Chapter 13, Problem 13.AE

Interpretation Introduction

Interpretation:

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

Explanation of Solution

The given solution containing $\text{0}\text{.010 mol}$ hydroxybenzene (HA), $\text{0}\text{.030 mol}$ dimethyamine (B), $\text{0}\text{.015 mol}\text{HCl}\text{.}$

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

${\text{Hydroxybenzcne = HA with pK}}_{\text{HA}}\text{ = 9}\text{.997}$

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

The chemical reaction that happened in the given solution can be shown as follows.

$\begin{array}{l}\text{<x-custom-btb-me data-me-id="211" class="microExplainerHighlight">Chemical reactions</x-custom-btb-me>: }\text{HA}\stackrel{}{\rightleftharpoons }{\text{A}}^{\text{- }}{\text{+ H}}^{\text{+}}{\text{K}}_{\text{HA}}{\text{ = 10}}^{\text{-9}\text{.997}}\\ {\text{BH}}^{\text{+}}\stackrel{}{\rightleftharpoons }{\text{B + H}}^{\text{+}}{\text{K}}_{{\text{BH}}^{\text{+}}}{\text{ = 10}}^{\text{-10}\text{.774}}\\ {\text{H}}_{\text{2}}\text{O}\stackrel{}{\rightleftharpoons }{\text{H}}^{\text{+}}{\text{ + OH}}^{\text{-}}{\text{K}}_w{\text{ = 10}}^{\text{-14}}\end{array}$

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

$\begin{array}{l}\text{Mass balances:}{\text{[Cl}}^{\text{-}}\text{] = 0}\text{.015 M}\\ {\text{[BH}}^{\text{+}}\text{] + [B] = 0}{\text{.030 M = F}}_{\text{B}}\\ {\text{[HA] + [A}}^{\text{-}}\text{] = 0}{\text{.010 M = F}}_{\text{A}}\end{array}$

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

For this we have to determine the fractional composition equations as follows.

${\text{[BH}}^{\text{+}}{\text{] = α}}_{\text{BH}}+{\text{F}}_{\text{B}}=\frac{{\text{[H}}^{\text{+}}{\text{]F}}_{\text{B}}}{{\text{[H}}^{\text{+}}{\text{] + K}}_{{\text{BH}}^{+}}}$

${\text{[B] = α}}_{\text{B}}{\text{F}}_{\text{B}}=\frac{{\text{K}}_{\text{BH}}{\text{+F}}_{\text{B}}}{{\text{[H}}^{\text{+}}{\text{] + K}}_{{\text{BH}}^{+}}}$

${\text{[HA] = α}}_{\text{HA}}{\text{F}}_{\text{A}}=\frac{{\text{[H}}^{\text{+}}{\text{]F}}_{\text{B}}}{{\text{[H}}^{\text{+}}{\text{] + K}}_{\text{HA}}}$

${\text{[A}}^{-}{\text{] = α}}_{\text{A}}{\text{-F}}_{\text{A}}=\frac{{\text{K}}_{\text{HA}}{\text{F}}_{\text{A}}}{{\text{[H}}^{\text{+}}{\text{] + K}}_{\text{HA}}}$

Let’s do the charge balance by substituting and we get the equation as follows and is denoted by A.

${\text{[H}}^{\text{+}}{\text{] + α}}_{\text{BH}}{\text{+ F}}_{\text{B}}{\text{ = K}}_{\text{w}}{\text{/[H}}^{\text{+}}{\text{] + α}}_{\text{A}}{\text{- F}}_{\text{A}}\text{+ [0}\text{.0 15 M]}\text{(A)}$

Now, we are going to solve this equation A for ${\text{[H}}^{\text{+}}\text{]}$ by using the Solver in the spreadsheet and have the detailed information in the following spreadsheet.

The $\text{pH}$ and concentrations of species in the given solution is mentioned in the following spreadsheet.

Figure 1

Conclusion

The $\text{pH}$ and concentrations of species in the given solution was determined.