Q 2643523443.     Give : `Zn(OH)_2(s) ⇋ Zn(OH)_2 (aq) ; K_1 = 10^-6`

`Zn (OH)_2 (aq) ⇋ [ Zn(OH)]^+ + OH^- ; K_2 = 10^-7`

`[Zn(OH)]^+ (aq) ⇋ Zn^(+2) + OH^-` `; K_3 = 10^-4`

`Zn(OH)_2 (aq) + OH^(-) ⇋ [Zn(OH)_3]^(-) ; K_4 = 10^3`

`[Zn(OH)_3]^(-) (aq) + OH^(-) ⇋ [Zn(OH)_4]^(2-) ; K_5 = 10`

Find out the negative of logarithm of the solubility of solid `Zn(OH)_2` at 25°C at `pH =6` .Consider `Zn(OH)_2` makes saturated solution at 25°C.

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