1.

The salt Zn(OH)_(2) is involved in the following two equilibria, Zn(OH)_(2)(s) hArr Zn^(+2) (aq) + 2OH^(-) (aq), K_(sp) Zn(OH)_(2) = 1.2 xx 10^(-17) Zn ( OH)_(2)(s) + 2OH^(-) ( aq) hArr [Zn(OH)_(4)]^(2-)(aq), K_(c ) = 0.13 Calculate the pH at which the solubility of Zn(OH)_(2) be minimum. Also find the solubility at this concentration.

Answer»

Solution :The solubility of `Zn(OH)_(2)` will be given as
`s = [ Zn^(+2) ] + [ Zn(OH)_(4) ] ^(2-)= ( K_(sp))/([OH^(-) ] ^(2) ) + K_(c ) [ OH^(-) ] ^(2-)`
The MINIMUM SOLUBILLITY will be obtained by setting
`( ds )/( d[OH^(-) ] ) = 0`
HENCE ` ( - 2K_(sp))/( [OH^(-) ] ^(3)) + 2K_(c ) [ OH^(-) ] = 0`
`[OH^(-) ] = ((2K_(sp))/( 2K_(c )))^((1)/(4)) = ((1.2 xx 10^(-17))/(0.13))^(1//4) = 9.8 xx 10^(-5)`
`PH = 14-4.01 = 9.99`
`s = ( 1.2 xx 10^(-17))/(( 9.8 xx 10^(-5))^(2)) + ( 0.13 ) ( 9.8 xx 10^(-5))^(2) = 2.5 xx 10^(-9) M`


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