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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

`8 xx 10^(-2) M`

B

`8 xx 10^(-11) M`

C

`1.6 xx 10^(-11) M`

D

`8 xx 10^(-5) M`

Text Solution

Verified by Experts

The correct Answer is:
2
`{:(CH_(3)NH_(2),+,HCl,to,CH_(3)NH_(3).^(+)Cl^(-)),(0.1,,0.08,,0),(0.02,,0,,0.08),(,,,,"(Basic buffer solution"):}`
`pOH = pK_(b) + log . (0.08)/(0.02) = pK_(b) + 0.602 = 3.30 + 0.602 = 3.902 `
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