`0.1` mole of `CH_(3)NH_(2) (K_(b)=5xx10^(-4))` is mixed with `0.08` mole of `HCl` and diluted to one litre. The `[H^(+)]` in solution is

Options

A. `8 xx 10^(-2)M`
B. `8 xx 10^(-11)M`
C. `1.6 xx 10^(-11)M`
D. `8 xx 10^(-5)M`

Answer

B

Explanation

`{:(CH_(3)NH_(2)+,HCIrarr,CH_(3)NH_(3)^(o+)CI^(Theta)),(0.1,0.08,0),((0.1-0.08),0,0.08),(=0.02,0,0.08):}`
This is basic buffer.
`[overset(Theta)OH] = (K_(b)["Base"])/(["Salt"])`
`=(K_(w))/([overset(Theta)OH]) = (K_(w)["Salt"])/(K_(b)["Base"])`
`= (10^(-14)xx0.08)/(5xx10^(-4)xx0.02) = (4)/(5) xx 10^(-10) = 8 xx 10^(-11)`.