
00:58
So, moles of acid is equal to molarity, multiply liter of solution.
01:14
So putting the value 0 .175 mole per liter into 0 .035 is equal to 0 .006 mole.
01:29
Therefore, mole of k -o -h, 0 .006 moll implies volume of k -o -h moul upon molarity, 0 .006 0 .0 .2 .0.
02:00
Mole per liter...