`0.1mol` of `RNH_(2)(K_(b) = 5 xx 10^(-5))` is mixed with `0.08mol` of `HC1` and diluted to `1L`. Calculate the `[H^(o+)]` in the solution.

Options

A. `8 xx 10^(-11)M`
B. `1.6 xx 10^(-11)M`
C. `8 xx 10^(-5)M`
D. `8 xx 10^(-2)M`

Answer

A

Explanation

`{:(,RNH_(2)+,HC1rarr,Roverset(o+)NH_(3)+,C1^(Theta)),("Initial mol",0.1,0.08,0,0),("Final moles",(0.1-0.08),(0.08-0.08),0.08,-),(,=0.02,=0,,):}`
`W_(B)` is left behind so basic buffer is formed.
Use Henderson's equation:
`[overset(Theta)OH] = K_(b) (["Base"])/(["Salt"])`
`= 5 xx 10^(-14) xx (0.02)/(0.08) = 1.25 xx 10^(-4)`.
`:. [H^(o+)] = (K_(w))/([overset(Theta)OH]) = (10^(-14))/(1.25 xx10^(-4))`
`= 8 xx 10^(-11)M`