Question

A 0.1000 M solution of a weak acid, HA, requires 50.00 mL of

0.1000 M NaOH to titrate it to its equivalence point. The pH of the solution is

4.50 when only 40.00 mL of the base has been added.

(a) Calculate the ionization constant Ka of the acid.

(b) Calculate the pH of the solution at the equivalence point.

Short answer

$\begin{array}{l}\mathrm{a}){\mathrm{K}}_{\mathrm{a}}=1.29\times {10}^{-4}\\ \mathrm{b})\mathrm{pH}=8.29\end{array}$

Step-by-step solution

The reaction

The reaction is:

$\mathrm{HA}+{\mathrm{H}}_2\mathrm{O}\to {\mathrm{A}}^{-}+{\mathrm{H}}_3{\mathrm{O}}^{+}$

The${\mathrm{K}}_{\mathrm{a}}$ is${\mathrm{K}}_{\mathrm{a}}=\frac{\left[{\mathrm{A}}^{-}\right]\left[{\mathrm{H}}_3\mathrm{O}\right]}{\left[\mathrm{HA}\right]}$

Neutralization of HA

We know that of 0.1 M NaOH is needed to neutralize 0.1 M HA. So the

volume of HA is

$\begin{array}{l}\left[\mathrm{HA}\right]·\mathrm{V}\left(\mathrm{HA}\right)=\left[\mathrm{NaOH}\right]·\mathrm{V}\left(\mathrm{NaOH}\right)\\ \mathrm{V}\left(\mathrm{HA}\right)=\frac{\left[\mathrm{NaOH}\right]·\mathrm{V}\left(\mathrm{NaOH}\right)}{\left[\mathrm{HA}\right]}\\ \mathrm{V}\left(\mathrm{HA}\right)=\frac{0.1\mathrm{M}·50·{10}^{-3}\mathrm{L}}{0.1\mathrm{M}}\\ \mathrm{V}\left(\mathrm{HA}\right)=0.05\mathrm{L}\end{array}$

The value of HA

When V(B)=40ml, the n of OH is

The$n$of is:

$\mathrm{n}\left(\mathrm{HA}\right)=\mathrm{n}{\left(\mathrm{HA}\right)}_{\mathrm{initial}}-\mathrm{n}\left(\mathrm{OH}\right)$

Where the initial$n$ of HA is:

$n{\left(HA\right)}_{initial}=\left[HA\right]·V\left(HA\right)$

$\begin{array}{l}n{\left(HA\right)}_{initial}=0.1M·0.05L\\ n{\left(HA\right)}_{initial}={5.10}^{-3}mol\\ n\left(HA\right)=({5.10}^{-3}-{4.10}^{-3})mol\\ n\left(HA\right)={1.10}^{-3}mol\end{array}$

While$n$ the$A^{-}$ofis:

$\begin{array}{l}n\left(A^{-}\right)=n{\left(A\right)}_{initial}+n\left(O{H}^{-}\right)\\ n\left(A^{-}\right)=0+{4.10}^{-3}mol\\ n\left(A^{-}\right)={4.10}^{-3}mol\end{array}$

The Hendersson-Hasselbatch equation

So, we calculate${\mathrm{pK}}_{\mathrm{a}}$ using the Henderson-Hasselbatch equation:

$\begin{array}{l}\mathrm{pH}={\mathrm{pK}}_{\mathrm{a}}+\log \left(\frac{{\mathrm{A}}^{-}}{\left[\mathrm{HA}\right]}\right)\\ 4.5={\mathrm{pK}}_{\mathrm{a}}+\log \left(\frac{4·{10}^{-3}}{1{.10}^{-3}}\right)\\ 4.5={\mathrm{pK}}_{\mathrm{a}}+0.6021\\ {\mathrm{pK}}_{\mathrm{a}}=3.89\end{array}$

The$K_a$ is

$\begin{array}{l}{\mathrm{K}}_{\mathrm{a}}={10}^{-{\mathrm{pK}}_{\mathrm{a}}}={10}^{-3.89}\\ {\mathrm{K}}_{\mathrm{a}}=1.29·{10}^{-4}\end{array}$

The reaction at the equivalence point

b) At the equivalence point, the quantity of${\mathrm{OH}}^{-}$ is exactly enough to consume the HA. So, the reaction is:

${\mathrm{A}}^{-}+{\mathrm{H}}_2\mathrm{O}\to \mathrm{HA}+{\mathrm{OH}}^{-}$

The concentration

The concentration of${\mathrm{A}}^{-}$ is:

$\begin{array}{l}\left[{\mathrm{A}}^{-}\right]=\frac{\mathrm{n}\left(\mathrm{HA}\right)}{\mathrm{V}\left(\mathrm{HA}\right)+\mathrm{V}\left(\mathrm{added}\right)}\\ \left[{\mathrm{A}}^{-}\right]=\frac{5·{10}^{-3}\mathrm{mol}}{50·{10}^{-3}\mathrm{L}+50·{10}^{-3}\mathrm{L}}\\ \left[{\mathrm{A}}^{-}\right]=0.05\mathrm{M}\end{array}$

${\mathrm{K}}_{\mathrm{b}}$is

$\begin{array}{l}{\mathrm{K}}_{\mathrm{b}}=\frac{{\mathrm{K}}_{\mathrm{w}}}{{\mathrm{K}}_{\mathrm{a}}}=\frac{1\times {10}^{-14}}{1.29\times {10}^{-4}}\\ {\mathrm{K}}_{\mathrm{b}}=7.75\cdot {10}^{-11}\end{array}$

So$K_b$ is

$\begin{array}{l}{K}_b=\frac{\left[HA\right]\left[O{H}^{-}\right]}{\left[A^{-}\right]}\\ K_b=\frac{x^2}{\left[A^{-}\right]}\\ x=\sqrt{K_b\times }\left[A^1\right]\\ x=\sqrt{7.75\times {10}^{-11}\times 0.05M}\\ x=1.97\times {10}^{-6}\end{array}$

The concentration of pOH  and pH

$x$is equal to the concentration. The is:

And the pH is