A0.24 m solution of a weak base a‑ is made. to a limited extent, the a‑ reacts with h2o to form some oh- and some of the corresponding weak acid ha. if the weak acid has a pka of 9.5, what will be the ph of the solution (to the nearest hundredths)? for a hint see the acids and bases handout on canvas.

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

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