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

Answer

B

Explanation

`{:(CH_(3)NH_(2)+,HCIrarr,CH_(3)NH_(3)^(+)CI),(0.1,0.08,0),(0.02,0,0.08):}`
This is basic buffer solution.
`[OH^(-)]= K_(b)xx(["Base"])/(["Salt"])=5xx10^(-4)xx(0.02)/(0.08)`
`= 1.25xx10^(-4)`
`:. [H^(+)]= (10^(-14))/([OH^(-)])= (10^(-14))/(1.25xx10^(-4))`
`= 8xx10^(-11)M`