Question

You have 75.0 mL of 0.10 M HA. After adding 30.0 mLof 0.10 M NaOH, the pH is 5.50. What is the K_a valueof HA?

Short answer

Thevalue of K_a is $2.1\times {10}^{-6}.$

Step-by-step solution

Given Information

The volume and concentration of HA are 75 mL and 0.1 M. The volume and concentration of NaOH are 30 mL, and 0.1 M. pH is 5.5.

Finding the concentration of each species

The major species of HA and NaOH are $\mathrm{HA},{\mathrm{OH}}^{-},{\mathrm{Na}}^{+},{\mathrm{H}}_2\mathrm{O}.$The reaction between strong base and acid can be denoted as:

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

The concentration of each species can be expressed as:

$\begin{array}{l}\left[\mathrm{HA}\right]=75\mathrm{mL}\times 0.01\mathrm{M}\\ =7.5\mathrm{mmol}\\ \left[{\mathrm{A}}^{-}\right]=30\mathrm{mL}\times 0.01\mathrm{M}\\ =3\mathrm{mmol}\end{array}$

HA dissociation and acid dissociation constant

The HA dissociation can be denoted as:

$\mathrm{HA}\left(\mathrm{aq}\right)\rightleftharpoons {\mathrm{H}}^{+}\left(\mathrm{aq}\right)+{\mathrm{A}}^{-}\left(\mathrm{aq}\right)$

The constant of acid dissociation can be denoted as:

$\begin{array}{l}{\mathrm{K}}_{\mathrm{a}}=\frac{\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{A}}^{-}\right]}{\left[\mathrm{HA}\right]}\\ \left[\mathrm{HA}\right]=\frac{4.5\mathrm{mmol}}{(75+30)\mathrm{mL}}\\ =0.043\mathrm{M}\\ \left[{\mathrm{A}}^{-}\right]=\frac{3\mathrm{mmol}}{(75+30)\mathrm{mL}}\\ =0.029\mathrm{M}\\ {\mathrm{K}}_{\mathrm{a}}=\frac{\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{A}}^{-}\right]}{\left[\mathrm{HA}\right]}\\ {\mathrm{K}}_{\mathrm{a}}=\frac{\mathrm{x}(0.029+\mathrm{x})}{(0.043-\mathrm{x})}\end{array}$

The value of x can be denoted as;

$\begin{array}{l}\mathrm{x}=\left[{\mathrm{H}}^{+}\right]\\ ={10}^{-\mathrm{pH}}\\ ={10}^{-5.5}\\ \mathrm{x}=3.16\times {10}^{-6}\\ {\mathrm{K}}_{\mathrm{a}}=\frac{\mathrm{x}(0.029+\mathrm{x})}{(0.043-\mathrm{x})}\\ =\frac{(3.16\times {10}^{-6})(0.029+3.16\times {10}^{-6})}{(0.043-3.16\times {10}^{-6})}\\ {\mathrm{K}}_{\mathrm{a}}=2.1\times {10}^{-6}\end{array}$