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

Answer

A

Explanation

The weak base `[CH_(3)NH_(2)]` on reaction with strong acid `[HCl]` gives a salt of weak base with strong acid `[CH_(3)NH_(3)Cl]`:
`{:(,CH_(3)NH_(2)+HClrarrCH_(3)overset(+)(N)H_(3)Cl^(-)),("Initial (moles)"," 0.1 0.08 0"),("Change (moles)"," -0.08 -0.08 +0.08"),("Final (moles)",bar(" 0.02 0 0.08 ")):}`
Note that HCI is the limiting reagent. The final solution acts as a basic buffer due to the presence of weak base and its salt with a strong acid. According to the Henderson equation,
`pOH=pK_(b)+log.(C_("salt"))/(C_("base"))`
Since the volume of solution is one litre, the number of moles is equal to the concentration. Thus,
`pOH= - log(5xx10^(-4))+log.(0.08)/(0.02)`
`=(-log 5)+4log 10+log 4`
`= -0.699+4+0.602`
`=3.903`
At 298 K,
`pH+pOH=pK_(w)=14`
`pH=14-pOH`
`= 14-3.903`
`=10.097`
`:. -log C_(H^(+))//mol L^(-1)= 10.097`
Taking antilog, we get
`C_(H^(+))=8.0xx10^(-11)`