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

Answer

D

Explanation

`CH_(3)NH_(2)+HCl hArr CH_(3)NH_(3)^(+)Cl^(-)`
When 0.1 mol of `CH_(3)NH_(2)` is mixed with 0.08 mole HCl, the resulting solution contains 0.08 mol of salt `(CH_(3)NH_(3)^(+)Cl^(-))` and `0.02 (0.1-0.08=0.02)` mol of base `(CH_(3)NH_(2))`
Using Henderson's equation,
`pOH=pK_(b)+log.(["Salt"])/(["Base"])`
`=-log 5 xx 10^(-4)+log. (0.08)/(0.02)`
`=-(bar(4).6990)+0.6021`
`=4-0.6990+0.6021=3.9031`
`pH =14-pOH =14-3.9031=10.0969`
`[H^(+)]=` antilog`(-10.09696)=8 xx 10^(-11)`