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0.1 mole of CH(3)NH(2)(K(b) = 5 xx 10^(-...

`0.1` mole of `CH_(3)NH_(2)(K_(b) = 5 xx 10^(-4))` is mixed with `0.08` mole of HCl and adiluted to one litre. What will be the `H^(+)` concentration in the solution ?

A

`8 xx 10^(-2)`

B

`2xx10^(-11)`

C

`1.23 xx10^(-4)`

D

`8xx10^(-11)`

Text Solution

Verified by Experts

The correct Answer is:
D
` {:(,CH_(3)NH_(2),+,HCl,to,CH_(3)NH_(3)^(+)Cl^(-)),("Intial",0.10,,0.08,,0),("Final",0.02,,0,,0.08):}`
`pOH = pK_(b) + log. (["Salt"])/(["Base"])`
` = - log 5 xx 10^(-4) + log . (0.08)/(0.02) = 3.903`
`pH = 10.0967`
`[H^(+)] ="antilog " (-10.0967) = 8 xx 10^(-11)`
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