1.

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

Answer»

`0.5xx10^(-15)`
`0.25xx10^(-10)`
`0.125xx10^(-15)`
`0.5xx10^(-10)`

Solution :`0.5xx10^(-15)`
`Ca(OH)_(2)hArr Ca^(2+)+2OH^(-)`
Given that `pH=9`
`pOH=14-9=5`
`[pOH=-log_(10)[OH^(-)]]`
`:.[OH^(-)]=10^(-pOH)`
`[OH^(-)]=10^(-5)M`
`K_(SP)=[Ca^(2+)][OH^(-)]^(2)`
`=(10^(-5))/(2)XX(10^(-5))^(2)=0.5xx10^(-15)`


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