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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

`8x 10^(-2)M`

B

`8x 10^(-11)M`

C

`1.6x 10^(-11)M`

D

`8x 10^(-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^(-)])=1.25xx10^(-4)`
`=8xx10^(-11)M`
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