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

Answer

D

Explanation

` {:(,CH_(3)NH_(2),+,HCl,to,CH_(3)NH_(3)^(+)Cl^(-)),("Intial",0.10,,0.08,,0),("Final",0.02,,0,,0.08):}`
`pOH = pK_(b) + log. (["Salt"])/(["Base"])`
` = - log 5 xx 10^(-4) + log . (0.08)/(0.02) = 3.903`
`pH = 10.0967`
`[H^(+)] ="antilog " (-10.0967) = 8 xx 10^(-11)`