Question

What is the pH at the equivalence point when 0.100M hydroxyacetic acid is titrated with 0.0500M KOH?

Short answer

The pH at the equivalence pointwhen 0.100M hydroxyacetic acid is titrated with 0.0500M KOH is 8.18.

Step-by-step solution

Step 1:Definethe equivalence point.

The stage of titration where the concentrations of titrate and titrant are chemically equivalent is known as the equivalence point. The equivalence point occurs a few milliseconds before or virtually simultaneously with the endpoint.

Find the new concentration of the acid.

At the equivalence point of the titration of a weak acid with a strong base, the base reacts with all the acid and only the conjugate base of the acid is present in the solution.

First, we need to write the reaction of the weak base with water:

$$\begin{gathered} {\mathrm{A}}^{-}+{\mathrm{H}}_2\mathrm{O}\rightleftharpoons \mathrm{HA}+{\mathrm{OH}}^{-} \\ \left[{\mathrm{A}}^{-}\right]=\mathrm{F}\text{'}-\mathrm{x} \\ \left[\mathrm{HA}\right]=\mathrm{x} \\ \left[{\mathrm{OH}}^{-}\right]=\mathrm{x} \end{gathered}$$

Since the acid is diluted with the base we need to find a new concentration of the acid.

We need twice the volume of the base to reach the equivalence point.

Use the formula,

localid="1663394027035" $$\begin{gathered} \mathrm{F}\text{'}=\mathrm{initial}\mathrm{concentration}\mathrm{of}\mathrm{HA}·\frac{\mathrm{initial}\mathrm{volume}\mathrm{of}\mathrm{HA}}{\mathrm{total}\mathrm{volume}\mathrm{of}\mathrm{the}\mathrm{solution}} \\ \mathrm{F}\text{'}=0.100\mathrm{M}·\frac{\mathrm{V}\left(\mathrm{HA}\right)}{\mathrm{V}\left(\mathrm{HA}\right)+2\mathrm{V}\left(\mathrm{HA}\right)}=0.100\mathrm{M}·1/3=0.033\mathrm{M} \end{gathered}$$

Hence the new concentration of the acid is $0.033\mathrm{M}$.

Find the pH.

Calculate the value of ${\mathrm{K}}_{\mathrm{b}}$.

${\mathrm{K}}_{\mathrm{b}}=\frac{{\mathrm{K}}_{\mathrm{w}}}{{\mathrm{K}}_{\mathrm{a}}}=\frac{{10}^{-14}}{1.48\times {10}^{-4}}=6.76\times {10}^{-11}$

Now substitute in the equilibrium equation.

$$\begin{gathered} {\mathrm{K}}_{\mathrm{b}}=\frac{\left[\mathrm{HA}\right]\left[{\mathrm{OH}}^{-}\right]}{{\mathrm{A}}^{-}}=\frac{{\mathrm{x}}^2}{0.033-\mathrm{x}} \\ {\mathrm{x}}^2+6.76\times {10}^{-11}\mathrm{x}-2.231\times {10}^{-12}=0 \end{gathered}$$

Solve the quadratic equation.

$$\begin{gathered} \mathrm{x}=\left[{\mathrm{OH}}^{-}\right]=1.50\times {10}^{-6} \\ \mathrm{pOH}=5.82 \\ \mathrm{pH}=14-5.82=8.18 \end{gathered}$$

Hence the pH is 8.18.