Chapter 13, Problem 13.21P

Interpretation Introduction

Interpretation:

Value of ${\text{pMn}}^{2+}$ for $0\text{ mL}$, $20.0\text{ mL}$, $40.0\text{ mL}$, $49.0\text{ mL}$, $49.9\text{ mL}$, $50.0\text{ mL}$, $50.1\text{ mL}$, $55.0\text{ mL}$ and $60.0\text{ mL}$ of EDTA has to be determined. Also, titration curve for the same has to be drawn.

Concept Introduction:

EDTA or ethylenediaminetetraacetic acid is used for titration of majority of metals with help of formation of strong metal complexes in the ratio of $1:1$. It is utilized to bind metals, most commonly used in industrial processes. It is also used for prevention of food oxidation and in environmental chemistry. It can be used either directly or indirectly for analysis of most of the elements present in periodic table. EDTA titrations can be performed in many ways. Some of these are mentioned below.

1. Direct titration

2. Back titration

3. Displacement titration

4. Masking

Explanation of Solution

Expression for $K_{\text{f}}^{\text{'}}$ is as follows:

  $K_{\text{f}}^{\text{'}}=\left({\alpha }_{{\text{Y}}^{4-}}\right)\left(K_{\text{f}}\right)$        (1)

Here,

$K_{\text{f}}^{\text{'}}$ is conditional rate constant.

$K_{\text{f}}$ is rate constant.

${\alpha }_{{\text{Y}}^{4-}}$ is fraction of free EDTA in form of $Y^{4-}$.

Substitute $7.41\times {10}^{13}$ for $K_{\text{f}}$ and $4.2\times {10}^{-3}$ for ${\alpha }_{{\text{Y}}^{4-}}$ in equation (1).

  $\begin{array}{c}{K}_{\text{f}}^{\text{'}}=\left(4.2\times {10}^{-3}\right)\left(7.41\times {10}^{13}\right)\\ =\text{31}\text{.1}\times {\text{10}}^{10}\end{array}$

At $0\text{ mL}$ of EDTA solution,

Expression to calculate ${\text{pMn}}^{2+}$ is as follows:

  ${\text{pMn}}^{2+}=-\log \left[{\text{Mn}}^{2+}\right]$        (2)

Substitute $0.0200\text{ M}$ for $\left[{\text{Mn}}^{2+}\right]$ in equation (2).

  $\begin{array}{c}{\text{pMn}}^{2+}=-\log \left(0.0200\text{ M}\right)\\ =1.698\\ \approx 1.70\end{array}$

When $0\text{ mL}$ of EDTA is added to ${\text{Mn}}^{2+}$ solution, ${\text{pMn}}^{2+}$ is 1.70.

At $20.0\text{ mL}$ of EDTA solution,

Expression to calculate total volume of solution is as follows:

  $\text{Total volume of solution}=\left(\begin{array}{c}{\text{Volume of Mn}}^{2+}+\\ \text{Volume of EDTA}\end{array}\right)$        (3)

Substitute $25.0\text{ mL}$ for volume of ${\text{Mn}}^{2+}$ and $20.0\text{ mL}$ for volume of EDTA in equation (3).

  $\begin{array}{c}\text{Total volume of solution}=\left(25.0+20.0\right)\text{ mL}\\ =45\text{ mL}\end{array}$

Since EDTA forms strong metal complexes in the ratio of $1:1$, moles of EDTA will be equivalent to moles of metal reacted.

Expression to calculate excess concentration of ${\text{Mn}}^{2+}$ is as follows:

  $\text{Excess }\left[{\text{Mn}}^{2+}\right]=\left(\frac{\begin{array}{l}\left(\left[{\text{Mn}}^{2+}\right]\right)\left({\text{Volume of Mn}}^{2+}\right)-\\ \left(\left[\text{EDTA}\right]\right)\left(\text{Volume of EDTA}\right)\end{array}}{\text{Total volume of solution}}\right)$        (4)

Substitute $25.0\text{ mL}$ for volume of ${\text{Mn}}^{2+}$, $0.0200\text{ M}$ for $\left[{\text{Mn}}^{2+}\right]$, $20.0\text{ mL}$ for volume of EDTA, $0.0100\text{ M}$ for $\left[\text{EDTA}\right]$ and $45\text{ mL}$ for total volume of solution in equation (4).

  $\begin{array}{c}\text{Excess }\left[{\text{Mn}}^{2+}\right]=\left(\frac{\left(\text{0}\text{.0200 M}\right)\left(25.0\text{ mL}\right)-\left(0.0100\text{ M}\right)\left(\text{20}\text{.0 mL}\right)}{\text{45 mL}}\right)\\ =0.00667\text{ M}\end{array}$

Substitute $0.00667\text{ M}$ for $\left[{\text{Mn}}^{2+}\right]$ in equation (2).

  $\begin{array}{c}{\text{pMn}}^{2+}=-\log \left(0.00667\text{ M}\right)\\ =2.18\end{array}$

At $40.0\text{ mL}$ of EDTA solution,

Substitute $25.0\text{ mL}$ for volume of ${\text{Mn}}^{2+}$ and $40.0\text{ mL}$ for volume of EDTA in equation (3).

  $\begin{array}{c}\text{Total volume of solution}=\left(25.0+40.0\right)\text{ mL}\\ =65\text{ mL}\end{array}$

Substitute $25.0\text{ mL}$ for volume of ${\text{Mn}}^{2+}$, $0.0200\text{ M}$ for $\left[{\text{Mn}}^{2+}\right]$, $40.0\text{ mL}$ for volume of EDTA, $0.0100\text{ M}$ for $\left[\text{EDTA}\right]$ and $65\text{ mL}$ for total volume of solution in equation (4).

  $\begin{array}{c}\text{Excess }\left[{\text{Mn}}^{2+}\right]=\left(\frac{\left(\text{0}\text{.0200 M}\right)\left(25.0\text{ mL}\right)-\left(0.0100\text{ M}\right)\left(\text{40}\text{.0 mL}\right)}{65\text{ mL}}\right)\\ =0.00154\text{ M}\end{array}$

Substitute $0.00154\text{ M}$ for $\left[{\text{Mn}}^{2+}\right]$ in equation (2).

  $\begin{array}{c}{\text{pMn}}^{2+}=-\log \left(0.00154\text{ M}\right)\\ =2.81\end{array}$

At $49.0\text{ mL}$ of EDTA solution,

Substitute $25.0\text{ mL}$ for volume of ${\text{Mn}}^{2+}$ and $49.0\text{ mL}$ for volume of EDTA in equation (3).

  $\begin{array}{c}\text{Total volume of solution}=\left(25.0+49.0\right)\text{ mL}\\ =74\text{ mL}\end{array}$

Substitute $25.0\text{ mL}$ for volume of ${\text{Mn}}^{2+}$, $0.0200\text{ M}$ for $\left[{\text{Mn}}^{2+}\right]$, $49.0\text{ mL}$ for volume of EDTA, $0.0100\text{ M}$ for $\left[\text{EDTA}\right]$ and $74\text{ mL}$ for total volume of solution in equation (4).

  $\begin{array}{c}\text{Excess }\left[{\text{Mn}}^{2+}\right]=\left(\frac{\left(\text{0}\text{.0200 M}\right)\left(25.0\text{ mL}\right)-\left(0.0100\text{ M}\right)\left(\text{49}\text{.0 mL}\right)}{74\text{ mL}}\right)\\ =0.000135\text{ M}\end{array}$

Substitute $0.000135\text{ M}$ for $\left[{\text{Mn}}^{2+}\right]$ in equation (2).

  $\begin{array}{c}{\text{pMn}}^{2+}=-\log \left(0.000135\text{ M}\right)\\ =3.87\end{array}$

At $49.9\text{ mL}$ of EDTA solution,

Substitute $25.0\text{ mL}$ for volume of ${\text{Mn}}^{2+}$ and $49.9\text{ mL}$ for volume of EDTA in equation (3).

  $\begin{array}{c}\text{Total volume of solution}=\left(25.0+49.9\right)\text{ mL}\\ =74.9\text{ mL}\end{array}$

Substitute $25.0\text{ mL}$ for volume of ${\text{Mn}}^{2+}$, $0.0200\text{ M}$ for $\left[{\text{Mn}}^{2+}\right]$, $49.9\text{ mL}$ for volume of EDTA, $0.0100\text{ M}$ for $\left[\text{EDTA}\right]$ and $74.9\text{ mL}$ for total volume of solution in equation (4).

  $\begin{array}{c}\text{Excess }\left[{\text{Mn}}^{2+}\right]=\left(\frac{\left(\text{0}\text{.0200 M}\right)\left(25.0\text{ mL}\right)-\left(0.0100\text{ M}\right)\left(\text{49}\text{.9 mL}\right)}{74.9\text{ mL}}\right)\\ =0.0000013\text{ M}\end{array}$

Substitute $0.0000013\text{ M}$ for $\left[{\text{Mn}}^{2+}\right]$ in equation (2).

  $\begin{array}{c}{\text{pMn}}^{2+}=-\log \left(0.0000013\text{ M}\right)\\ =5.89\end{array}$

At $50.0\text{ mL}$ of EDTA solution,

Formula to calculate molarity of solution is as follows:

  $\text{Molarity of solution}\left(\text{M}\right)=\frac{\text{Moles of solute}}{\text{Volume }\left(\text{L}\right)\text{ of solution}}$        (5)

Rearrange equation (5) for moles of solute.

  $\text{Moles of solute}=\left[\begin{array}{c}\left(\text{Molarity of solution}\right)\\ \left(\text{Volume of solution}\right)\end{array}\right]$        (6)

Substitute $0.0200\text{ M}$ for molarity and $25.0\text{ mL}$ for volume of solution in equation (6) to calculate millimoles of ${\text{Mn}}^{2+}$.

  $\begin{array}{c}{\text{Millimoles of Mn}}^{2+}=\left(\text{0}\text{.0200 M}\right)\left(\text{25}\text{.0 mL}\right)\\ =0.5\text{ mmol}\end{array}$

Substitute $0.0100\text{ M}$ for molarity and $50.0\text{ mL}$ for volume of solution in equation (6) to calculate millimoles of EDTA.

  $\begin{array}{c}\text{Millimoles of EDTA}=\left(0.0100\text{ M}\right)\left(\text{50}\text{.0 mL}\right)\\ =0.5\text{ mmol}\end{array}$

Substitute $25.0\text{ mL}$ for volume of ${\text{Mn}}^{2+}$ and $50.0\text{ mL}$ for volume of EDTA in equation (3).

  $\begin{array}{c}\text{Total volume of solution}=\left(25.0+50.0\right)\text{ mL}\\ =75\text{ mL}\end{array}$

Chemical reaction occurs as follows:

  ${\text{Mn}}^{2+}+{\text{Y}}^{2-}\leftrightarrow {\text{MnY}}^{2-}$

Concentration of ${\text{MnY}}^{2-}$ at equivalence point is calculated as follows:

  $\begin{array}{c}\left[{\text{MnY}}^{2-}\right]=\frac{0.5\text{ mmol}}{75\text{ mL}}\\ =0.0067\text{ M}\end{array}$

Consider change in concentrations of ionic species to be negligible and amount of ${\text{Mn}}^{2+}$ and EDTA to be $x$.

Expression to calculate $K_{\text{f}}^{\text{'}}$ at equivalence point is as follows:

  $K_{\text{f}}^{\text{'}}=\frac{\left[{\text{MnY}}^{2-}\right]}{\left[{\text{Mn}}^{2+}\right]\left[\text{EDTA}\right]}$        (7)

Substitute $0.0067\text{ M}$ for $\left[{\text{MnY}}^{2-}\right]$, $x$ for $\left[{\text{Mn}}^{2+}\right]$, $x$ for $\left[\text{EDTA}\right]$ and $\text{31}\text{.1}\times {\text{10}}^{10}$ for $K_{\text{f}}^{\text{'}}$ in equation (7).

  $\text{31}\text{.1}\times {\text{10}}^{10}=\frac{0.0067\text{ M}}{\left(x\right)\left(x\right)}$

Solve for $x$,

  $x=1.27\times {10}^{-7}\text{ M}$

Substitute $1.27\times {10}^{-7}\text{ M}$ for $\left[{\text{Mn}}^{2+}\right]$ in equation (2).

  $\begin{array}{c}{\text{pMn}}^{2+}=-\log \left(1.27\times {10}^{-7}\text{ M}\right)\\ =6.89\end{array}$

At $50.0\text{ mL}$ of EDTA solution,

Substitute $25.0\text{ mL}$ for volume of ${\text{Mn}}^{2+}$ and $50.1\text{ mL}$ for volume of EDTA in equation (3).

  $\begin{array}{c}\text{Total volume of solution}=\left(25.0+50.1\right)\text{ mL}\\ =75.1\text{ mL}\end{array}$

Chemical reaction occurs as follows:

  ${\text{Mn}}^{2+}+{\text{Y}}^{2-}\leftrightarrow {\text{MnY}}^{2-}$

Concentration of ${\text{MnY}}^{2-}$ at equivalence point is calculated as follows:

  $\begin{array}{c}\left[{\text{MnY}}^{2-}\right]=\frac{0.5\text{ mmol}}{75.1\text{ mL}}\\ =0.00066\text{ M}\end{array}$

Concentration of excess EDTA is calculated as follows:

$\begin{array}{c}\text{Excess }\left[\text{EDTA}\right]=\frac{\left(0.0100\text{ M}\right)\left(0.1\text{ mL}\right)}{75.1\text{ mL}}\\ =1.33\times {10}^{-5}\text{ M}\end{array}$

Rearrange equation (7) for $\left[{\text{Mn}}^{2+}\right]$.

  $\left[{\text{Mn}}^{2+}\right]=\frac{\left[{\text{MnY}}^{2-}\right]}{K_{\text{f}}^{\text{'}}\left[\text{EDTA}\right]}$        (8)

Substitute $0.00066\text{ M}$ for $\left[{\text{MnY}}^{2-}\right]$, $1.33\times {10}^{-5}\text{ M}$ for $\left[\text{EDTA}\right]$ and $\text{31}\text{.1}\times {\text{10}}^{10}$ for $K_{\text{f}}^{\text{'}}$ in equation (8).

  $\begin{array}{c}\left[{\text{Mn}}^{2+}\right]=\frac{0.00066\text{ M}}{\left(\text{31}\text{.1}\times {\text{10}}^{10}\right)\left(1.33\times {10}^{-5}\text{ M}\right)}\\ =1.60\times {10}^{-10}\text{ M}\end{array}$

Substitute $1.60\times {10}^{-10}\text{ M}$ for $\left[{\text{Mn}}^{2+}\right]$ in equation (2).

  $\begin{array}{c}{\text{pMn}}^{2+}=-\log \left(1.60\times {10}^{-10}\text{ M}\right)\\ =9.80\end{array}$

At $55.0\text{ mL}$ of EDTA solution,

Substitute $25.0\text{ mL}$ for volume of ${\text{Mn}}^{2+}$ and $55.0\text{ mL}$ for volume of EDTA in equation (3).

  $\begin{array}{c}\text{Total volume of solution}=\left(25.0+55.1\right)\text{ mL}\\ =80\text{ mL}\end{array}$

Chemical reaction occurs as follows:

  ${\text{Mn}}^{2+}+{\text{Y}}^{2-}\leftrightarrow {\text{MnY}}^{2-}$

Concentration of ${\text{MnY}}^{2-}$ at equivalence point is calculated as follows:

  $\begin{array}{c}\left[{\text{MnY}}^{2-}\right]=\frac{0.5\text{ mmol}}{80\text{ mL}}\\ =6.25\times {10}^{-3}\text{ M}\end{array}$

Concentration of excess EDTA is calculated as follows:

  $\begin{array}{c}\text{Excess }\left[\text{EDTA}\right]=\frac{\left(0.0100\text{ M}\right)\left(0.1\text{ mL}\right)}{80\text{ mL}}\\ =6.25\times {10}^{-4}\text{ M}\end{array}$

Substitute $6.25\times {10}^{-3}\text{ M}$ for $\left[{\text{MnY}}^{2-}\right]$, $6.25\times {10}^{-4}\text{ M}$ for $\left[\text{EDTA}\right]$ and $\text{31}\text{.1}\times {\text{10}}^{10}$ for $K_{\text{f}}^{\text{'}}$ in equation (8).

  $\begin{array}{c}\left[{\text{Mn}}^{2+}\right]=\frac{6.25\times {10}^{-3}\text{ M}}{\left(\text{31}\text{.1}\times {\text{10}}^{10}\right)\left(6.25\times {10}^{-4}\text{ M}\right)}\\ =3.215\times {10}^{-11}\text{ M}\end{array}$

Substitute $3.215\times {10}^{-11}\text{ M}$ for $\left[{\text{Mn}}^{2+}\right]$ in equation (2).

  $\begin{array}{c}{\text{pMn}}^{2+}=-\log \left(3.215\times {10}^{-11}\text{ M}\right)\\ =10.49\end{array}$

At $60.0\text{ mL}$ of EDTA solution,

Substitute $25.0\text{ mL}$ for volume of ${\text{Mn}}^{2+}$ and $60.0\text{ mL}$ for volume of EDTA in equation (3).

  $\begin{array}{c}\text{Total volume of solution}=\left(25.0+60.1\right)\text{ mL}\\ =85\text{ mL}\end{array}$

Chemical reaction occurs as follows:

  ${\text{Mn}}^{2+}+{\text{Y}}^{2-}\leftrightarrow {\text{MnY}}^{2-}$

Concentration of ${\text{MnY}}^{2-}$ at equivalence point is calculated as follows:

  $\begin{array}{c}\left[{\text{MnY}}^{2-}\right]=\frac{0.5\text{ mmol}}{85\text{ mL}}\\ =5.88\times {10}^{-3}\text{ M}\end{array}$

Concentration of excess EDTA is calculated as follows:

  $\begin{array}{c}\text{Excess }\left[\text{EDTA}\right]=\frac{\left(0.0100\text{ M}\right)\left(0.1\text{ mL}\right)}{85\text{ mL}}\\ =1.18\times {10}^{-3}\text{ M}\end{array}$

Substitute $5.88\times {10}^{-3}\text{ M}$ for $\left[{\text{MnY}}^{2-}\right]$, $1.18\times {10}^{-3}\text{ M}$ for $\left[\text{EDTA}\right]$ and $\text{31}\text{.1}\times {\text{10}}^{10}$ for $K_{\text{f}}^{\text{'}}$ in equation (8).

  $\begin{array}{c}\left[{\text{Mn}}^{2+}\right]=\frac{5.88\times {10}^{-3}\text{ M}}{\left(\text{31}\text{.1}\times {\text{10}}^{10}\right)\left(1.18\times {10}^{-3}\text{ M}\right)}\\ =1.602\times {10}^{-11}\text{ M}\end{array}$

Substitute $1.602\times {10}^{-11}\text{ M}$ for $\left[{\text{Mn}}^{2+}\right]$ in equation (2).

  $\begin{array}{c}{\text{pMn}}^{2+}=-\log \left(1.602\times {10}^{-11}\text{ M}\right)\\ =10.80\end{array}$

Titration curve for the same is drawn as follows: