Chapter 13, Problem 13.CE

(a)

Interpretation Introduction

Interpretation:

The $\text{pH}$ and concentrations of species in the given solution has to be determined by neglecting the activity coefficients.

Explanation of Solution

Given information,

The given $\text{1}\text{.00}\text{L}$ of solution containing $\text{0}\text{.040 mol}$ 2-aminobenzoic acid (HA), $\text{0}\text{.020 mol}$ dimethylamine (B), $\text{0}\text{.015 mol}\text{HCl}\text{.}$

The value of $\text{pH}$ and concentrations of the species can be determined as follows,

$\text{2-aminobenzoic}\text{acid = HA from}\text{diprotic}{\text{H}}_{\text{2}}{\text{A}}^{\text{+}}\text{,}{\text{pK}}_{\text{1}}\text{ = 2}\text{.08}\text{,}{\text{pK}}_{\text{2}}\text{=}\text{4}\text{.96}\text{.}$

${\text{Dimethylamine = B from monoprotic BH}}^{\text{+}}{\text{ with pK}}_{\text{a}}\text{ = 10}\text{.774}$

The chemical reaction that happened in the given solution is given by

$\begin{array}{l}{\text{<x-custom-btb-me data-me-id="211" class="microExplainerHighlight">Chemical reactions</x-custom-btb-me>: H}}_{\text{2}}{\text{A}}^{\text{+}}\stackrel{}{\rightleftharpoons }\text{HA}\text{+}\text{}{\text{H}}^{\text{+}}{\text{K}}_{\text{1}}\text{=}{\text{10}}^{\text{-2}\text{.08}}\\ \text{HA}\stackrel{}{\rightleftharpoons }{\text{A}}^{\text{- }}{\text{+ H}}^{\text{+}}{\text{K}}_{\text{2}}{\text{= 10}}^{\text{-4}\text{.96}}\\ {\text{BH}}^{\text{+}}\stackrel{}{\rightleftharpoons }{\text{B + H}}^{\text{+}}{\text{K}}_{\text{a}}{\text{= 10}}^{\text{-10}\text{.774}}\\ {\text{H}}_{\text{2}}\text{O}\stackrel{}{\rightleftharpoons }{\text{H}}^{\text{+}}{\text{ + OH}}^{\text{-}}{\text{K}}_{\text{w}}{\text{ = 10}}^{\text{-14}\text{.00}}\end{array}$

$\text{Charge balance:}{\text{[H}}^{\text{+}}\text{] +}\left[{\text{H}}_{\text{2}}{\text{A}}^{\text{+}}\right]{\text{+ [BH}}^{\text{+}}{\text{] = [OH}}^{\text{-}}{\text{]+ [A}}^{\text{-}}{\text{] + [Cl}}^{\text{- }}\text{]}$

$\begin{array}{l}\text{Mass balances:}{\text{[Cl}}^{\text{-}}\text{] = 0}\text{.015 M}\\ {\text{[BH}}^{\text{+}}\text{] + [B] = 0}\text{.020 M }\equiv {\text{ F}}_{\text{B}}\\ {\text{[H}}_{\text{2}}{\text{A}}^{\text{+}}\text{] + }\left[\text{HA}\right]\text{+}{\text{[A}}^{\text{-}}\text{] = 0}\text{.040 M }\equiv {\text{ F}}_{\text{A}}\end{array}$

From, the above information we have eight equations and eight chemical species.

Write the fractional composition expressions

$\begin{array}{l}{\text{[BH}}^{\text{+}}{\text{] = α}}_{{\text{BH}}^{\text{+}}}{\text{F}}_{\text{B}}\text{=}\frac{{\text{[H}}^{\text{+}}{\text{]F}}_{\text{B}}}{{\text{[H}}^{\text{+}}{\text{] + K}}_{\text{a}}}\\ \\ {\text{[B] = α}}_{\text{B}}{\text{F}}_{\text{B}}\text{=}\frac{{\text{K}}_{\text{a}}{\text{F}}_{\text{B}}}{{\text{[H}}^{\text{+}}{\text{] + K}}_{\text{a}}}\end{array}$

${\text{[H}}_{\text{2}}{\text{A}}^{\text{+}}{\text{] = α}}_{{\text{H}}_{\text{2}}{\text{A}}^{\text{+}}}{\text{F}}_{\text{A}}\text{=}\frac{{{\text{[H}}^{\text{+}}\text{]}}^{\text{2}}{\text{F}}_{\text{A}}}{{{\text{[H}}^{\text{+}}\text{]}}^{\text{2}}\text{+}\left[{\text{H}}^{\text{+}}\right]{\text{K}}_{\text{1}}\text{ +}{\text{K}}_{\text{1}}{\text{K}}_{\text{2}}}$

$\begin{array}{l}{\text{[HA] = α}}_{\text{HA}}{\text{F}}_{\text{A}}\text{=}\frac{{\text{K}}_{\text{1}}{\text{[H}}^{\text{+}}{\text{]F}}_{\text{A}}}{{{\text{[H}}^{\text{+}}\text{]}}^{\text{2}}\text{+}\left[{\text{H}}^{\text{+}}\right]{\text{K}}_{\text{1}}{\text{ + K}}_{\text{1}}{\text{K}}_{\text{2}}}\\ \\ {\text{[A}}^{\text{-}}{\text{] = α}}_{{\text{A}}^{\text{-}}}{\text{F}}_{\text{A}}\text{=}\frac{{\text{K}}_{\text{1}}{\text{K}}_{\text{2}}{\text{F}}_{\text{A}}}{{{\text{[H}}^{\text{+}}\text{]}}^{\text{2}}\text{+}\left[{\text{H}}^{\text{+}}\right]{\text{K}}_{\text{1}}{\text{ + K}}_{\text{1}}{\text{K}}_{\text{2}}}\end{array}$

Substitute the above values into the charge balance equation

$\left[{\text{H}}^{\text{+}}\right]\text{+}\text{}{\text{α}}_{{\text{H}}_{\text{2}}{\text{A}}^{\text{+}}}{\text{F}}_{\text{A}}\text{+}{\text{α}}_{{\text{BH}}^{\text{+}}}{\text{F}}_{\text{B}}\text{=}{\text{K}}_{\text{W}}\text{/}\left[{\text{H}}^{\text{+}}\right]\text{+}\text{}{\text{α}}_{{\text{A}}^{\text{-}}}{\text{F}}_{\text{A}}\text{+}\left[\text{0}\text{.015}\text{M}\right]\text{(A)}$

The value of $\text{pH}$ and concentrations of species in the given solution are mentioned in the following spreadsheet in figure 1.

Figure 1

The value of $\text{pH}$ is $\text{4}\text{.153}\text{.}$

The concentration of species in the charge balance equation are given below:

$\begin{array}{l}\left[{\text{H}}^{\text{+}}\right]\text{=}\text{7}\text{.03}\text{×}{\text{10}}^{\text{-5}}\text{M}\text{;}\left[{\text{BH}}^{\text{+}}\right]\text{=}\text{2}\text{.00}\text{×}{\text{10}}^{\text{-2}}\text{M}\text{;}\left[{\text{A}}^{\text{-}}\right]\text{=}\text{5}\text{.36}\text{×}{\text{10}}^{\text{-3}}\text{M}\\ \\ \left[{\text{Cl}}^{\text{-}}\right]\text{=}\text{0}\text{.015}\text{M}\text{;}\left[{\text{OH}}^{\text{-}}\right]\text{=}\text{1}\text{.42}\text{×}{\text{10}}^{\text{-10}}\text{M}\text{;}\left[{\text{H}}_{\text{2}}{\text{A}}^{\text{+}}\right]\text{=}\text{2}\text{.90}\text{×}{\text{10}}^{\text{-4}}\text{M}\end{array}$

The other concentrations are,

$\left[\text{HA}\right]\text{=}\text{3}\text{.43}\text{×}{\text{10}}^{\text{-2}}\text{M}\text{;}\left[\text{B}\right]\text{=}\text{4}\text{.79}\text{×}{\text{10}}^{\text{-9}}\text{M}\text{.}$

(b)

Interpretation Introduction

Interpretation:

The fraction of $\text{HA}$ is in each of its three forms and fraction of $\text{B}$ is in each of its two forms has to be given.

Explanation of Solution

Given information,

Figure 1

The value of $\text{pH}$ is $\text{4}\text{.153}\text{.}$

The concentration of species are given below:

$\begin{array}{l}\left[{\text{H}}^{\text{+}}\right]\text{=}\text{7}\text{.03}\text{×}{\text{10}}^{\text{-5}}\text{M}\text{;}\left[{\text{BH}}^{\text{+}}\right]\text{=}\text{2}\text{.00}\text{×}{\text{10}}^{\text{-2}}\text{M}\text{;}\left[{\text{A}}^{\text{-}}\right]\text{=}\text{5}\text{.36}\text{×}{\text{10}}^{\text{-3}}\text{M}\\ \\ \left[{\text{Cl}}^{\text{-}}\right]\text{=}\text{0}\text{.015}\text{M}\text{;}\left[{\text{OH}}^{\text{-}}\right]\text{=}\text{1}\text{.42}\text{×}{\text{10}}^{\text{-10}}\text{M}\text{;}\left[{\text{H}}_{\text{2}}{\text{A}}^{\text{+}}\right]\text{=}\text{2}\text{.90}\text{×}{\text{10}}^{\text{-4}}\text{M}\end{array}$

The other concentrations are,

$\left[\text{HA}\right]\text{=}\text{3}\text{.43}\text{×}{\text{10}}^{\text{-2}}\text{M}\text{;}\left[\text{B}\right]\text{=}\text{4}\text{.79}\text{×}{\text{10}}^{\text{-9}}\text{M}\text{.}$

The calculated fraction of 2-aminobenzoic acid $\left(\text{HA}\right)$ is in each of its three forms are given below:

${\text{H}}_{\text{2}}{\text{A}}^{\text{+}}\text{=}\text{0}\text{.7\%}\text{;}\text{HA}\text{=}\text{85}\text{.9}\text{\%}\text{;}{\text{A}}^{\text{-}}\text{=13}\text{.4}\text{\%}\text{.}$

The calculated fraction of methylamine $\left({\text{BH}}^{\text{+}}\right)$ is in each of its two forms are given below:

${\text{BH}}^{\text{+}}\text{=}\text{100}\text{\%}\text{;}\text{B}\text{=}\text{0}\text{.0}\text{\%}\text{.}$

The simple prediction is that $\text{HCl}$ would consume $\text{B}$, giving $\text{100}\text{\%}{\text{BH}}^{\text{+}}$.  The remaining $\text{5}\text{mmol}\text{B}$ consumes $\text{5}\text{mmol}\text{HA}$, making $\text{5}\text{mmol}{\text{A}}^{\text{-}}$ and leaving $\text{35}\text{mmol}\text{HA}$.

$\begin{array}{l}\text{predicted}\text{fractions}\text{are:}\\ {\text{A}}^{\text{-}}\text{=}\frac{\text{5}}{\text{40}}\text{=}\text{12}\text{.5\%}\text{;}\text{HA}\text{=}\frac{\text{35}}{\text{40}}\text{=}\text{87}\text{.5}\text{\%}\end{array}$

The estimated $\text{pH}$ is calculated as follows

$\begin{array}{l}\text{pH}\text{=}{\text{pK}}_{\text{2}}\text{+}\text{log}\left(\left[{\text{A}}^{\text{-}}\right]\text{/}\left[\text{HA}\right]\right)\\ \\ \text{=}\text{4}\text{.96}\text{+}\text{log}\left(\text{5/35}\right)\text{=}\text{4}\text{.11}\text{.}\end{array}$