Chapter 8, Problem 8.29P

(a)

Interpretation Introduction

Interpretation:

The equations needed to find the solubility of $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$ 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

Pertinent reactions are:

$\begin{array}{l}\text{Ca}{\left(\text{OH}\right)}_{\text{2}}{}_{\left(\text{S}\right)}\stackrel{{\text{K}}_{\text{SP}}}{\rightleftharpoons }{\text{Ca}}^{\text{2+}}\text{+}\text{2}\text{}{\text{OH}}^{\text{-}}{\text{K}}_{\text{SP}}\text{=}\left[{\text{Ca}}^{\text{2+}}\right]{\text{γ}}_{{\text{Ca}}^{\text{2+}}}{\left[{\text{OH}}^{\text{-}}\right]}^{\text{2}}{\text{γ}}^{\text{2}}{}_{{\text{OH}}^{\text{-}}}\text{=}\text{6}\text{.5}\text{×}{\text{10}}^{\text{-6}}\\ \\ {\text{Ca}}^{\text{2+}}\text{+}\text{}{\text{OH}}^{\text{-}}\stackrel{{\text{K}}_{\text{1}}}{\rightleftharpoons }{\text{CaOH}}^{\text{+}}{\text{K}}_{\text{1}}\text{=}\frac{\left[{\text{CaOH}}^{\text{+}}\right]{\text{γ}}_{{\text{CaOH}}^{\text{+}}}}{\left[{\text{Ca}}^{\text{2+}}\right]{\text{γ}}_{{\text{Ca}}^{\text{2+}}}\left[{\text{OH}}^{\text{-}}\right]{\text{γ}}_{{\text{OH}}^{\text{-}}}}\text{=}\text{2}\text{.0}\text{×}{\text{10}}^{\text{1}}\\ \\ {\text{H}}_{\text{2}}\text{O}\stackrel{{\text{K}}_{\text{W}}}{\rightleftharpoons }{\text{H}}^{\text{+}}\text{+}\text{}{\text{OH}}^{\text{-}}{\text{K}}_{\text{W}}\text{=}\left[{\text{H}}^{\text{+}}\right]{\text{γ}}_{{\text{H}}^{\text{+}}}\left[{\text{OH}}^{\text{-}}\right]{\text{γ}}_{{\text{OH}}^{\text{-}}}\text{=}\text{1}\text{.0}\text{×}{\text{10}}^{\text{-14}}\end{array}$

Write the charge balance equation

$\text{2}\left[{\text{Ca}}^{\text{2+}}\right]\text{+}\text{}\left[{\text{CaOH}}^{\text{+}}\right]\text{+}\text{}\left[{\text{H}}^{\text{+}}\right]\text{=}\left[{\text{OH}}^{\text{-}}\right]$

Write the mass balance equation

$\underset{\text{Species}\text{containing}{\text{OH}}^{\text{-}}}{\underbrace{\left[{\text{OH}}^{\text{-}}\right]\text{+}\text{}\left[{\text{CaOH}}^{\text{+}}\right]}}\text{=}\text{2}\underset{\text{Species}\text{containing}{\text{Ca}}^{\text{2+}}}{\underbrace{\left\{\left[{\text{Ca}}^{\text{2+}}\right]\text{+}\text{}\left[{\text{CaOH}}^{\text{+}}\right]\right\}}}\text{+}\text{}\left[{\text{H}}^{\text{+}}\right]$

There are 4 equations (3 equilibria and charge balance) and 4 unknowns:

$\left[{\text{Ca}}^{\text{2+}}\right]\text{,}\left[{\text{CaOH}}^{\text{+}}\right]\text{,}\left[{\text{H}}^{\text{+}}\right]\text{and}\left[{\text{OH}}^{\text{-}}\right]\text{.}$

(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

Pertinent reactions are:

$\begin{array}{l}\text{Ca}{\left(\text{OH}\right)}_{\text{2}}{}_{\left(\text{S}\right)}\stackrel{{\text{K}}_{\text{SP}}}{\rightleftharpoons }{\text{Ca}}^{\text{2+}}\text{+}\text{2}\text{}{\text{OH}}^{\text{-}}{\text{K}}_{\text{SP}}\text{=}\left[{\text{Ca}}^{\text{2+}}\right]{\text{γ}}_{{\text{Ca}}^{\text{2+}}}{\left[{\text{OH}}^{\text{-}}\right]}^{\text{2}}{\text{γ}}^{\text{2}}{}_{{\text{OH}}^{\text{-}}}\text{=}\text{6}\text{.5}\text{×}{\text{10}}^{\text{-6}}\\ \\ {\text{Ca}}^{\text{2+}}\text{+}\text{}{\text{OH}}^{\text{-}}\stackrel{{\text{K}}_{\text{1}}}{\rightleftharpoons }{\text{CaOH}}^{\text{+}}{\text{K}}_{\text{1}}\text{=}\frac{\left[{\text{CaOH}}^{\text{+}}\right]{\text{γ}}_{{\text{CaOH}}^{\text{+}}}}{\left[{\text{Ca}}^{\text{2+}}\right]{\text{γ}}_{{\text{Ca}}^{\text{2+}}}\left[{\text{OH}}^{\text{-}}\right]{\text{γ}}_{{\text{OH}}^{\text{-}}}}\text{=}\text{2}\text{.0}\text{×}{\text{10}}^{\text{1}}\\ \\ {\text{H}}_{\text{2}}\text{O}\stackrel{{\text{K}}_{\text{W}}}{\rightleftharpoons }{\text{H}}^{\text{+}}\text{+}\text{}{\text{OH}}^{\text{-}}{\text{K}}_{\text{W}}\text{=}\left[{\text{H}}^{\text{+}}\right]{\text{γ}}_{{\text{H}}^{\text{+}}}\left[{\text{OH}}^{\text{-}}\right]{\text{γ}}_{{\text{OH}}^{\text{-}}}\text{=}\text{1}\text{.0}\text{×}{\text{10}}^{\text{-14}}\end{array}$

The charge balance equation

$\text{2}\left[{\text{Ca}}^{\text{2+}}\right]\text{+}\text{}\left[{\text{CaOH}}^{\text{+}}\right]\text{+}\text{}\left[{\text{H}}^{\text{+}}\right]\text{=}\left[{\text{OH}}^{\text{-}}\right]$

The mass balance equation

$\underset{\text{Species}\text{containing}{\text{OH}}^{\text{-}}}{\underbrace{\left[{\text{OH}}^{\text{-}}\right]\text{+}\text{}\left[{\text{CaOH}}^{\text{+}}\right]}}\text{=}\text{2}\underset{\text{Species}\text{containing}{\text{Ca}}^{\text{2+}}}{\underbrace{\left\{\left[{\text{Ca}}^{\text{2+}}\right]\text{+}\text{}\left[{\text{CaOH}}^{\text{+}}\right]\right\}}}\text{+}\text{}\left[{\text{H}}^{\text{+}}\right]$

First neglect activity coefficients and neglect $\left[{\text{H}}^{\text{+}}\right]$ in the charge balance equation because $\left[{\text{H}}^{\text{+}}\right]\text{<<}\left[{\text{OH}}^{\text{-}}\right]$ in basic solution.

Now the charge balance equation becomes

$\text{2}\left[{\text{Ca}}^{\text{2+}}\right]\text{+}\text{}\left[{\text{CaOH}}^{\text{+}}\right]\text{=}\left[{\text{OH}}^{\text{-}}\right]\text{(A)}$

Substitute $\left[{\text{CaOH}}^{\text{+}}\right]\text{=}{\text{K}}_{\text{1}}\left[{\text{Ca}}^{\text{2+}}\right]\left[{\text{OH}}^{\text{-}}\right]$ in the above equation

$\begin{array}{l}\text{2}\left[{\text{Ca}}^{\text{2+}}\right]\text{+}{\text{K}}_{\text{1}}\left[{\text{Ca}}^{\text{2+}}\right]\left[{\text{OH}}^{\text{-}}\right]\text{=}\left[{\text{OH}}^{\text{-}}\right]\\ \\ \left[{\text{Ca}}^{\text{2+}}\right]\text{=}\frac{\left[{\text{OH}}^{\text{-}}\right]}{\text{2}\text{+}{\text{K}}_{\text{1}}\left[{\text{OH}}^{\text{-}}\right]}\end{array}$

Substitute the above expression for $\left[{\text{Ca}}^{\text{2+}}\right]$ into the solubility product expression:

${\text{K}}_{\text{SP}}\text{=}\left[{\text{Ca}}^{\text{2+}}\right]{\left[{\text{OH}}^{\text{-}}\right]}^{\text{2}}\text{=}\frac{{\left[{\text{OH}}^{\text{-}}\right]}^{\text{3}}}{\text{2}\text{+}\text{}{\text{K}}_{\text{1}}\left[{\text{OH}}^{\text{-}}\right]}\text{(B)}$

The spreadsheet to calculate the concentration of species is given in figure 1

Figure 1

The concentration of species is given below

$\begin{array}{l}\left[{\text{Ca}}^{\text{2+}}\right]\text{=}\text{0}\text{.0101}\text{M}\text{;}\left[{\text{CaOH}}^{\text{+}}\right]\text{=}\text{0}\text{.0051}\text{M}\\ \\ \left[{\text{OH}}^{\text{-}}\right]\text{=}\text{0}\text{.0254}\text{M}\text{;}\left[{\text{H}}^{\text{+}}\right]\text{=}\frac{{\text{K}}_{\text{W}}}{\left[{\text{OH}}^{\text{-}}\right]}\text{=}\text{3}\text{.9}\text{×}{\text{10}}^{\text{-13}}\text{M}\end{array}$

Total dissolved $\text{Ca}\text{=}\text{0}\text{.0101}\text{+}\text{0}\text{.0051}\text{=}\text{0}\text{.0152}\text{M}\text{.}$

The solubility is given by

$\begin{array}{l}\text{Formula}\text{mass}\text{of}\text{Ca}{\left(\text{OH}\right)}_{\text{2}}\text{is}\text{74}\text{.09}\text{g/mol}\text{.}\\ \text{so,}\text{0}\text{.0152}\text{M}\text{is}\text{1}\text{.13}\text{g/L}\text{.}\end{array}$

Fraction of hydrolysis is given by

$\begin{array}{l}\text{Fraction}\text{of}\text{hydrolysis}\text{=}\left[{\text{CaOH}}^{\text{+}}\right]\text{/}\left(\left[{\text{Ca}}^{\text{2+}}\right]\text{+}\left[{\text{CaOH}}^{\text{+}}\right]\right)\\ \\ \text{=}\frac{\text{0}\text{.0051}\text{M}}{\text{0}\text{.0152}\text{M}\text{+}\text{0}\text{.0051}\text{M}}\text{=}\text{0}\text{.2512}\text{=}\text{25\%}\text{.}\end{array}$