`a) Fe^{3+}+H_2O->Fe(OH)^{2+}+H^+`

CB: `0,01-x.      -> x.   -> x(M)`

`->x^2/(0,01-x)=6,3.10^{-3}->x=0,0054->pH=2,27`

`b) ` để bắt đầu kết tủa `Fe(OH)_3` thì `K_s=[Fe^{3+}].[OH^-]^3`

`C_{Fe^{3+}}=[Fe^{3+}]+[Fe(OH)^{2+}]`

`=> 0,01=[Fe^{3+}].(1+K_a/[H^+])`

`=> 0,01=(6,3.10^{-38})/[OH^-]^3.(1+6,3.10^{-3}.10^{14}.[OH^-])`

`=> [OH^-]=2,54.10^{-12}(M)=>pH=2,4`

c) kết tủa `Fe(OH)_3` hoàn toàn khi `[Fe^{3+}]+[Fe(OH)^{2+}]=10^{-6}`

`=> (6,3.10^{-38})/[OH^-]^{3}.(1+6,3.10^{-3}.10^{14}.[OH^-])=10^{-6}`

`=> [OH^-]=2.10^{-10}(M)=> pH=4,3`