`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. `8 xx 10^(-2)M`
B. `8 xx 10^(-11) M`
C. `1.6 xx 10^(-11) M`
D. `8 xx 10^(-5) M`

Answer

B

Explanation

`{:(CH_(3)NH_(2),+,HCl,rarr,CH_(3)NH_(3)^(+)Cl^(-)),(0.1,,0.08,,0),(0.02,,0,,0.08):}`
As it is a basic buffer solution.
`pOH = pK_(b) + log.(0.08)/(0.02) = - log5 xx 10^(-4) + log 4`
`= 3.30 + 0.602 = 3.902`
`pH = 14 - 3.92 = 10.09`,
`[H^(+)] = 7.99 xx 10^(-11) = 8 xx 10^(-11) M`