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0.1 mole of CH(3)NH(2) (K(b)=5xx10^(-4))...

`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

A

`1.6xx10^(-11)`

B

`8xx10^(-11)`

C

`5xx10^(-5)`

D

`8xx10^(-2)`

Text Solution

Verified by Experts

The correct Answer is:
B
`{:(,CH_(3)NH_(2),+,HCl,rarr,CH_(3)N overset(+)H_(3)Cl^(-),,,),("Initial conc.",0.1,,0.08,,0 "mol" L^(-1),,,),("Conc. in solution after mixing",0.1-0.08 ,,0,,0.08 "mol " L^(-1),,,):}`
`=0.02` (HCl being limiting reactant)
Thus, the final solution is a basic buffer of `CH_(3)NH_(2) and CH_(3)NH_(3)^(+)Cl`
`:. pOH = pK_(b) + log. (["Salt"])/(["Base"])`
i.e., `-log [OH^(-)]=-log K_(b) + log. (["Salt"])/(["Base"])`
or `[OH^(-)]=K_(b) (["Salt"])/(["Base"]) = 5xx10^(-4)xx(0.02)/(0.08) `
`=(5)/(4) xx 10^(-4)M = 1.25 xx 10^(-4)M`
`:. [H^(+)]=(K_(w))/([OH^(-)])=(10^(-14))/(1.25xx10^(-4))=8xx10^(-11)`
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