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0.1 mole of CH(3)CH (K(b) = 5 xx 10^(-4)...

`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

A

`8xx10^(-2)` M

B

`8xx10^(-11)` M

C

`1.6xx10^(-11)` M

D

`8xx10^(-5)` M

Text Solution

Verified by Experts

The correct Answer is:
B
`{:(CH_(3)NH_(2),,+,,HCl,,to,,CH_(3)NH_(3)^(+)Cl^(-)),(0.1,,,,0.08,,,,0),(0.02,,,,0,,,,0.08):}`
(Basic buffer solution)
`[OH^(-)]=K_(b)xx("Base")/("Salt")=5xx10^(-4)xx(0.02)/(0.08)=1.25xx10^(-4)`
`:.[H^(+)]=(10^(-14))/([OH^(-)])=(10^(-14))/(1.25xx10^(-4))=8xx10^(-11)M`.
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