1.

Solubility product of M(OH), is 10^(-14). What should be the concentration of M^(2+) in 0.1 M solution of NH_(4)OH, if NH_(4)OH gets 10% ionised?

Answer»

`10^(-10)`
`10^(-5)`
`10^(-12)`
`10^(-4)`

Solution :`K_(sp)=[M^(+2)][OH^(-)]^(2)`
`[OH^(-)]=(10)/(100)xx0.1=0.01`
`[M^(+2)]=(K_(sp))/([OH^(-)]^(2))=(10^(-14))/([0.01]^(2))=10^(-10)`


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