1.

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

Answer»

`3.3 xx 10^(-7)`
`5.0 xx 10^(-7)`
`4.0 xx 10^(-6)`
`5.0 xx 10^(-6)`

Solution :`{:(Ba(OH)_(2),HARR,Ba^(2+)+,2OH^(-)),(,,s,2S):}`
`[OH^(-)] = 10^(-2)`
`2s = 10^(-2)`
`s = (10^(-2))/(2) K_(sp) = 4s^(3) = 4 xx ((10^(-2))/(2))^(3) = 5 xx 10^(-7)`


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