Experiment 9-11

Experiment 9: Determination of Magnesium Using EDTA Titration (Complexometric Titration)

Complexometric Titration with EDTA

• Classic method for determining magnesium and other suitable cations involves titration with a standardized solution of ethylenediaminetetraacetic acid (EDTA).

• EDTA's structure is represented as "H4Y" to avoid repeatedly drawing it or writing out the chemical formula.

• Each acid hydrogen on EDTA can be removed, producing HY3-, HY2-, HY-3, and Y4- ions.

• EDTA 4- forms very stable complexes with most transition metals such as Al3+, Ca2+, Cu2+, Fe2+, Fe3+, Mg2+, and Zn2+.

EDTA as a Chelating Agent

• EDTA is a hexadentate ligand, meaning it can bind to a metal ion through six donor atoms: four carboxylate groups and two amine groups.

• This multi-dentate binding capability forms highly stable complexes with metal ions.

• The structure of EDTA changes with pH due to the sequential deprotonation of its acidic hydrogen ions.

• At a pH of about 10, EDTA exists primarily in its fully deprotonated Y4- form, maximizing its binding efficiency with metal ions.

• The general complexation reaction of EDTA with a divalent metal ion (M2+) is represented as:
  $M^{2+} + Y^{4-} \rightarrow MY^{2-}$
  Where M2+ represents a divalent metal ion such as Ca2+ or Mg2+, and Y4- is the fully deprotonated form of EDTA. The resulting MY2- complex is highly stable.

The Role of pH and Buffers in EDTA Titration

• The pH of the solution is crucial in EDTA titrations.

• Buffering with an ammonia-ammonium buffer is used to maintain a pH of around 10.

• At this pH, EDTA is in its Y4- form, which is ideal for binding Mg2+ ions.

Indicators in EDTA Titration

• Eriochrome Black T (EBT) is a common indicator used in EDTA titrations.

• EBT binds weakly with metal ions, producing a wine-red color.

• When all Mg2+ ions are complexed with EDTA, the EBT indicator is released, resulting in a blue color change at the endpoint.

• The reactions with EBT:

  • Before titration:
    $M^{2+} + EBT \rightarrow M-EBT \text{ complex (wine-red)}$

  • During titration:
    $M-EBT \text{ complex } + Y^{4-} \rightarrow MY^{2-} + EBT \text{ (blue)}$

• This color change from wine-red to blue signals that the EDTA has complexed all metal ions present in the sample, indicating the endpoint.

• $Mg^{2+} + HIn^{2-} \rightleftharpoons MgIn^{-} + H^{+}$

  • Analyte + Free indicator (Blue) -> Metal ion-indicator complex (Wine-red)

• $MgIn^{-} + HY^{3-} \rightleftharpoons MgY^{2-} + HIn^{2-}$

  • Metal ion indicator complex (Wine-red) + titrant (EDTA) -> Metal ion-EDTA complex (Colorless) + Free indicator (Blue)

Experiment 10: Standardization of I2 Against Previously Standardized Na2S2O3 (Iodometry)

Materials:

• potassium iodide (KI) crystals, iodine (I2) crystals, starch indicator, 3M H2SO4, distilled water, standardized sodium thiosulfate (Na2S2O3) solution

Redox Titration

• Redox titration is a titration in which the reaction between the analyte and titrant is an oxidation/reduction reaction.

• The titrant can either be an oxidizing agent or a reducing agent.

• Conversely, the analyte can either be a reducing agent or an oxidizing agent respectively.

• Using iodine (I2) in a titration is a classic example of oxidation-reduction (redox) titration.

• Iodine is a weak oxidizing agent and is useful only for the analysis of analytes that are strong reducing agents.

• This apparent limitation, however, makes I2 a more selective titrant for the analysis of a strong reducing agent in the presence of weaker reducing agents.

Calculations

*Titration result is calculated by considering the stoichiometry involved in the reaction.

• ppm Mg2+ in Sample:

  • Since the stoichiometric ratio between Mg2+ and EDTA is 1:1, the formula to find ppm Mg2+:
    $ppm \; Mg^{2+} = M \; (EDTA) \; \times \; V \; of \; EDTA \; \times \; 2431$
    where,
    $1 \; mol \; Mg^{2+} = 24.31 \; g/mol$
    $2431 = \frac{24.31 \; g/mol \; \times \; 1000 \; mg/g}{1000}$

  • Example:
    M = 0.0100 mol/L;
    V = 12.00 mL;
    1 mg/L = 1 ppm
    $ppm \; Mg^{2+} = \frac{0.0100 \; mol}{L} \times 12.00 \; mL \times \frac{2431 \; mg}{mol \cdot mL} = 291.72 \; mg/L = 291.72 \; ppm$

Types of Titrations Using I2

1. Iodimetry:

   • A direct method that makes use of a standard I2 solution in titrating an easily oxidized substance.

   • The indicator used is a starch indicator, and the endpoint is indicated by the appearance of a blue color.

2. Iodometry:

   • An indirect method used for analyzing oxidizing agents.

   • The substance to be analyzed is brought into contact with excess iodide ion.

   • This results in liberating a quantity of iodine that is chemically equivalent to the amount of oxidizing agent present.

   • The amount of liberated iodine is determined by titration of a standard sodium thiosulfate (Na2S2O3) solution.

• For this experiment, you are going to prepare I2 solution and standardize it using iodometric titration with previously standardized Na2S2O3 solution.

• Iodine is not very soluble in water (0.001 M).

• To prepare solutions having analytically useful concentrations of the element, iodine is usually dissolved in moderately concentrated solutions of potassium iodide.

• In this medium, iodine is reasonably soluble as a consequence of the reaction that produces triiodide (I3-):
  $I<em>2 + I^{-} \rightleftharpoons I</em>3^{-}$

• Reaction of triiodide (I3-) with thiosulfate (S2O32-) is as follows:
  $I<em>3^{-} + 2S</em>2O<em>3^{2-} \rightarrow 3I^{-} + S</em>4O<em>6^{2-}$ where $S</em>4O_6^{2-}$ is tetrathionate

• The prepared iodine solution, which is initially dark brown, is titrated with a standard sodium thiosulfate solution to a pale straw color, and then the starch indicator is added.

• The mixture will turn blue due to the formation of a starch/iodine complex.

• Titrate further with sodium thiosulfate solution to the disappearance of the blue color.

• The blue color of the starch/iodine complex may reappear after titration has been completed because of the air oxidation of iodide ion.

Calculations

• Based on the reaction, the stoichiometric ratio between I2 and Na2S2O3 is 1 mol I2 / 2 mol Na2S2O3.

• The formula to find the molarity of the I2 titrant:
  $M(I<em>2) = \frac{M \; of \; Na</em>2S<em>2O</em>3 \times V \; of \; Na<em>2S</em>2O<em>3 \; (in \; mL)}{V \; of \; I</em>2 \; (in \; mL) \times 2}$
  *Note that M of Na2S2O3 will be provided by your instructor. After calculating the individual molarities, take the average. Calculate the precision of your experiment as standard deviation and as % relative standard deviation.

• Example:

  • Vol of Na2S2O3 = 10.00 mL = 0.0100 L

  • Molarity of Na2S2O3 = 0.0200 M

  • Vol. of Iodine solution = 25.00 mL = 0.0250 L

  • The reaction:
    $I<em>2 + 2Na</em>2S<em>2O</em>3 \rightarrow 2NaI + Na<em>2S</em>4O_6$

  • $mol \; I<em>2 = 0.0200 \; \frac{mol \; Na</em>2S<em>2O</em>3}{L} \times 0.0100 \; L \times \frac{1 \; mol \; I<em>2}{2 \; mol \; Na</em>2S<em>2O</em>3} = 0.0001 \; mol \; I_2$

  • $M(I<em>2) = \frac{mol \; I</em>2}{L<em>{I</em>2}} = \frac{0.0001 \; mol}{0.025 \; L} = 3.98 \times 10^{-3} \; M = 0.0040 \; mol/L$

Experiment 11: Determination of %Ascorbic Acid in an Impure Substance (Iodometry)

Materials:

• standardized iodine (I2) solution, starch indicator, distilled water

• Iodine has been used as an oxidizing titrant for a number of compounds of pharmaceutical interest.
  *One application is for the analysis of ascorbic acid (vitamin C; MW= 176.12 g/mol) by oxidizing the enediol functional group to an alpha diketone

*Ascorbic acid reacts with I2 in the following stoichiometric equation:
$ascorbic \; acid + I_2 \rightarrow dehydroascorbic \; acid + 2I^{-}$

*Due to this reaction, the iodine is immediately reduced to iodide if there is any ascorbic acid present.
*Once all the ascorbic acid has been oxidized, the excess iodine is free to react with starch indicator, forming a blue starch/iodine complex. This is the endpoint of the titration.

Calculations:

• Based on the equation, the stoichiometric ratio between ascorbic acid and I2 is 1 mol ascorbic acid / 1 mol I2.

• The formula to find the % ascorbic acid in your sample:
  $\% \; ascorbic \; acid = \frac{M(I<em>2) \times V \; of \; I</em>2 \; used \times 176.12}{wt. \; of \; Sx \times 1000} \times 100$

*After calculating the individual % ascorbic acid, take the average. Calculate the precision of your experiment as standard deviation and as % relative standard deviation.

• Reaction Equation:
  $C<em>6H</em>8O<em>6 + I</em>2 \rightarrow C<em>6H</em>6O_6 + 2HI$
  1 mol ascorbic acid reacts with 1 mol of I2

• Steps:

  1. Calculate moles of I2 used.
     $mol \; I_2 = M \times V = 0.0050 \; mol/L \times 0.01850 \; L = 9.25 \times 10^{-5} \; mol$

  2. Use mole ratio to get moles of ascorbic acid.
     $1 \; mol \; I_2 = 1 \; mol \; ascorbic \; acid$
     $mol \; Ascorbic \; Acid = 9.25 \times 10^{-5} \; mol$

  3. Convert moles to grams.
     $g \; Ascorbic \; Acid = mol \times molar \; mass = 9.25 \times 10^{-5} \; mol \times 176.12 \; g/mol = 0.01269 \; g$

  4. Calculate % by mass.
     $\% \; Ascorbic \; acid = \frac{0.01269 \; g}{15.0 \; g} \times 100 = 0.0846 \%$

Example:

A 10.00 mL juice sample is titrated with standardized iodine solution. It required 18.50 mL of 0.00500 M I2 to reach the endpoint. Calculate the percent by mass of ascorbic acid (C6H8O6) in the juice sample if the juice has a mass of 15.0 g.

Given:

• vol. of I2 = 18.50 mL = 0.01850 L

• molarity of I2 = 0.0050 M

• molar mass of ascorbic acid = 176. 12 g/mol

• mass of juice sample = 15.0 g