What is the `pH` at which `0.01MCo^(2+)` ions in solution precipiate down as `Co(OH)_(2)? K_(sp)` of `Co(OH)_(2)` is `2.5 xx 10^(-16)`.

Explanation

`Co(OH)_(2) hArr Co^(2+) (aq) + 2OH^(Θ) (aq)`
`[Co^(2+)] [overset(Θ)OH]^(2) = K_(sp) = 2.5 xx 10^(-16)`
`(0.01) [overset(Θ)OH]^(2) = 2.5 xx 10^(-16)`
`[overset(Θ)OH]^(2) = (2.5 xx 10^(-16))/(0.01) = 2.5 xx 10^(-14)`
`[overset(Θ)OH] = sqrt(2.5 xx 10^(-14)) = 1.58 xx 10^(-7)M`
`pOH =- log 1.58 xx 10^(-7) = 6.8`
`pH + pOH = 14` and `pH = 14 - pOH = 14 - 6.8 = 7.2`