`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.

Options

A. The concentration of `H^(o+)` ion is `8 xx 10^(-11)M`.
B. The concentration of `H^(o+)` ion is `8 xx 10^(-5)M`.
C. The `pH` of solution is `9.8`
D. The `pOH` of solution is `10.2`.

Answer

A, C

Explanation

`{:(,CH_(3)NH_(2)+,HCIrarr,CH_(3)overset(o+)NH_(3)+,CI^(Theta)),("Initial",0.1,0.08,,),("moles",(0.1-0.08),(0.008-0.008),0.008,-),(,=0.02,=0,,):}`
Since `W_(B)` is left, so it forms basic buffer solution
`[overset(Theta)OH] = K_(b) (["Base"])/(["Salt"]) = 5xx 10^(-4) xx (0.02)/(0.08) = 1.25 xx 10^(-14)`
`[H^(o+)] = (K_(w))/([overset(Theta)OH]) = (10^(-14))/(1.25 xx 10^(-4)) = 8 xx 10^(-11)M`
`pH =- log (8 xx 10^(-11))`
`=- log2^(4) +11`
`=- 0.3 xx 4+11 = 9.8`
`pOH = 14 - 9.8 = 4.2`