Compute the titration curve for Demonstration 16-1, in which 400.0mLof $\mathbf{37}\mathbf{.}\mathbf{5}{\mathbf{mMFe}}^{\mathbf{2}\mathbf{+}}$ are titrated with $\mathbf{20}\mathbf{.}\mathbf{0}{\mathbf{mMMnO}}_{\mathbf{4}}^{\mathbf{-}}$ at a fixed pH of 0.00 in $\mathbf{1}{\mathbf{MH}}_{\mathbf{2}}{\mathbf{SO}}_{\mathbf{4}}$ . Calculate the potential versus S.C.E. at titrant volumes of 1.0,7.5,14.0,15.0,16.0, and 30.0 mL and sketch the titration curve.

• Titrations are frequently graphed as titration curves, with the volume of the titrant as the independent variable and the pH of the solution as the dependent variable (because it changes depending on the composition of the two solutions).
• Titration is a typical quantitative chemical analysis procedure used in laboratories to quantify the concentration of a specified analyte. The titrant or titrator is a reagent that is made as a standard solution with a known concentration and volume.

The potential vs. S.C.E. at titrant quantities of a particular volume must be computed, and a titration curve created.

Potential of the electrode (E): Electrode potential is the electromotive force between two electrodes. Two electrodes make up a cell: one is a standard electrode (such as calomel electrode or standard hydrogen electrode) and the other is a provided electrode.

$\mathrm{E}={\mathrm{E}}_{+}{\mathrm{E}}_{-}$

Where,

${\mathrm{E}}_{+}$is the electrode's potential when it's linked to the positive terminal

${\mathrm{E}}_{-}$is the electrode's potential when it's linked to the negative terminal

Titration:

$$\begin{gathered} {\mathrm{MnO}}_4^{-}+5{\mathrm{Fe}}^{2+}+8{\mathrm{H}}^{+}\to {\mathrm{Mn}}^{2+}+5\mathrm{Fe}3++4{\mathrm{H}}_2\mathrm{O} \\ \rightleftharpoons {\mathrm{Fe}}^{3+}{\mathrm{e}}^{-}{\mathrm{E}}^0=0.68\mathrm{Vin}1{\mathrm{MH}}_2{\mathrm{SO}}_4 \\ {\mathrm{MnO}}_4^{-}+8{\mathrm{H}}^{+}+5\mathrm{e}\to {\mathrm{Mn}}^{2+}+4{\mathrm{H}}_2\mathrm{O}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{\mathrm{E}}^0=1.507\mathrm{V} \end{gathered}$$

At 15.0 mL, the equivalency point is reached.

Prior to the moment of equivalency,

$$\begin{gathered} \mathrm{E}={\mathrm{E}}_{+}+{\mathrm{E}}_{-} \\ =(0.68-0.05916\log \frac{\left|{\mathrm{Fe}}^{2+}\right|}{[{\mathrm{Fe}}^{2+}\mid }-0.241 \\ 1.00\mathrm{mL}: \\ [{\mathrm{Fe}}^{2+}]/[{\mathrm{Fe}}^{3+}]=14.0/1.0 \\ =0.371\mathrm{V} \\ 7.5\mathrm{mL}: \\ [{\mathrm{Fe}}^{2+}]/[{\mathrm{Fe}}^{3+}]=7.5/7.5 \\ =0.439\mathrm{V} \\ 14.0\mathrm{mL}: \\ [{\mathrm{Fe}}^{2+}]/[{\mathrm{Fe}}^{3+}]=1.0/4.0 \\ =0.507\mathrm{V} \end{gathered}$$

Use the electrode potential equation at the equivalence point,

$6E_4=8.215-0.05916log\frac{1}{[H^{+}{]}^{\pi }}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}pH=0$

$$\begin{gathered} {\mathrm{E}}_4=1.369\mathrm{V} \\ \mathrm{E}={\mathrm{E}}_{+}{\mathrm{E}}_{-} \\ =1.369\mathrm{V}-0.241 \\ =1.128\mathrm{V} \end{gathered}$$

After the equivalence point,

$$\begin{gathered} \mathrm{E}={\mathrm{E}}_{+}+{\mathrm{E}}_{-} \\ =\left(1.507-\frac{0.05916}{5}\log \frac{\left|{\mathrm{Mn}}^{2+}\right|}{\left|{\mathrm{MnO}}_4^2\right||{\mathrm{H}}^{+}{|}^{\mathrm{N}}}\right)-0.241 \\ 16.0\mathrm{mL}: \\ \left[{\mathrm{Mn}}^{2+}\right]/\left[{\mathrm{MnO}}_4\right]=15.0/1.0\mathrm{and}\left[{\mathrm{H}}^{+}\right]=1\mathrm{M} \\ =1.252\mathrm{V} \\ 30.0\mathrm{mL}: \\ \left[{\mathrm{Mn}}^{2+}\right)]/[{\mathrm{MnO}}_4^{-}]=15.0/15.0\mathrm{and}[{\mathrm{H}}^{+}]=1\mathrm{M} \\ =1.266\mathrm{V} \end{gathered}$$

titration curve must be drawn