91影视

Titration is a method or procedure for estimating the concentration of a dissolved material by calculating the least quantity of known-concentration reagent necessary to produce a particular effect in interaction with a known volume of the test solution.

a)

Now, in Appendix H , write two reduction half-reactions and determine their standard reduction potentials

$$\begin{gathered} 2{\mathrm{Fe}}^{3+}+{\mathrm{e}}^{-}鈫�{\mathrm{Fe}}^{2+}E掳=0.767V \\ dehydroascorbicacid+2{\mathrm{H}}^{+}+2{\mathrm{e}}^{-}鈫�ascorbicacid+E掳=0.390V \end{gathered}$$

To create the balanced equation, multiply the first equation by 2 and the second equation backwards:

$$\begin{gathered} 2{\mathrm{Fe}}^{3+}+{\mathrm{e}}^{-}鈫�2{\mathrm{Fe}}^{2+} \\ dehydroascorbicacid+{\mathrm{H}}_2\mathrm{O}鈫�ascorbicacid+2H^{+}+2e^{-} \end{gathered}$$

The sum of these two reactions is:

$$\begin{gathered} 2{\mathrm{Fe}}^{3+}+2{\mathrm{e}}^{-}+\mathrm{ascorbic}\mathrm{acid}+{\mathrm{H}}_2\mathrm{O}鈫�2{\mathrm{Fe}}^{2+}\mathrm{dehydroascorbic}\mathrm{acid}+2{\mathrm{H}}^{+}+2{\mathrm{e}}^{-} \\ 2{\mathrm{Fe}}^{3+}+\mathrm{ascorbic}\mathrm{acid}+{\mathrm{H}}_2\mathrm{O}鈫�2{\mathrm{Fe}}^{2+}\mathrm{dehydroascorbic}\mathrm{acid}+2{\mathrm{H}}^{+} \end{gathered}$$

b)

Calculate the equivalency point using data from the job before proceeding with potential calculations for each extra volume:

$$\begin{gathered} n\left(\mathrm{ascorbic}\mathrm{acid}\right)=2\mathrm{n}\left({\mathrm{Fe}}^{2+}\right) \\ V\left(\mathrm{ascorbic}\mathrm{acid}\right)=2路\frac{\mathrm{c}\left({\mathrm{Fe}}^{2+}\right)路\mathrm{V}\left({\mathrm{Fe}}^{2+}\right)}{\mathrm{c}\left(\mathrm{ascorbic}\mathrm{acid}\right)} \\ V\left(\mathrm{ascorbic}\mathrm{acid}\right)=2路\frac{0.02\mathrm{MH}路10\mathrm{mL}}{0.01,\mathrm{M}} \\ V\left(\mathrm{ascorbic}\mathrm{acid}\right)=10\mathrm{mL} \end{gathered}$$

Now determine three titration regions:

1. Before the equivalence point 5.00mL.

2. At the equivalence point $-10.00\mathrm{mL}$.

3. After the equivalence point $-15.00\mathrm{mL}$.

The Nernst equations for each reduction reaction are:

$$\begin{gathered} \mathrm{E}=0.767\mathrm{V}鈭�0.05916\log \frac{\left[{\mathrm{Fe}}^{2+}\right]}{\left[{\mathrm{Fe}}^{3+}\right]}鈭�0.197\mathrm{V}鈥�鈥�.\left(1\right) \\ \mathrm{E}=0.390\mathrm{V}鈭�\frac{0.05916}{2}\log \frac{[\mathrm{ascorbicacid}]}{[\mathrm{dehydroascorbicacid}]{\left[{\mathrm{H}}^{+}\right]}^2}鈭�0.197\mathrm{V}.......\left(2\right) \end{gathered}$$

Because there is an excess of Fe³⁺ in the solution before the equivalence point, we can utilise equation (1) to determine the potential by simply inserting Fe ion concentrations with their associated volume. The volume of ascorbic acid is one-half of the quantity required for the equivalence point at 5.00mL, therefore the log term is 0. In addition, E_ of $\mathrm{Ag}-\mathrm{AgCI}$may be found in appendix H. The potential equation is as follows:

$E=E掳-E_{-}$

For the task that means:

$$\begin{gathered} E=E掳\left(F{e}^{3+}\right|F{e}^{2+})-E_{-} \\ E=0.767V-0.197V \\ E=0.57V \end{gathered}$$

To compute potential at equivalent volume, multiply the ascorbic acid equation by 2 so that the logarithm components may be added on:

$$\begin{gathered} 2E_{+}=2\left(0.390\mathrm{V}鈭�\frac{0.05916}{2}\log \frac{\left[\mathrm{ascorbicacid}\right]}{\left[\mathrm{dehydroascorbicacid}]{\left[{\mathrm{H}}^{+}\right]}^2\right.}\right) \\ 2E_{+}=0.78\mathrm{V}鈭�0.05916\log \frac{\left[\mathrm{ascorbicacid}\right]}{\left[\mathrm{dehydroascorbicacid}]{\left[{\mathrm{H}}^{+}\right]}^2\right.} \end{gathered}$$

For each reduction reaction, the Nernst equations are as follows:

$$\begin{gathered} \mathrm{E}=0.767\mathrm{V}鈭�0.05916\log \frac{\left[{\mathrm{Fe}}^{2+}\right]}{\left[{\mathrm{Fe}}^{3+}\right]}鈭�0.197\mathrm{V}鈥�鈥�鈥�.\left(1\right) \\ \mathrm{E}=0.390\mathrm{V}鈭�\frac{0.05916}{2}\log \frac{[\mathrm{ascorbicacid}]}{[\mathrm{dehydroascorbicacid}]{\left[{\mathrm{H}}^{+}\right]}^2}鈭�0.197\mathrm{V}........\left(2\right) \end{gathered}$$

At the equivalence point, and dehydroascorbic acid, as well as and ascorbic acid, are at equilibrium, therefore,

$$\begin{gathered} \left[{\mathrm{Fe}}^{2+}\right]=2\left[\mathrm{dehydroascorbic}\mathrm{acid}\right] \\ \left[{\mathrm{Fe}}^{3+}\right]=2\left[\mathrm{ascorbic}\mathrm{acid}\right] \end{gathered}$$

By inserting the concentrations in the sum of two reactions,

$3{\mathrm{E}}_{+}=1.547\mathrm{V}鈭�0.05916\log \frac{\left[\mathrm{ascorbicacid}\right]\left[\left[{\mathrm{Fe}}^{2+}\right]\right.}{\left[\mathrm{dehydroascorbicacid}]{\left[{\mathrm{H}}^{+}\right]}^2\left[{\mathrm{Fe}}^{3+}\right]\right.}$

By substituting iron concentrations,

$3{\mathrm{E}}_{+}=1.547\mathrm{V}鈭�0.05916\log \frac{2\left[\mathrm{ascorbicacid}\right]\mathrm{dehydroascorbicacid}]}{2\left[\mathrm{dehydroascorbicacid}][\mathrm{ascorbicacid}]{\left[{\mathrm{H}}^{+}\right]}^2\right.}$

The equation now looks like this:

localid="1667808874918" $3E_{+}=1.547V-0.05916\log \left(\frac{1}{{\left[{\mathrm{H}}^{+}\right]}^2}\right)$

Concentration of can be calculated from pH :

$$\begin{gathered} \left[{\mathrm{H}}^{+}\right]={10}^{-\mathrm{pH}} \\ \left[{\mathrm{H}}^{+}\right]={10}^{-0.30} \\ \left[{\mathrm{H}}^{+}\right]=0.501\mathrm{M} \end{gathered}$$

Now calculate :

$$\begin{gathered} E_{+}=\frac{1.547V-0.05916\log \left({\displaystyle \frac{1}{0.{501}^2}}\right)}{3} \\ {E}_{+}=0.504V \end{gathered}$$

The total potential at is:

$$\begin{gathered} E=E_{+}-E_{-} \\ E=\left(0.504-0.197\right)V \\ E=0.307V \end{gathered}$$

There is an excess of ascorbic acid in the solution after equivalence point at 15mL of additional ascorbic acid, therefore use equation (2) to determine the potential by simply replacing concentrations of ions with their corresponding volume.

For example, if 15mL of ascorbic acid was added, the volume of formed dehydroascorbic acid is 10/10 and the volume of excess ascorbic acid is 5/10.

Now use equation (2) to calculate potential:

$$\begin{gathered} \mathrm{E}=0.390\mathrm{V}鈭�\frac{0.05916}{2}\log \frac{[\mathrm{ascorbicacid}]}{\left[\mathrm{dehydroascorbicacid}]{\left[{\mathrm{H}}^{+}\right]}^2\right.}鈭�0.197\mathrm{V} \\ \mathrm{E}=0.390\mathrm{V}鈭�\frac{0.05916}{2}\log \frac{\frac{5}{10}}{\frac{10}{10}鈰�0{.501}^2}鈭�0.197\mathrm{V} \\ \mathrm{E}=0.184\mathrm{V} \end{gathered}$$

The cell voltage is 0.184V .