Determine the degree of ionization and `pH` of `0.05M` of ammonia solution. The ionization constant of ammonia can be taken from Table `7.7`. Also, calculate the ionization constant of the conjugate acid of ammonia.

Explanation

The ionization of `NH_(3)` in water is represented by equation :
`NH_(3)+H_(2)OhArrNH_(4)^(+)+OH^(-)`
We use equation `(7.33)` to calculate hydroxyl ion concentration.
`[OH^(-)]=c alpha=0.05 alpha`
`K_(b)=0.05alpha^(2)//(1-alpha)`
The value of `alpha` is small, therefore the quadratic equation can be simplified by neglecting `alpha` in comparison to `1` in the denominator on right hand side of the equation,
Thus,
`K_(b)=c alpha^(2)` or `alpha=(1.77xx10^(-5)//0.05)`
`=0.018`
`[OH^(-)]=calpha=0.05xx0.018=9.4xx10^(-4)M`
`[H^(+)]=K_(w)//[OH^(-)]=10^(-14)//(9.4xx10^(-4))`
`=1.06xx10^(-11)`
`pH=-log(1.06xx10^(-11))=10.97`
Now, using the relation for conjugate acid-base pair.
`K_(a)xxk_(b)=K_(w)`
using the value of `K_(b)` of `NH_(3)` from Table `7.7`
We can determine the concentration of conjugate acid `NH_(4)^(+)`
`K_(a)=K_(w)//K_(b)=10^(-14)//1.77xx10^(-5)`
`=5.64xx10^(-10)`