1.

0.1 mol of CH_(3)NH_(2)(K_(b)=5.0xx10^(-4)) is mixed with 0.08 mol of HCl and diluted to one litre. What will be the H^(+) ion concentration in the solution ?

Answer»

`8xx10^(-2)M`
`8xx10^(-11)M`
`1.6xx10^(-11)M`
`8xx10^(-5)M`

SOLUTION :`underset(0.02)underset(0.1)(CH_(3)NH_(2))+underset(0)underset(0.08)(HCl)RARR underset(0.08)underset(0)(CH_(3)NH_(s)^(+)Cl^(-))`
For basic buffer
`pOH=pK_(b)+"LOG"(0.08)/(0.02)`
`pK_(b)=-log(5xx10^(-4))=3.30`
`therefore pOH=pK_(b)+0.602`
`=3.30+0.602=3.902`
`pH=-log(3.902)=10.09`
`[H^(+)]=-log(10.09)=7.99xx10^(-11)`
or`=8xx10^(-11)M`


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