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`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`

1 Answer

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Best answer
Correct Answer - B
`CH_(3)NH_(2) + HCl rarr CH_(3)NH_(3)^(+)Cl^(-)`
`{:(0.1,0.08,0),(0.02,0,0.08):}`
(Basic buffer solution)
`[OH^(-)] = K_(b) xx ("Base")/("Salt")`
`= 5 xx 10^(-4) xx (0.02)/(0.08) x= (10^(-14))/(1.25 xx 10^(-4))`
`= 8 xx 10^(-11) M`

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