`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. `8xx 10^(-2)`
B. `2xx10^(11)`
C. `1.23 xx 10^(-4)`
D. `8xx 10^(-11)`

Answer

D

Explanation

`{:(CH_(3)NH_(2),+,HCIto ,CH_(3)NH, +CI^(-)),(0.1"mol",,0.08"mol",,0),(0.02"mol",,0,,0.08):}`
`pOh =pK_(b) + log .("salt")/("base")`
`=- log 5 xx 10^(-4) + log .(0.08)/(0.02)`
`=(4 -log 5) +log 4`
`=4- 0.6989 + 0.6020 =3.903`
`pOH =3.903, pH =10.0967`
`[H^(+)] = 8xx 10^(-11)`