1.

pH of a saturated solution of Ca(OH)_2 is 9. The solubility product (K_(sp)) of Ca(OH)_2……..

Answer»

`0.5 times 10^-15`
`0.25 times 10^-10`
`0.125 times 10^-15`
`0.5 times 10^-10`

Solution :`Ca(OH)_2 leftrightarrowCa^(2+)+2OH^-`
Given that pH=9
`POH=14-9=5`
`[pOH=-log_10[OH^-]]`
`therefore[OH^-]=10^(-pOH)`
`[OH^-]=10^-5 M`
`K_(SP)=[Ca^(2+)][OH^-]^2`
`=10^-5/2 times (10^-5)^2=0.5 times 10^-15`


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