Question

Chromium(III) hydroxide has Ksp 1.6 x 10-30. What is the molar solubility of Cr(OH)3 in a solution whose pH is maintained at 6.00?

Answer 1

Step-by-step explanation:

Chemical reaction equation for
`Cr(OH)_(3)` is as follows.

`Cr(OH)_(3) \rightarrow Cr^(3+) + 3OH^(-)`

As it is given that pH is 6. So, the concentration of hydrogen ions will be as follows.

pH =
`-log [H^(+)]`

antilog (-6) =
`[H^(+)]`

`[H^(+)] = 10^(-6)`

and
`[OH^(-)] = (10^(-14))/([H^(+)])`

=
`(10^(-14))/(10^(-6))`

=
`10^(-8)`

Now, let us assume that the solubility is "s". Therefore,

`Cr(OH)_(3) \rightarrow Cr^(3+) + 3OH^(-)`

s
`10^(-8)`

Therefore, calculate the value of
`K_(sp)` for this reaction as follows.

`K_(sp) = [Cr^(3+)][10^(-8)]^(3)`

`1.6 * 10^(-30) = s * (10^(-8))^(-30)`

s =
`1.6 * 10^(-6) mol/L`

Thus, we can conclude that the molar solubility of
`Cr(OH)_(3)` is
`1.6 * 10^(-6) mol/L`.