In a solution of `0.04M FeCl_(2)` and `0.01 M FeCl_(3)`, how large may be its pH of without being precipitation of either `Fe(OH)_(2)` or `Fe(OH)_(3)` ?
[Given `K_(sp) Fe(OH)_(2) = 16 xx 10^(-6)` and `K_(sp) Fe(OH)_(3) = 8 xx 10^(-26)`]

Options

A. `5.7`
B. `6.3`
C. `8.3`
D. `10.7`

Answer

B

Explanation

For the precipitation of `Fe(OH)_(2)`
`K_(sp) = [0.04][OH^(-)]^(2) = 16 xx 10^(-6)`
`[OH^(-)] = 2 xx 10^(-2)`
and for `Fe[OH]_(3)`
`K_(sp) = 0.02 xx [OH^(-)]^(2) = 16 xx 10^(-6)`
`[OH^(-)] = 2 xx 10^(-2)`
For `Fe(OH)_(3)` the `[OH^(-)]` ions are required less
`pOH = 8 - "log" 2`
`pH = 6 + "log" 2 = 6.3`