The `pK_(b)` of `NH_(3)` is `4.75`. Calculate the concentration of `H^(+)` ions in solution formed by mixing `0.2 M NH_(4)CI` and `0.1M NH_(3)`.

Options

A. `0.88xx10^(-5)`
B. `1.12xx10^(-9)`
C. `1.12xx10^(-5)`
D. `0.88xx10^(-9)`

Answer

B

Explanation

`pK_(b)= -log K_(b)`
`:. K_(b) = "antilog (-pK_(b))`
`= "antilog" (-4.75)`
`1.77xx10^(-5)`
`{:(,N_(3)H(aq.)+H_(2)O(I)hArrNH_(4)^(+)(aq.)+OH^(-)(aq.)),("Inital (M)"," 0.1 0.20 0"),("Change (M)"," -x +x +x"),("Equilibrium (M)",bar(" 0.1-x 0.20+x x ")):}`
`K_(b)=(C_(NH_(4)^(+))C_OH^(-))/(CNH_(3))`
`1.77xx10^(-5)=((0.20+x)(x))/((0.1-x))`
Neglecting x in comparison to `0.20` and `0.1`, we have
`1.77xx10^(-5)=((0.20)(x))/((0.1))`
`x=C_(OH^(-))=0.88xx10^(-5)`
`CH^(+)C_(OH^(-)=K_(w))`
`:. CH^(+)=(K_(w))/(C_(OH^(-)))=(1.0xx10^(-14))/(0.88xx10^(-5))`
`=1.12xx10^(-9)`