`0.1M NH_(3)` solution is found to have a `[overset(Θ)OH]` of `.133 xx 10^(-3)M`.
a. What is the `pH` of the solution?
b. What will be the `pH` of the solution after `0.1M NaOh` is added to it?
c. Calculate `K_(b)` and `pK_(b)` for `NH_(3)`?
d. How will `NaOh` added to the solution affect the extent of dissociation of `NH_(3)`?

Explanation

a. `[overset(Θ)OH] = 1.33 xx 10^(-3), pOH = 2.88`
`:. pH = 14 - 2.88 = 11.12`
b. `[overset(Θ)OH] = 0.1M, pOH = 1:. pH = 14 - 1 = 13`.
The `[overset(Θ)OH]` from `NH_(3)` is negligible, since `NaOH` supresses the dissociation of the weak base.
c. `NH_(3)+H_(2)O hArr overset(oplus)(N)H_(4)+ overset(Theta)(O)H`
`:. K_(b) = ([overset(o+)NH_(4)][overset(Θ)OH])/([NH_(3)])`
`= ((1.33 xx 10^(-3))(1.33 xx 10^(-3)))/(0.1 - 0.00133) [0.1 - 0.00133~~ 01]`
`= ((1.33 xx 10^(-3))^(2))/(0.1)`
`= 1.79 xx 10^(-5) ~~ 1.8 xx 10^(-5)`
`:. pK_(b) =- log (1.8 xx 10^(-5)) = 4.7447`