Question

Iron (III) hydroxide is a form of rust. What is its solubility, (s) in a solution at pH 6.50?

Answer 1

solubility of
`Fe(OH)_(3)` is
`2.0* 10^(-15)(M)`

Step-by-step explanation:

Solubility equilibrium of Iron(III) hydroxide:

`Fe(OH)_(3)\rightleftharpoons Fe^(3+)+3OH^(-)`

If solubility of
`Fe(OH)_(3)` is S (M) then concentration of
`Fe^(3+)`,
`[Fe^(3+)]` is S (M)

We know, solubility product,
`K_(sp)=[Fe^(3+)][OH^(-)]^(3)`

pH=6.50

or, pOH= 14-6.50

or, pOH = 7.50

or,
`[OH^(-)]=10^(-7.50)`

So,
`S = [Fe^(3+)]=(K_(sp))/([OH^(-)]^(3))` =
`(6.3* 10^(-38))/((10^(-7.50))^(3))(M)=2.0* 10^(-15)(M)`

Hence solubility of
`Fe(OH)_(3)` is
`2.0* 10^(-15)(M)`