Question

Novocain, the commonly used local anesthetic, is a weak base

with${\mathbf{K}}_b\text{=}\mathbf{5}.\mathbf{7}\times {\mathbf{10}}^{26}$ M.

(a) If you had a 0.0200 M solution of Novocain in water, what would be the approximate concentration ofand the pH?

(b) Suppose that you wanted to determine the concentration of Novocain in a solution that is about 0.020 M by titration with 0.020 M HCl. Calculate the expected pH at the equivalence point.

Short answer

$\begin{array}{l}\mathrm{a})\mathrm{pH}=10.57\\ \mathrm{b})\mathrm{pH}=5.42\end{array}$

Step-by-step solution

The reaction of weak base with water

a) The reaction of weak base with water is:

$\mathrm{B}+{\mathrm{H}}_2\mathrm{O}\to {\mathrm{BH}}^{+}+{\mathrm{OH}}^{-}$

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

$\begin{array}{l}{\mathrm{K}}_{\mathrm{b}}=\frac{\left[{\mathrm{BH}}^{+}\right]\left[{\mathrm{OH}}^{-}\right]}{\mathrm{B}}\\ 7{.10}^{-6}=\frac{{\mathrm{x}}^2}{0.02-\mathrm{x}}\\ {\mathrm{x}}^2+7{.10}^{-6}\mathrm{x}-1.4·{10}^{-7}=0\\ \mathrm{x}=3.71·{10}^{-4}\mathrm{M}=\left[\mathrm{OH}\right]\end{array}$

The reaction of Ka

The pOH is

$\begin{array}{l}\mathrm{pOH}=-\log \left[{\mathrm{OH}}^{-}\right]\\ \mathrm{pOH}=-\log (3.71·{10}^{-4})\\ \mathrm{pOH}=3.43\end{array}$

The pH is

$\begin{array}{l}\mathrm{pH}=14-\mathrm{pOH}=14-3.43\\ \mathrm{pH}=10.57\end{array}$

The reaction is${\mathrm{BH}}^{+}+{\mathrm{H}}_2\mathrm{O}\to \mathrm{B}+{\mathrm{H}}_3{\mathrm{O}}^{+}$

The${\mathrm{K}}_{\mathrm{a}}$ is

$\begin{array}{l}{\mathrm{K}}_{\mathrm{a}}=\frac{{\mathrm{K}}_{\mathrm{w}}}{{\mathrm{K}}_{\mathrm{b}}}=\frac{1{.10}^{-14}}{7{.10}^{-6}}\\ {\mathrm{K}}_{\mathrm{a}}=1.43·{10}^{-9}\end{array}$

The Ka at equivalence point

At the equivalence point, the solution contains$0.01\mathrm{M}$of${\mathrm{BH}}^{+}$. The${\mathrm{K}}_{\mathrm{a}}$is:

$\begin{array}{l}{\mathrm{K}}_{\mathrm{a}}=\frac{\left[\mathrm{B}\right]\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]}{{\mathrm{BH}}^{+}}\\ 1.43·{10}^{-9}=\frac{{\mathrm{x}}^2}{0.01-\mathrm{x}}\\ {\mathrm{x}}^2+1.43·{10}^{-9}\mathrm{x}-1.43·{10}^{-11}=0\\ \end{array}$

The pH value is found

The pH is

$\begin{array}{l}{\text{pH=-log[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}\\ \text{pH=-log[3.78}\times {\text{10}}^{\text{-6}}\text{]}\\ \text{pH=5.42}\end{array}$