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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)`. |
Answer» `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` |
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