Chapter 15, Problem 15.84QA

Interpretation Introduction

To calculate:

The $\mathrm{p}\mathrm{H}$ of $1.25\times {10}^{-2}\mathrm{ }\mathrm{M}$ ephedrine hydrochloride

Answer to Problem 15.84QA

Solution (final answer):

The $\mathrm{p}\mathrm{H}$ of $1.25\times {10}^{-2}\mathrm{ }\mathrm{M}$ ephedrine hydrochloride is $6.02$.

Explanation of Solution

1) Concept:

We are asked to determine the $\mathrm{p}\mathrm{H}$ of $1.25\times {10}^{-2}\mathrm{ }\mathrm{M}$ of ephedrine hydrochloride. It dissociates into water and produces ${{\mathrm{H}}_3\mathrm{O}}^{+}$ and ephedrine ions

${\mathrm{E}\mathrm{p}\mathrm{h}\mathrm{e}\mathrm{d}\mathrm{r}\mathrm{i}\mathrm{n}\mathrm{e}\mathrm{ }\mathrm{H}\mathrm{C}\mathrm{l}}_{\left(\mathrm{a}\mathrm{q}\right)}+{{\mathrm{H}}_2\mathrm{O}}_{\left(\mathrm{a}\mathrm{q}\right)}\rightleftharpoons {{{\mathrm{H}}_3\mathrm{O}}^{+}}_{\left(\mathrm{a}\mathrm{q}\right)}+{\mathrm{E}\mathrm{p}\mathrm{h}\mathrm{e}\mathrm{d}\mathrm{r}\mathrm{i}\mathrm{n}\mathrm{e}}_{\left(\mathrm{a}\mathrm{q}\right)}{\mathrm{p}\mathrm{K}}_{\mathrm{b}}=3.86$

We first determine the $\mathrm{p}{\mathrm{K}}_{\mathrm{a}}$ value and ${\mathrm{K}}_{\mathrm{a}}$. Then, we find the equilibrium concentration of ${{\mathrm{H}}_3\mathrm{O}}^{+}$ ions using the RICE table. From ${{[\mathrm{H}}_3\mathrm{O}}^{+}]$, we can find $\mathrm{p}\mathrm{H}$ of $1.25\times {10}^{-2}\mathrm{ }\mathrm{M}$ ephedrine hydrochloride by the given formula

2) Formula:

i) $\mathrm{p}\mathrm{H}=-\log {{[\mathrm{H}}_3\mathrm{O}}^{+}]\mathrm{ }$

ii) $\mathrm{p}{\mathrm{K}}_{\mathrm{a}}+\mathrm{p}{\mathrm{K}}_{\mathrm{b}}=14$

3) Given:

i) We are given concentration of ephedrine hydrochloride that is $1.25\times {10}^{-2}\mathrm{ }\mathrm{M}$

ii) $\mathrm{p}{\mathrm{K}}_{\mathrm{b}}=3.86$ of ephedrine.

4) Calculations:

To find ${\mathrm{p}\mathrm{K}}_{\mathrm{a}}$ from ${\mathrm{p}\mathrm{K}}_{\mathrm{b}}$ we know the formula

$\mathrm{p}{\mathrm{K}}_{\mathrm{a}}+\mathrm{p}{\mathrm{K}}_{\mathrm{b}}=14$

$\mathrm{p}{\mathrm{K}}_{\mathrm{a}}+3.86=14$

$\mathrm{p}{\mathrm{K}}_{\mathrm{a}}=10.14$

And to calculate ${\mathrm{K}}_{\mathrm{a}}$ we have

$\mathrm{p}{\mathrm{K}}_{\mathrm{a}}=-\mathrm{l}\mathrm{o}\mathrm{g}{\mathrm{K}}_{\mathrm{a}}$

${\mathrm{K}}_{\mathrm{a}}={10}^{-{\mathrm{p}\mathrm{K}}_{\mathrm{a}}}$

${\mathrm{K}}_{\mathrm{a}}={10}^{-\left(10.14\right)}$

${\mathrm{K}}_{\mathrm{a}}=7.24\times {10}^{-11}$

Now we frame up the RICE table for the reaction.

$\mathrm{R}\mathrm{e}\mathrm{a}\mathrm{c}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}{\mathrm{E}\mathrm{p}\mathrm{h}\mathrm{e}\mathrm{d}\mathrm{r}\mathrm{i}\mathrm{n}\mathrm{e}\mathrm{ }\mathrm{H}\mathrm{C}\mathrm{l}}_{\left(\mathrm{a}\mathrm{q}\right)}+{{\mathrm{H}}_2\mathrm{O}}_{\left(\mathrm{a}\mathrm{q}\right)}\rightleftharpoons {{{\mathrm{H}}_3\mathrm{O}}^{+}}_{\left(\mathrm{a}\mathrm{q}\right)}+{\mathrm{E}\mathrm{p}\mathrm{h}\mathrm{e}\mathrm{d}\mathrm{r}\mathrm{i}\mathrm{n}\mathrm{e}}_{\left(\mathrm{a}\mathrm{q}\right)}$
	$\left[\mathrm{E}\mathrm{p}\mathrm{h}\mathrm{e}\mathrm{d}\mathrm{r}\mathrm{i}\mathrm{n}\mathrm{e}\mathrm{ }\mathrm{H}\mathrm{C}\mathrm{l}\right]\mathrm{ }\left(\mathrm{M}\right)$	$\left[{{\mathrm{H}}_3\mathrm{O}}^{+}\right]\left(\mathrm{M}\right)$	$\left[\mathrm{E}\mathrm{p}\mathrm{h}\mathrm{e}\mathrm{d}\mathrm{r}\mathrm{i}\mathrm{n}\mathrm{e}\right]\mathrm{ }\left(\mathrm{M}\right)$
$\mathrm{I}\mathrm{n}\mathrm{i}\mathrm{t}\mathrm{i}\mathrm{a}\mathrm{l}\mathrm{ }\left(\mathrm{I}\right)$	$1.25\times {10}^{-2}$	$0$	$0$
$\mathrm{C}\mathrm{h}\mathrm{a}\mathrm{n}\mathrm{g}\mathrm{e}\mathrm{ }\left(\mathrm{C}\right)$	$-\mathrm{x}$	$+\mathrm{x}$	$+\mathrm{x}$
$\mathrm{E}\mathrm{q}\mathrm{u}\mathrm{i}\mathrm{l}\mathrm{i}\mathrm{b}\mathrm{r}\mathrm{i}\mathrm{u}\mathrm{m}\mathrm{ }\left(\mathrm{E}\right)$	$1.25\times {10}^{-2}-\mathrm{x}$	$\mathrm{x}$	$\mathrm{x}$

Now we can set up ${\mathrm{K}}_{\mathrm{a}}$ expression as

${\mathrm{K}}_{\mathrm{a}}=\frac{\left[{{\mathrm{H}}_3\mathrm{O}}^{+}\right]\left[\mathrm{E}\mathrm{p}\mathrm{h}\mathrm{e}\mathrm{d}\mathrm{r}\mathrm{i}\mathrm{n}\mathrm{e}\right]\mathrm{ }}{\left[\mathrm{E}\mathrm{p}\mathrm{h}\mathrm{e}\mathrm{d}\mathrm{r}\mathrm{i}\mathrm{n}\mathrm{e}\mathrm{ }\mathrm{H}\mathrm{C}\mathrm{l}\right]\mathrm{ }}$

$7.24\times {10}^{-11}=\frac{\left(\mathrm{x}\right)\left(\mathrm{x}\right)}{\left[\right(1.25\times {10}^{-2})-\mathrm{x}]}$

The initial concentration of ephedrine HCl is  $(1.25\times {10}^{-2})/\left(7.22\times {10}^{-11}\right)=1.7\times {10}^8$ times the value of ${\mathrm{K}}_{\mathrm{a}}$, so to simplify the calculation we can neglect $–\mathrm{x}$ term in the denominator

$7.24\times {10}^{-11}=\frac{\left(\mathrm{x}\right)\left(\mathrm{x}\right)}{[1.25\times {10}^{-2}]}$

${\mathrm{x}}^2=(7.24\times {10}^{-11})\times (1.25\times {10}^{-2})$

${\mathrm{x}}^2=9.055\times {10}^{-13}$

$\mathrm{x}={{[\mathrm{H}}_3\mathrm{O}}^{+}]=9.51\times {10}^{-7}$

Now to calculate $\mathrm{p}\mathrm{H}$

$\mathrm{p}\mathrm{H}=-\log {{[\mathrm{H}}_3\mathrm{O}}^{+}]\mathrm{ }$

$\mathrm{p}\mathrm{H}=-\log (9.51\times {10}^{-7})\mathrm{ }$

$\mathrm{p}\mathrm{H}=6.02$

Conclusion:

The pH of the ephedrine HCl is calculated from the initial concentration and RICE table.