00:01
Here in this problem given ph of a solution of magnesium hydroxide is 11 .96 and ksp, that is solubility product constant of m .g .o .h.
00:14
Twice is given air.
00:15
And we have to find the molar solubility of m .g .o .h.
00:19
Whole twice.
00:20
Now here, ph of the solution given 11 .96.
00:27
So we can find out the poh.
00:31
Since ph plus poh is equal to 14.
00:37
And we have the ph.
00:39
So we can find out poh 14 minus ph, which is 11 .96.
00:46
So the poh of this solution is 2 .04.
00:56
Now we can find out which minus ion concentration from here.
01:01
What is poh is that is negative log oh minus ion concentration and that is 2 .04.
01:11
So we can write log of concentration of o .h minus ion is negative 2 .04.
01:21
So concentration of o .h minus ion 10 to the power negative 2 .04 and that is 0 .0 .0 .0.
01:31
0 .091 .2 molar.
01:36
Now, we have the concentration of o -h -1 -2 -ions.
01:40
Now, let's write the solubility equilibrium for m -g -o -h -to -i in its saturated solution.
01:47
M -g -o -h -h -2 -2 -1 -ions.
01:51
In -aquoise solution produces m .g2 plus and 2 -o -h -1 -ions.
02:00
For each mole of m .g...