Question #885cc

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Question #885cc

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Answer 1 · 2015-04-27T10:03:22

$[O H^{-}]$ $[OH^-]$ = $10^{- 5}$ $10^-5$

From $K s p$ $Ksp$ of $C u {(O H)}_{2}$ $Cu(OH)_2$ , it starts precipitating at:
$2.2 \cdot 10^{- 20}$ $2.2*10^-20$ = $[C u^{2 +}]$ $[Cu^(2+)]$ * ${[O H^{-}]}^{2}$ $[OH^-]^2$
$[O H^{-}]$ $[OH^-]$ = $\sqrt{0.22 \cdot 10^{- 20}}$ $sqrt (0.22*10^-20)$ = $0.46 \cdot 10^{- 10}$ $0.46*10^(-10)$

and from $K s p$ $Ksp$ of $M {g (O H)}_{2}$ $Mg(OH)_2$ , it starts precipitating at
$6.3 \cdot 10^{- 10}$ $6.3*10^-10$ = $[M g^{2 +}]$ $[Mg^(2+)]$ * ${[O H^{-}]}^{2}$ $[OH^-]^2$
$[O H^{-}]$ $[OH^-]$ = $\sqrt{1.26 \cdot 10^{- 10}}$ $sqrt (1.26*10^-10)$ = $1.12 \cdot 10^{- 5}$ $1.12*10^(-5)$

therefore at $[O H^{-}]$ $[OH^-]$ = $10^{- 5}$ $10^(-5)$ all $C u {(O H)}_{2}$ $Cu(OH)_2$ is precipitated and all $M {g (O H)}_{2}$ $Mg(OH)_2$ is in solution

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Question #885cc

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