Determine the degree if ionization and pH of `0.05 M` 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)O hArr NH_(4)^(+) + OH^(-)`
We use equation `(7.33)` to calculate hydroxyl ion concentration.
`[OH^(-)] = c alpha = 0.05 alpha`
`K_(b) = 0.05 alpha^(2) //(1-alpha)`
The value of `alpha` is small, therefore the quadratic equation can be simplified by neglecting `alpha` in comparsion to 1 in the denominator on right hand side of the equation.
Thus,
`K_(b) = c alpha^(2)` or `alpha = sqrt((1.77 xx 10^(-5)//0.05))`
`= 0.018`.
`[OH^(-)] = c alpha = 0.05 xx 0.018 = 9.4 xx 10^(-4) M`
`[H^(+)] = K_(w) // [OH^(-)] = 10^(-14) //(9.4 xx 10^(-4))`
`= 1.06 xx 10^(-11)`
`pH = - "log" (1.06 xx 10^(-11)) = 10.97`
Now, using the relation for conjugate acid-base pair,
`K_(a) xx K_(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) //1.77 xx 10^(-5)`
`= 5.64 xx 10^(-10)`.