00:01
Collate solubility of aluminum hydroxide in three different media.
00:04
The first one is pure water, and here just solubility is dictated by the solubility product only, and the solubility product only equals to concentration of aluminum times concentration of oh -h -minus power by 3.
00:20
So let's rewrite that in terms of solubility, the saturation of aluminum 3 -plus is s, concentration of oh minus is 3 s and it's also powered by 3 so the overall solubility is 27 times s powered by 4 then s equals to solubility product divided by 4 and power by oh sorry divided by 27 and power by 0 .25 let's calculate this number it equals to 5 .2 times 10 power by negative 9 molar.
01:37
Okay, that's answer to question a.
01:39
Now let's move on to question b.
01:41
Here we have to calculate in a solution with a buffer ph of 6 .0.
01:48
So if the ph is 6 .0, then poh is 14 minus 6, which is 8.
02:01
8 .0 and then the concentration of h minus equals to 10 power by negative logarithm of 8.
02:15
Let's calculate this number.
02:18
Oh sorry, 10 power by negative 8.
02:26
Yeah, we don't need to calculate anything.
02:30
We already know the concentration of oh h minus.
02:34
Now let's calculate the solubility of aluminum here.
02:41
The solubility of alumina is dictated by the similar equation, so it's alumina plus plus oh h minus...