`0.1` mole of `CH_(3)CH (K_(b) = 5 xx 10^(-4))` is mixed with `0.08` mole of HCl and diluted to one litre. What will be the `H^(+)` concentration in the solution

Options

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

Answer

B

Explanation

`{:(CH_3NH_2 +, HCl to , CH_3NH_3^(+),+Cl^(-)),(0.1,0.08,0,),(0.02,0,0.08,):}`
Basic buffer
pOH=pK_b+log `0.08/0.02 pK_b=-log (5xx10^(-4))`
=3.30
`therefore pOH=pK_b+0.602`
=3.30+0.602
=3.902
pH=-log (3.902 )
=10.09
`[H^+]` =-log (10.09)
`=7.99xx10^(-11)`
`=8xx10^(-11)` M