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 ?

Options

A. `8xx10^(-2)M`
B. `8xx10^(-11)M`
C. `1.6xx10^(-11)M`
D. `8xx10^(-5)M`

Answer

B

Explanation

`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`