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0.1mol of RNH(2)(K(b) = 5 xx 10^(-4)) is...

`0.1mol` of `RNH_(2)(K_(b) = 5 xx 10^(-4))` is mixed with `0.08mol` of `HC1` and diluted to `1L`. Calculate the `[H^(o+)]` 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:
B
`{:(CH_(3)NH_(2)+,HCIrarr,CH_(3)NH_(3)^(o+)CI^(Theta)),(0.1,0.08,0),((0.1-0.08),0,0.08),(=0.02,0,0.08):}`
This is basic buffer.
`[overset(Theta)OH] = (K_(b)["Base"])/(["Salt"])`
`=(K_(w))/([overset(Theta)OH]) = (K_(w)["Salt"])/(K_(b)["Base"])`
`= (10^(-14)xx0.08)/(5xx10^(-4)xx0.02) = (4)/(5) xx 10^(-10) = 8 xx 10^(-11)`.
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