Chapter 8, Problem 8.28P

(a)

Interpretation Introduction

Interpretation:

The equilibria and charge and mass balances needed to find the composition of $\text{0}\text{.01}\text{M}$ sodium acetate $\left({\text{Na}}^{\text{+}}{\text{A}}^{\text{-}}\right)$ has to be given.

Concept introduction:

Charge balance:

The overall positive charges in solution equals the overall negative charges in solution.

$\begin{array}{l}{\text{n}}_{\text{1}}\left[{\text{C}}_{\text{1}}\right]\text{+}{\text{n}}_{\text{2}}\left[{\text{C}}_{\text{2}}\right]\text{+}\mathrm{......}\text{=}{\text{m}}_{\text{1}}\left[{\text{A}}_{\text{1}}\right]\text{+}{\text{m}}_{\text{2}}\left[{\text{A}}_{\text{2}}\right]\text{+}\mathrm{......}\\ \text{where,}\\ \left[\text{C}\right]\text{-}\text{concentration}\text{of}\text{a}\text{cation}\text{and}\text{n-}\text{charge}\text{of}\text{the}\text{cation}\text{.}\\ \left[\text{A}\right]\text{-}\text{concentration}\text{of}\text{a}\text{anion}\text{and}\text{m}\text{-}\text{magnitude}\text{of}\text{the}\text{charge}\text{of}\text{the}\text{anion}\text{.}\end{array}$

Mass balance:

The amount of all the species in a solution containing a particular atom (or a group of atoms) must equal the amount of that atom (or group) delivered to the solution.

Explanation of Solution

Given information,

$\text{0}\text{.01}\text{M}\text{sodium}\text{acetate}\text{is}\text{given}\text{.}$

Pertinent reactions are:

$\begin{array}{l}{\text{A}}^{\text{-}}{\text{+H}}_{\text{2}}\text{O}\stackrel{{\text{K}}_{\text{b}}}{\rightleftharpoons }\text{HA}\text{+}\text{}{\text{OH}}^{\text{-}}{\text{K}}_{\text{b}}\text{=}\frac{\left[\text{HA}\right]{\text{γ}}_{\text{HA}}\left[{\text{OH}}^{\text{-}}\right]{\text{γ}}_{{\text{OH}}^{\text{-}}}}{\left[{\text{A}}^{\text{-}}\right]{\text{γ}}_{{\text{A}}^{\text{-}}}}\text{=}\text{5}\text{.7}\text{×}{\text{10}}^{\text{-10}}\\ \\ {\text{H}}_{\text{2}}\text{O}\stackrel{{\text{K}}_{\text{W}}}{\rightleftharpoons }{\text{H}}^{\text{+}}\text{+}\text{}{\text{OH}}^{\text{-}}{\text{K}}_{\text{W}}\text{=}\left[{\text{H}}^{\text{+}}\right]{\text{γ}}_{{\text{H}}^{\text{+}}}\left[{\text{OH}}^{\text{-}}\right]{\text{γ}}_{{\text{OH}}^{\text{-}}}\text{=}\text{1}\text{.0}\text{×}{\text{10}}^{\text{-14}}\end{array}$

Write the charge balance equation

$\left[{\text{H}}^{\text{+}}\right]\text{+}\text{}\left[{\text{Na}}^{\text{+}}\right]\text{=}\left[{\text{OH}}^{\text{-}}\right]\text{+}\left[{\text{A}}^{\text{-}}\right]$

Write the mass balance equation

$\left[{\text{Na}}^{\text{+}}\right]\text{=}\text{0}\text{.01}\text{M}\equiv \text{F}$

$\begin{array}{l}\left[\text{HA}\right]\text{+}\left[{\text{A}}^{\text{-}}\right]\text{=}\text{0}\text{.01}\text{M}\equiv \text{F}\\ \left(\text{where,}\text{F}\text{-formal}\text{concentration}\right)\end{array}$

(b)

Interpretation Introduction

Interpretation:

The concentration of species in $\text{0}\text{.01}\text{M}$ sodium acetate $\left({\text{Na}}^{\text{+}}{\text{A}}^{\text{-}}\right)$ has to be given and also the value of pH and fraction of hydrolysis has to be given.

Concept introduction:

Charge balance:

The overall positive charges in solution equals the overall negative charges in solution.

$\begin{array}{l}{\text{n}}_{\text{1}}\left[{\text{C}}_{\text{1}}\right]\text{+}{\text{n}}_{\text{2}}\left[{\text{C}}_{\text{2}}\right]\text{+}\mathrm{......}\text{=}{\text{m}}_{\text{1}}\left[{\text{A}}_{\text{1}}\right]\text{+}{\text{m}}_{\text{2}}\left[{\text{A}}_{\text{2}}\right]\text{+}\mathrm{......}\\ \text{where,}\\ \left[\text{C}\right]\text{-}\text{concentration}\text{of}\text{a}\text{cation}\text{and}\text{n-}\text{charge}\text{of}\text{the}\text{cation}\text{.}\\ \left[\text{A}\right]\text{-}\text{concentration}\text{of}\text{a}\text{anion}\text{and}\text{m}\text{-}\text{magnitude}\text{of}\text{the}\text{charge}\text{of}\text{the}\text{anion}\text{.}\end{array}$

Mass balance:

The amount of all the species in a solution containing a particular atom (or a group of atoms) must equal the amount of that atom (or group) delivered to the solution.

Explanation of Solution

Given information,

Pertinent reactions are:

$\begin{array}{l}{\text{A}}^{\text{-}}{\text{+H}}_{\text{2}}\text{O}\stackrel{{\text{K}}_{\text{b}}}{\rightleftharpoons }\text{HA}\text{+}\text{}{\text{OH}}^{\text{-}}{\text{K}}_{\text{b}}\text{=}\frac{\left[\text{HA}\right]{\text{γ}}_{\text{HA}}\left[{\text{OH}}^{\text{-}}\right]{\text{γ}}_{{\text{OH}}^{\text{-}}}}{\left[{\text{A}}^{\text{-}}\right]{\text{γ}}_{{\text{A}}^{\text{-}}}}\text{=}\text{5}\text{.7}\text{×}{\text{10}}^{\text{-10}}\text{(A)}\\ \\ {\text{H}}_{\text{2}}\text{O}\stackrel{{\text{K}}_{\text{W}}}{\rightleftharpoons }{\text{H}}^{\text{+}}\text{+}\text{}{\text{OH}}^{\text{-}}{\text{K}}_{\text{W}}\text{=}\left[{\text{H}}^{\text{+}}\right]{\text{γ}}_{{\text{H}}^{\text{+}}}\left[{\text{OH}}^{\text{-}}\right]{\text{γ}}_{{\text{OH}}^{\text{-}}}\text{=}\text{1}\text{.0}\text{×}{\text{10}}^{\text{-14}}\text{(B)}\end{array}$

Write the charge balance equation

$\left[{\text{H}}^{\text{+}}\right]\text{+}\text{}\left[{\text{Na}}^{\text{+}}\right]\text{=}\left[{\text{OH}}^{\text{-}}\right]\text{+}\left[{\text{A}}^{\text{-}}\right]\text{(C)}$

Write the mass balance equation

$\left[{\text{Na}}^{\text{+}}\right]\text{=}\text{0}\text{.01}\text{M}\equiv \text{F}\text{(D)}$

$\begin{array}{l}\left[\text{HA}\right]\text{+}\left[{\text{A}}^{\text{-}}\right]\text{=}\text{0}\text{.01}\text{M}\equiv \text{F}\text{(E)}\\ \left(\text{where,}\text{F}\text{-formal}\text{concentration}\right)\end{array}$

First neglect activity coefficients. Make the following substitutions in the charge balance equation:

$\begin{array}{l}\left[{\text{OH}}^{\text{-}}\right]\text{=}{\text{K}}_{\text{W}}\text{/}\left[{\text{H}}^{\text{+}}\right]\\ \\ \left[\text{HA}\right]\text{=}\frac{{\text{K}}_{\text{b}}\left[{\text{A}}^{\text{-}}\right]}{\left[{\text{OH}}^{\text{-}}\right]}\text{(F)}\\ \\ \left[\text{HA}\right]\text{=}\text{F}\text{-}\left[{\text{A}}^{\text{-}}\right]\text{(G)}\end{array}$

Equate the expressions for $\left[\text{HA}\right]$ from equations F and G to solve for $\left[{\text{A}}^{\text{-}}\right]$

$\begin{array}{l}\frac{{\text{K}}_{\text{b}}\left[{\text{A}}^{\text{-}}\right]}{\left[{\text{OH}}^{\text{-}}\right]}\text{=}\text{F}\text{-}\left[{\text{A}}^{\text{-}}\right]\\ \\ \left[{\text{A}}^{\text{-}}\right]\left(\frac{{\text{K}}_{\text{b}}}{\left[{\text{OH}}^{\text{-}}\right]}\text{+}\text{}\text{1}\right)\text{=}\text{F}\\ \\ \left[{\text{A}}^{\text{-}}\right]\text{=}\frac{\text{F}\left[{\text{OH}}^{\text{-}}\right]}{{\text{K}}_{\text{b}}\text{+}\left[{\text{OH}}^{\text{-}}\right]}\end{array}$

Now substitute ${\text{K}}_{\text{W}}\text{/}\left[{\text{H}}^{\text{+}}\right]$ for each $\left[{\text{OH}}^{\text{-}}\right]$ in the above equation to get

$\left[{\text{A}}^{\text{-}}\right]\text{=}\frac{{\text{FK}}_{\text{W}}\text{/}\left[{\text{H}}^{\text{+}}\right]}{{\text{K}}_{\text{b}}\text{+}\text{}{\text{K}}_{\text{W}}\text{/}\left[{\text{H}}^{\text{+}}\right]}\text{=}\frac{{\text{FK}}_{\text{W}}}{{\text{K}}_{\text{b}}\left[{\text{H}}^{\text{+}}\right]\text{+}\text{}{\text{K}}_{\text{W}}}\text{(H)}$

We can now substitute for all terms in the charge balance using $\left[{\text{A}}^{\text{-}}\right]$ from the equation H

$\left[{\text{OH}}^{\text{-}}\right]\text{=}{\text{K}}_{\text{W}}\text{/}\left[{\text{H}}^{\text{+}}\right]\text{,}\text{and}\left[{\text{Na}}^{\text{+}}\right]\text{=}\text{F}$

Charge balance:

$\begin{array}{l}\left[{\text{H}}^{\text{+}}\right]\text{+}\left[{\text{Na}}^{\text{+}}\right]\text{=}\left[{\text{OH}}^{\text{-}}\right]\text{+}\left[{\text{A}}^{\text{-}}\right]\\ \\ \left[{\text{H}}^{\text{+}}\right]\text{+}\text{}\text{F}\text{=}\frac{{\text{K}}_{\text{W}}}{\left[{\text{H}}^{\text{+}}\right]}\text{+}\text{}\frac{{\text{FK}}_{\text{W}}}{{\text{K}}_{\text{b}}\left[{\text{H}}^{\text{+}}\right]\text{+}\text{}{\text{K}}_{\text{W}}}\text{(I)}\end{array}$

Rearrange the above equation

$\text{0}\text{=}\frac{{\text{K}}_{\text{W}}}{\left[{\text{H}}^{\text{+}}\right]}\text{+}\text{}\frac{{\text{FK}}_{\text{W}}}{{\text{K}}_{\text{b}}\left[{\text{H}}^{\text{+}}\right]\text{+}\text{}{\text{K}}_{\text{W}}}\text{-}\left[{\text{H}}^{\text{+}}\right]\text{-}\text{}\text{F}\text{(J)}$

The spreadsheet for the calculation of species concentrations is given in figure 1

$\left[{\text{OH}}^{\text{-}}\right]\text{=}\frac{{\text{K}}_{\text{W}}}{\left[{\text{H}}^{\text{+}}\right]}\left[{\text{A}}^{\text{-}}\right]\text{=}\frac{{\text{FK}}_{\text{W}}}{{\text{K}}_{\text{b}}\left[{\text{H}}^{\text{+}}\right]\text{+}\text{}{\text{K}}_{\text{W}}}\left[\text{HA}\right]\text{=}\frac{{\text{K}}_{\text{b}}\left[{\text{A}}^{\text{-}}\right]}{\left[{\text{OH}}^{\text{-}}\right]}$

Figure 1

The concentrations are given below

$\begin{array}{l}\left[{\text{A}}^{\text{-}}\right]\text{=}\text{1}\text{.00}\text{×}{\text{10}}^{\text{-2}}\text{M}\\ \left[{\text{OH}}^{\text{-}}\right]\text{=}\text{2}\text{.39}\text{×}{\text{10}}^{\text{-6}}\text{M}\\ \left[\text{HA}\right]\text{=}\text{2}\text{.39}\text{×}{\text{10}}^{\text{-6}}\text{M}\\ \left[{\text{H}}^{\text{+}}\right]\text{=}\text{4}\text{.19}\text{×}{\text{10}}^{\text{-9}}\text{M}\end{array}$

The value of ionic strength is $\text{0}\text{.0100}\text{M}\text{.}$

The value of pH is $\text{8}\text{.38}$.

Fraction of hydrolysis

$\begin{array}{l}\text{Fraction}\text{of}\text{hydrolysis}\text{=}\frac{\left[\text{HA}\right]}{\text{F}}\\ \\ \text{=}\frac{\text{2}\text{.39}\text{×}{\text{10}}^{\text{-6}}\text{M}}{\text{0}\text{.01}\text{M}}\text{=}\text{0}\text{.000239}\text{=}\text{0}\text{.024}\text{\%}\text{.}\end{array}$