Chapter 8, Problem 8.IE

(a)

Interpretation Introduction

Interpretation:

The concentration of species in $\text{0}\text{.10}\text{M}{\text{NaClO}}_{\text{4}}$ saturated with $\text{Mn}{\left(\text{OH}\right)}_{\text{2}}$ has to be calculated.

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,

$\begin{array}{l}\text{0}\text{.10}\text{M}{\text{NaClO}}_{\text{4}}\text{saturated}\text{with}\text{Mn}{\left(\text{OH}\right)}_{\text{2}}\text{.}\\ \text{Value}\text{of}\text{µ}\text{=}\text{0}\text{.1}\text{.}\\ \text{Size}\text{of}{\text{MnOH}}^{\text{+}}\text{is}\text{400}\text{pm}\text{.}\end{array}$

$\begin{array}{l}\text{Mn}{\left(\text{OH}\right)}_{\text{2}}{}_{\left(\text{s}\right)}\stackrel{{\text{K}}_{\text{SP}}}{\rightleftharpoons }{\text{Mn}}^{\text{2+}}\text{+}{\text{2OH}}^{\text{-}}{\text{K}}_{\text{SP}}\text{=}{\text{10}}^{\text{-12}\text{.8}}\\ {\text{Mn}}^{\text{2+}}\text{+}{\text{OH}}^{\text{-}}\stackrel{{\text{K}}_{\text{1}}}{\rightleftharpoons }{\text{MnOH}}^{\text{+}}{\text{K}}_{\text{1}}\text{=}{\text{10}}^{\text{3}\text{.4}}\\ {\text{H}}_{\text{2}}\text{O}\stackrel{}{\rightleftharpoons }{\text{H}}^{\text{+}}\text{+}{\text{OH}}^{\text{-}}{\text{K}}_{\text{W}}\text{=}{\text{10}}^{\text{-14}\text{.00}}\end{array}$

Write the charge balance equation

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

Write the mass balance equation

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

Mass balance is equivalent to charge balance.

The equilibrium constant expression is given by

$\begin{array}{l}{\text{K}}_{\text{SP}}\text{=}\left[{\text{Mn}}^{\text{2+}}\right]{\text{γ}}_{{\text{Mn}}^{\text{2+}}}{\left[{\text{OH}}^{\text{-}}\right]}^{\text{2}}{\text{γ}}^{\text{2}}{}_{{\text{OH}}^{\text{-}}}\\ \\ {\text{K}}_{\text{1}}\text{=}\frac{\left[{\text{MnOH}}^{\text{+}}\right]{\text{γ}}_{{\text{MnOH}}^{\text{+}}}}{\left[{\text{Mn}}^{\text{2+}}\right]{\text{γ}}_{{\text{Mn}}^{\text{2+}}}\left[{\text{OH}}^{\text{-}}\right]{\text{γ}}_{{\text{OH}}^{\text{-}}}}\\ \\ {\text{K}}_{\text{W}}\text{=}\left[{\text{H}}^{\text{+}}\right]{\text{γ}}_{{\text{H}}^{\text{+}}}\left[{\text{OH}}^{\text{-}}\right]{\text{γ}}_{{\text{OH}}^{\text{-}}}\end{array}$

Figure 1

The concentration of all the species is given below:

$\begin{array}{l}\left[{\text{Mn}}^{\text{2+}}\right]\text{=}\text{5}\text{.349}\text{×}{\text{10}}^{\text{-5}}\text{M}\\ \left[{\text{OH}}^{\text{-}}\right]\text{=}\text{1}\text{.130}\text{×}{\text{10}}^{\text{-4}}\text{M}\\ \left[{\text{MnOH}}^{\text{+}}\right]\text{=}\text{6}\text{.016}\text{×}{\text{10}}^{\text{-6}}\text{M}\\ \left[{\text{H}}^{\text{+}}\right]\text{=}\text{1}\text{.408}\text{×}{\text{10}}^{\text{-10}}\text{M}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The concentration of species in $\text{0}\text{.10}\text{M}{\text{NaClO}}_{\text{4}}$ saturated with $\text{Mn}{\left(\text{OH}\right)}_{\text{2}}$ has to be calculated if there is no ${\text{NaClO}}_{\text{4}}$ in the solution.

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,

$\begin{array}{l}\text{0}\text{.10}\text{M}{\text{NaClO}}_{\text{4}}\text{saturated}\text{with}\text{Mn}{\left(\text{OH}\right)}_{\text{2}}\text{.}\\ \text{Value}\text{of}\text{µ}\text{=}\text{0}\text{.1}\text{.}\\ \text{Size}\text{of}{\text{MnOH}}^{\text{+}}\text{is}\text{400}\text{pm}\text{.}\end{array}$

$\begin{array}{l}\text{Mn}{\left(\text{OH}\right)}_{\text{2}}{}_{\left(\text{s}\right)}\stackrel{{\text{K}}_{\text{SP}}}{\rightleftharpoons }{\text{Mn}}^{\text{2+}}\text{+}{\text{2OH}}^{\text{-}}{\text{K}}_{\text{SP}}\text{=}{\text{10}}^{\text{-12}\text{.8}}\\ {\text{Mn}}^{\text{2+}}\text{+}{\text{OH}}^{\text{-}}\stackrel{{\text{K}}_{\text{1}}}{\rightleftharpoons }{\text{MnOH}}^{\text{+}}{\text{K}}_{\text{1}}\text{=}{\text{10}}^{\text{3}\text{.4}}\\ {\text{H}}_{\text{2}}\text{O}\stackrel{}{\rightleftharpoons }{\text{H}}^{\text{+}}\text{+}{\text{OH}}^{\text{-}}{\text{K}}_{\text{W}}\text{=}{\text{10}}^{\text{-14}\text{.00}}\end{array}$

Write the charge balance equation

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

Write the mass balance equation

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

Mass balance is equivalent to charge balance.

The equilibrium constant expression is given by

$\begin{array}{l}{\text{K}}_{\text{SP}}\text{=}\left[{\text{Mn}}^{\text{2+}}\right]{\text{γ}}_{{\text{Mn}}^{\text{2+}}}{\left[{\text{OH}}^{\text{-}}\right]}^{\text{2}}{\text{γ}}^{\text{2}}{}_{{\text{OH}}^{\text{-}}}\\ \\ {\text{K}}_{\text{1}}\text{=}\frac{\left[{\text{MnOH}}^{\text{+}}\right]{\text{γ}}_{{\text{MnOH}}^{\text{+}}}}{\left[{\text{Mn}}^{\text{2+}}\right]{\text{γ}}_{{\text{Mn}}^{\text{2+}}}\left[{\text{OH}}^{\text{-}}\right]{\text{γ}}_{{\text{OH}}^{\text{-}}}}\\ \\ {\text{K}}_{\text{W}}\text{=}\left[{\text{H}}^{\text{+}}\right]{\text{γ}}_{{\text{H}}^{\text{+}}}\left[{\text{OH}}^{\text{-}}\right]{\text{γ}}_{{\text{OH}}^{\text{-}}}\end{array}$

The spreadsheet is the same as part (a), except that cell $\text{B5}$ contains the formula

$\text{"}\text{= 0}{\text{.5*(E8}}^{\text{Λ}}{\text{2*C8+E9}}^{\text{Λ}}{\text{2*C9+E10}}^{\text{Λ}}{\text{2*C10+E11}}^{\text{Λ}}\text{2*C11)"}\text{.}$

Figure 1

The concentration of the all species was found as

$\begin{array}{l}\left[{\text{Mn}}^{\text{2+}}\right]\text{=}\text{3}\text{.30}\text{×}{\text{10}}^{\text{-5}}\text{M}\text{;}\left[{\text{OH}}^{\text{-}}\right]\text{=}\text{7}\text{.18}\text{×}{\text{10}}^{\text{-5}}\text{M}\text{.}\\ \left[{\text{MnOH}}^{\text{+}}\right]\text{=}\text{5}\text{.68}\text{×}{\text{10}}^{\text{-6}}\text{M}\text{;}\left[{\text{H}}^{\text{+}}\right]\text{=}\text{1}\text{.43}\text{×}{\text{10}}^{\text{-10}}\text{M}\text{.}\\ \text{The}\text{value}\text{of}\text{µ}\text{is}\text{1}\text{.05}\text{×}{\text{10}}^{\text{-4}}\text{M}\text{.}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The solubility of $\text{Mn}{\left(\text{OH}\right)}_{\text{2}}$ is greater when ${\text{NaClO}}_{\text{4}}$ is present. Reason for this observation has to be given.

Explanation of Solution

Given information,

The solubility of $\text{Mn}{\left(\text{OH}\right)}_{\text{2}}$ is greater when ${\text{NaClO}}_{\text{4}}$ is present.

Reason for the observation:

The solubility of $\text{Mn}{\left(\text{OH}\right)}_{\text{2}}$ depends on the attractive force between ${\text{Mn}}^{\text{2+}}$ and ${\text{OH}}^{\text{-}}$ ions.

${\text{Na}}^{\text{+}}$ and ${\text{ClO}}_{\text{4}}{}^{\text{-}}$ ions create ionic atmosphere, this ionic atmospheres reduces the attraction between ${\text{Mn}}^{\text{2+}}$ and ${\text{OH}}^{\text{-}}$ ions.  Hence, the solubility of $\text{Mn}{\left(\text{OH}\right)}_{\text{2}}$ is greater when ${\text{NaClO}}_{\text{4}}$ is present.

The quotient is calculated by

$\begin{array}{l}\left(\left[{\text{Mn}}^{\text{2+}}\right]\text{in}\text{0}\text{.1}\text{M}{\text{NaClO}}_{\text{4}}\right)\text{/}\left(\left[{\text{Mn}}^{\text{2+}}\right]\text{without}{\text{NaClO}}_{\text{4}}\right)\\ \\ \frac{\text{5}\text{.35}\text{×}{\text{10}}^{\text{-5}}\text{M}}{\text{3}\text{.30}\text{×}{\text{10}}^{\text{-5}}\text{M}}\text{=}\text{1}\text{.6212}\text{=}\text{1}\text{.6}\text{.}\end{array}$

The value of quotient is $\text{1}\text{.6}$.