Chapter 19, Problem 19.150P

(a)

Interpretation Introduction

Interpretation:

Value of $\left[{\text{Co}}^{\text{2+}}\right]$ and $\left[{\text{EDTA}}^{\text{4-}}\right]$ have to be calculated when $\text{25}\text{.0}\text{ml}$ $\text{0}\text{.050}\text{M}$ of $\left[{\text{EDTA}}^{\text{4-}}\right]$ is added to $\text{50}\text{.0}\text{ml}$ of $\text{0}\text{.048}\text{M}$ $\left[{\text{Co}}^{\text{2+}}\right]$ during titration.

Concept Introduction:

Chemical Equilibrium:

Chemical equilibrium is the process where the rate of forward reaction and the rate of backward reaction are equal.

Titration:

Titration is a quantitative chemical analysis to determine the concentration of an identified analyte.  The titrant is the reagent which is prepared as a standard solution of known concentration volume.  The titrant reacts with the analyte to determine the analyte’s concentration.  The volume of the titrant reacting with analyte is called the titration volume.

Formation constant:

A stability constant or formation constant is an equilibrium constant for the formation of a complex ion in the solution and it measures the strength of interaction between the reactants that forms the complex.

Answer to Problem 19.150P

The concentrations calculated are $\left[{\text{Co}}^{\text{2+}}\right]\text{=}0.0153\text{M}$ and $\left[{\text{EDTA}}^{\text{4-}}\right]\text{=}\text{5}{\text{.346×10}}^{\text{-17}}\text{M}$.

Explanation of Solution

$\text{EDTA}$ binds with metal ions to form complex ions and therefore it is used to determine the concentration of the metal ions.

In the problem it is given that the $\text{EDTA}$ forms complex with Cobalt.

  ${\text{Co}}^{\text{2+}}\text{+}{\text{EDTA}}^{\text{4-}}\stackrel{}{\to }\left[\text{Co}{\left(\text{EDTA}\right)}^{\text{2-}}\right]$

Given that solution contains $\text{50}\text{.0}\text{ml}$ of $\text{0}\text{.048}\text{M}$${\text{Co}}^{\text{2+}}$.

Also $\text{25}\text{.0}\text{ml}$ of $\text{0}\text{.050}\text{M}$ ${\text{EDTA}}^{\text{4-}}$ is added.

As ${\text{EDTA}}^{\text{4-}}$ is added in limiting amount so excess ${\text{Co}}^{\text{2+}}$ will remain in the solution.

So the number of moles for ${\text{Co}}^{\text{2+}}$ and ${\text{EDTA}}^{\text{4-}}$ are calculated as,

  $\begin{array}{l}\text{No}\text{.}\text{of}\text{moles}\text{of}{\text{Co}}^{\text{2+}}\text{=}\text{volume}\text{×}\text{concentration}\\ \\ \text{=}\frac{\left[\text{50}\text{.0}\text{ml}\text{×}\text{0}\text{.048M}\right]}{\text{1000}\text{ml}}\\ \\ \text{=}\text{0}\text{.0024}\text{moles}\end{array}$

  $\begin{array}{l}\text{No}\text{.}\text{of}\text{moles}\text{of}{\text{EDTA}}^{\text{4-}}\text{=}\text{volume}\text{×}\text{concentration}\\ \\ \text{=}\frac{\left[\text{25}\text{.0}\text{ml}\text{×}\text{0}\text{.050M}\right]}{\text{1000}\text{ml}}\\ \\ \text{=}\text{0}\text{.00125}\text{moles}\end{array}$

Number of moles of $\text{Co}{\left(\text{EDTA}\right)}^{\text{2-}}$ formed is equal to number of moles of ${\text{EDTA}}^{\text{4-}}$ as it is the limiting reactant.

  $\text{No}\text{.}\text{of}\text{moles}\text{of}\text{Co}{\left(\text{EDTA}\right)}^{\text{2-}}\text{=}\text{0}\text{.00125}\text{moles}$

  $\begin{array}{l}\text{Excess no}\text{.}\text{of}\text{moles}\text{of}{\text{Co}}^{\text{2+}}\text{=}(0.0024-0.00125)\text{moles}\\ \\ \text{=}\text{0}\text{.00115}\text{moles}\end{array}$

  $\begin{array}{l}{\text{The total volume of solution = (25 ml of EDTA}}^{\text{4-}}\text{+}\text{50}\text{.0}\text{ml}\text{of}{\text{Co}}^{\text{2+}}\text{)}\\ \\ \text{=}7\text{5}\text{ml}\text{of}\text{solution}\end{array}$

Hence, concentration of ${\text{Co}}^{\text{2+}}$ calculated as,

  $\begin{array}{l}\left[{\text{Co}}^{\text{2+}}\right]\text{=}\frac{\text{0}\text{.00115mol}}{\left({\text{75×10}}^{\text{-3}}\right)\text{L}}\\ \\ \text{=}\text{0}\text{.0153}\text{M}\end{array}$

Hence, concentration of $\text{Co}{\left(\text{EDTA}\right)}^{\text{2-}}$ calculated as,

  $\begin{array}{l}\left[\text{Co}{\left(\text{EDTA}\right)}^{\text{2-}}\right]\text{=}\frac{\text{0}\text{.00125mol}}{\left({\text{75×10}}^{\text{-3}}\right)\text{L}}\\ \\ =0.0167M\end{array}$

Now, the formation constant for $\text{Co}{\left(\text{EDTA}\right)}^{\text{2-}}$ is calculated as,

  ${\text{K}}_{\text{f}}\text{=}\frac{\left[\text{Co}{\left(\text{EDTA}\right)}^{\text{2-}}\right]}{\left[{\text{Co}}^{\text{2+}}\right]\text{×}\left[{\text{EDTA}}^{\text{4-}}\right]}$

Given as that $\log {\text{K}}_{\text{f}}$ of $\left[\text{Co}{\left(\text{EDTA}\right)}^{\text{2-}}\right]$ is $16.31$.

${\text{K}}_{\text{f}}$ For $\left[\text{Co}{\left(\text{EDTA}\right)}^{\text{2-}}\right]\text{=}2.0417\times {10}^{16}$

To attain the equilibrium, there should be some amount of limiting reactant ${\text{EDTA}}^{\text{4-}}$ present in the solution and that is x (say).

If some amount ${\text{EDTA}}^{\text{4-}}$ is present then $\left[{\text{Co}}^{\text{2+}}\right]$ should be more and product complex $\left[\text{Co}{\left(\text{EDTA}\right)}^{\text{2-}}\right]$ should be less as some amount of ${\text{EDTA}}^{\text{4-}}$ is not taking part in equilibrium.

$\left[{\text{Co}}^{\text{2+}}\right]\text{=}\text{0}\text{.0153}\text{+}\text{x}\text{M}$

$\left[\text{Co}{\left(\text{EDTA}\right)}^{\text{2-}}\right]\text{=}\text{0}\text{.0167}\text{-}\text{x}\text{M}$

Thus from equation of ${\text{K}}_{\text{f}}$,

  $\left[{\text{EDTA}}^{\text{4-}}\right]\text{=}\frac{\left[\text{Co}{\left(\text{EDTA}\right)}^{\text{2-}}\right]}{\left[{\text{Co}}^{\text{2+}}\right]{\text{×K}}_{\text{f}}}$

  $\begin{array}{l}\therefore \text{x}\text{=}\frac{\text{(0}\text{.0167-x)}\text{M}}{\left[\text{(0}\text{.0153+x)}\text{M}\text{×}\text{2}{\text{.0417×10}}^{\text{16}}\right]}\\ \\ \text{=}\frac{\text{0}\text{.0167}\text{M}}{\left[\text{0}\text{.0153M}\text{×}\text{2}{\text{.0417×10}}^{\text{16}}\right]}\\ \\ \therefore \text{x}\text{=}\text{5}{\text{.346×10}}^{\text{-17}}\text{M}\end{array}$

Hence $\left[{\text{Co}}^{\text{2+}}\right]\text{=}0.0153\text{M}$ ($\text{x<<1}$, therefore x is neglected) and $\left[{\text{EDTA}}^{\text{4-}}\right]\text{=}\text{5}{\text{.346×10}}^{\text{-17}}\text{M}$.

(b)

Interpretation Introduction

Interpretation:

Value of $\left[{\text{Co}}^{\text{2+}}\right]$ and $\left[{\text{EDTA}}^{\text{4-}}\right]$ have to be calculated when $\text{75}\text{.0}\text{ml}\text{0}\text{.050}\text{M}$ of $\left[{\text{EDTA}}^{\text{4-}}\right]$ is added to $\text{50}\text{.0}\text{ml}$ of  $\text{0}\text{.048}\text{M}$ $\left[{\text{Co}}^{\text{2+}}\right]$ during titration.

Concept Introduction:

Chemical Equilibrium:

Chemical equilibrium is the process where the rate of forward reaction and the rate of backward reaction are equal.

Titration:

Titration is a quantitative chemical analysis to determine the concentration of an identified analyte. The titrant is the reagent which is prepared as a standard solution of known concentration volume. The titrant reacts with the analyte to determine the analyte’s concentration. The volume of the titrant reacting with analyte is called the titration volume.

Formation constant:

A stability constant or formation constant is an equilibrium constant for the formation of a complex ion in the solution and it measures the strength of interaction between the reactants that forms the complex.

Answer to Problem 19.150P

The value of $\left[{\text{Co}}^{\text{2+}}\right]$ and $\left[{\text{EDTA}}^{\text{4-}}\right]$ are respectively $8.707\times {10}^{-17}\text{M}$ and $\text{0}\text{.0108}\text{M}$.

Explanation of Solution

$\text{EDTA}$ binds with metal ions to form complex ions and therefore it is used to determine the concentration of the metal ions.

In the problem it is given that the $\text{EDTA}$ forms complex with Cobalt.

  ${\text{Co}}^{\text{2+}}\text{+}{\text{EDTA}}^{\text{4-}}\stackrel{}{\to }\left[\text{Co}{\left(\text{EDTA}\right)}^{\text{2-}}\right]$

Given that solution contains  $\text{50}\text{.0}\text{ml}$ of $\text{0}\text{.048}\text{M}$ ${\text{Co}}^{\text{2+}}$.

$\text{75}\text{.0}\text{ml}\text{0}\text{.050}\text{M}$ of ${\text{EDTA}}^{\text{4-}}$ is added.

As ${\text{EDTA}}^{\text{4-}}$ is added in excess amount so excess ${\text{EDTA}}^{\text{4-}}$ will remain in the solution and ${\text{Co}}^{\text{2+}}$ will finish in the solution.

Hence the number of moles of ${\text{Co}}^{\text{2+}}$ and ${\text{EDTA}}^{\text{4-}}$ are calculated as,

  $\begin{array}{l}\text{No}\text{. of moles of}{\text{Co}}^{\text{2+}}\text{= }\text{volume}\text{×}\text{concentration}\\ \\ \text{= }\frac{\left[\text{50}\text{.0}\text{ml}\text{×}\text{0}\text{.048M}\right]}{\text{1000}\text{ml}}\\ \\ \text{=}\text{0}\text{.0024 moles}\end{array}$

  $\begin{array}{l}\text{No}{\text{. of moles of EDTA}}^{\text{4-}}\text{=}\text{volume}\text{×}\text{concentration }\\ \\ \text{= }\frac{\left[\text{75}\text{.0}\text{ml}\text{×}\text{0}\text{.050M}\right]}{\text{1000}\text{ml}}\\ \\ \text{=}\text{0}\text{.00375 moles}\end{array}$

Number of moles of $\text{Co}{\left(\text{EDTA}\right)}^{\text{2-}}$ formed is equal to number of moles of ${\text{Co}}^{\text{2+}}$ as it is the Limiting reactant.

  $\text{No}\text{. of moles of Co}{\left(\text{EDTA}\right)}^{\text{2-}}\text{ = 0}\text{.0024 moles}\text{. }$

  $\begin{array}{l}\text{Excess no}{\text{. of moles of EDTA}}^{\text{4-}}\text{ =}\text{(0}\text{.00375-0}\text{.0024)moles }\\ \\ \text{=}\text{0}\text{.00135 moles}\end{array}$

  $\begin{array}{l}{\text{The total volume of solution = (75 ml of EDTA}}^{\text{4-}}\text{+}\text{50}\text{.0}\text{ml}\text{of}{\text{Co}}^{\text{2+}}\text{)}\\ \\ \text{=}\text{125}\text{ml}\text{of}\text{solution}\end{array}$

Hence, concentration of ${\text{EDTA}}^{\text{4-}}$,

  $\begin{array}{l}\left[{\text{EDTA}}^{\text{4-}}\right]\text{=}\frac{\text{0}\text{.00135mol}}{\left({\text{125×10}}^{\text{-3}}\right)\text{L}}\\ \\ \text{=}\text{0}\text{.0108}\text{M}\end{array}$

Hence, concentration of $\text{Co}{\left(\text{EDTA}\right)}^{\text{2-}}$,

  $\begin{array}{l}\left[\text{Co}{\left(\text{EDTA}\right)}^{\text{2-}}\right]\text{=}\frac{\text{0}\text{.0024mol}}{\left({\text{125×10}}^{\text{-3}}\right)\text{L}}\\ \\ \text{=0}\text{.0192}\text{M}\end{array}$

Now, formation constant of $\text{Co}{\left(\text{EDTA}\right)}^{\text{2-}}$ is calculated as,

  ${\text{K}}_{\text{f}}\text{=}\frac{\left[\text{Co}{\left(\text{EDTA}\right)}^{\text{2-}}\right]}{\left[{\text{Co}}^{\text{2+}}\right]\text{×}\left[{\text{EDTA}}^{\text{4-}}\right]}$

Given that $\log {\text{K}}_{\text{f}}$ for $\left[\text{Co}{\left(\text{EDTA}\right)}^{\text{2-}}\right]\text{=}16.31$.

${\text{K}}_{\text{f}}$ for $\left[\text{Co}{\left(\text{EDTA}\right)}^{\text{2-}}\right]\text{=}2.0417\times {10}^{16}$

To attain the equilibrium some amount of the limiting reactant ${\text{Co}}^{\text{2+}}$ must be present as unreacted in the solution and that is x (say).

Then the $\left[{\text{EDTA}}^{\text{4-}}\right]$ must be more and the forming complex $\left[\text{Co}{\left(\text{EDTA}\right)}^{\text{2-}}\right]$ must be ales as the whole ${\text{Co}}^{\text{2+}}$ is not reacting.

Hence the concentration of $\left[{\text{EDTA}}^{\text{4-}}\right]$ and $\left[\text{Co}{\left(\text{EDTA}\right)}^{\text{2-}}\right]$ can be considered as,

  $\left[{\text{EDTA}}^{\text{4-}}\right]\text{=}\text{0}\text{.0108}\text{+}\text{x}$

  $\left[\text{Co}{\left(\text{EDTA}\right)}^{\text{2-}}\right]\text{=}\text{0}\text{.0192}\text{-}\text{x}$

Hence the concentration of ${\text{Co}}^{\text{2+}}$ can be calculated as,

  $\left[{\text{Co}}^{\text{2+}}\right]\text{=}\frac{\left[\text{Co}{\left(\text{EDTA}\right)}^{\text{2-}}\right]}{\left[{\text{EDTA}}^{\text{4-}}\right]{\text{×K}}_{\text{f}}}$

  $\therefore \text{x=}\frac{\text{(0}\text{.0192}\text{+}\text{x)}\text{M}}{\left[\text{(0}\text{.0108}\text{-}\text{x)M}\text{×}\text{2}{\text{.0417×10}}^{\text{16}}\right]}$

$\begin{array}{l}\therefore \text{x=}\frac{\text{0}\text{.0192}\text{M}}{\left[\text{0}\text{.0108M}\text{×}\text{2}{\text{.0417×10}}^{\text{16}}\right]}\\ \\ =8.707\times {10}^{-17}\text{M}\end{array}$ $\left(x<<1\right)$

Thus the values are respectively,

  $\left[{\text{Co}}^{\text{2+}}\right]\text{=}8.707\times {10}^{-17}\text{M}$

  $\left[{\text{EDTA}}^{\text{4-}}\right]\text{=}\text{0}\text{.0108}\text{M}$ (‘x’ is very small)

Hence, the value of $\left[{\text{Co}}^{\text{2+}}\right]$ and $\left[{\text{EDTA}}^{\text{4-}}\right]$ are respectively $8.707\times {10}^{-17}\text{M}$ and $\text{0}\text{.0108}\text{M}$.