Solution: The Molarity of solution of weak base is 0.24 M and pK_{a} of weak acid formed by its reaction with water is 9.5.
Let's initially all the weak base reacts with water to form hydroxide ion and weak acid as follows:
A^{-}+H_{2}O\rightarrow OH^{-}+HA
thus, concentration of A^{-}, OH^{-} and HA will become 0, 0.24 and 0.24 respectively.
Reverse reaction will take place as follows:
OH^{-}+HA\rightarrow A^{-}+H_{2}O
The change in concentration takes place, let's the change be x thus, concentration of OH^{-}, HA and A^{-} will become 0.24-x, 0.24-x and x respectively.
For the reaction, expression for acid dissociation constant will be:
K_{a}=\frac{[A^{-}]}{[HA][OH^{-}]}
Putting the values of concentration,
K_{a}=\frac{[x]}{[0.24-x][0.24-x]}....... (1)
From the pK_{a} of weak acid, acid dissociation constant can be calculated as follows:
pK_{a}=-logK_{a}
thus,
K_{a}=10^{-pK_{a}}
=10^{-9.5}
=3.16\times 10^{-10}
Putting the value in equation (1),
3.16\times 10^{-10}=\frac{[x]}{[0.24-x][0.24-x]}
Here, the value of K_{a} is very small so x can be neglected from denominator, thus,
3.16\times 10^{-10}=\frac{[x]}{[0.24][0.24]}
On solving,
x=4.26\times 10^{-6}
Concentration of hydroxide ion will be:
[OH^{-}]=0.24-x=0.24-4.26\times 10^{-6}=0.2399
now, pOH will be:
pOH=-log[OH^{-}]
=-log(0.2399)
=0.62
Use the following equation to calculate pH,
pH+pOH=14
thus,
pH=14-pOH
=14-0.62
=13.38
Thus, pH of solution is 13.38
