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pH of saturated solution of Ba(OH)(2) is...

`pH` of saturated solution of `Ba(OH)_(2)` is `12`. The value of solubility product `(K_(sp))` of `Ba(OH)_(2)` is

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pH of solution `=12 , [H^(+)] = 10^(-12) M`
`[OH^(-)] = (K_(w))/[[H^(+)]]= (10^(-14) M^(2))/(10^(-12)M)= 10^(-2)M`
`Ba(OH)_(2) hArr underset([s]](Ba^(2+)) + underset[[2s]](2OH^(-))`
`2S = 10^(-2) M, S= 0.5 xx 10^(-2) M`
`K_(sp) =[S] [2S]^(2) =4S^(3) =4 xx (0.5 xx 10^(-2))^(3)`
`=5.0 xx 10^(-7) M^(3) (Mol L^(-1))^(3)`
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