GREEN CRYSTALLINE COMPOUND PART III: DETERMINATION OF PERCENT POTASSIUM AND PERCENT IRON

PURPOSE: In this third of four experiments involving the synthesis and analysis of a green crystalline compound with the formula, KxFey(C2O4)z· a H2O, you will determine both the percent potassium and percent iron. The % K and % Fe will be obtained by carrying out a pH titration after passing a solution containing a known mass of the compound through an ion exchange column to prepare for analysis.

The ion exchange column contains a resin. Resins are solids which are insoluble in water.

They usually have a granular texture and when they are added to water, they form a slurry. Ion exchange resins are made up of large molecules containing ionizable groups. The ionizable portion of the resin molecule to be used in this experiment is a sulfonic acid group (-SO3-H). This group ionizes in the presence of water.

R-SO3-H ------> R-SO3-1 + H+1

R symbolizes the rest of the resin molecule. H3O+ is the hydronium ion which is bound to the sulfonateion on the resin. This is a cation resin, so called because other cations can be substituted for H3O+ (theacid form of the resin). When a slurry of the resin and water is poured into a glass tube, plastic columnor buret, with a cotton plug at the bottom to hold the resin in and let excess water flow out, a workableion exchange column is obtained. When a massed sample of the green crystalline compound is dissolved in water, the saltdissociates by the equation:

KxFe(C2O4)y · zH2O ------> xK+(aq) + Fe(C2O4)yx-(aq) + zH2O

If this solution is passed through the cation exchange resin in the acid form, each of the K+ ions willbecome attached to the resin in place of an H3O+ ion. The equation is:

xR-SO3- H3O+ + xK+ + Fe(C2O4)yx------> xR-SO3-K+ + xH3O+ + Fe(C2O4)yx-

The number of moles of K+ in the solution of the green crystalline compound is equal to the number ofmoles of H3O+ in the solution (the eluate) that comes off the resin column. The number of moles ofH3O+ and thus the number of moles of K+ can be determined by doing an acid-base titration using a standardized solution of NaOH:

mol K+ = mol H3O+ = mol NaOH used in titration = VNaOH(in liters) · MNaOH

The mass of K+ can be obtained from the number of moles:

mass K+ in sample = mol K+ · 39.1 g/mol

The percent potassium can then be calculated:

%K = [(mass of K+ in sample)/(mass of sample of crystal)] x 100

Notice that iron comes through the ion exchange column in the form of Fe(C2O4)yx-. When thesolution from the ion exchange column is titrated with standardized NaOH solution, the reaction thattakes place first is given by the equation:

H3O+ + OH------> 2H2O

Once all of the acid (H3O+) has reacted,further addition of NaOH results in the reaction of Fe(C2O4)yx- by the following equation:

Fe(C2O4)yx-(aq) + 3OH-(aq) ------> Fe(OH)3(s) + yC2O42-(aq)

The Fe(OH)3 precipitates from solution as a rust colored precipitate. The equation shows that threemoles of OH- from the NaOH are needed to react with one mole of Fe in Fe(C2O4)yx-.

mol Fe = VNaOH(in liters) ·MNaOH·1 mol Fe

3 mol NaOH

The mass of Fe can be calculated from the number of moles.

mass Fe = mol Fe · 55.8 g/mol

Since the mass of the green crystalline compound is known, the percent iron can be determined.

%Fe = [(mass Fe)/(mass of crystalline compound)]x100

The solution obtained from the ion exchange column is to be titrated with a standard solution ofNaOH. A pH meter is used to follow the reaction. A titration curve of pH versus number of mL ofNaOH solution added will be graphed. The curve will have two equivalence points as shown in the figurebelow:

The volume of NaOH solution needed to reach the first equivalence point (V1) along with the molarity of thesolution can be used to calculate the number of moles of OH- needed to neutralize the H3O+. Thevolume of NaOH solution required to go from the first equivalence point to the second equivalence point (V2-V1) alongwith the molarity of the solution is used to calculate the number of moles of OH- needed to combinewith iron.

PRELAB PROBLEMS:

1. Suppose 0.240 gram of the green crystalline compound is measured out for this

experiment (instead of the traditional 0.160 gram). What are the mass percentages of K and Fe if the equivalencepoints are at V1 = 14.10 mL and V2 = 28.88 mL when the salt solution that has been passed through theion exchange column is titrated with 0.100 M NaOH?

MATERIALS:

Buret and clamp.100 M NaOHgreen crystalline compound

Dowex exchange resin150 mL beakerComputer and labpro device

10 mL graduated cylinderDistilled waterpH probe

8M nitric acidcotton ball

PROCEDURE:

  1. CONSULT YOUR INSTRUCTOR ON HOW TO PREPARE YOUR ION EXCHANGE COLUMN.
  2. Record the precise concentration of the 0.100 M NaOH solution on your data sheet.
  3. Measure out between 0.160 and 0.165 gram of the green crystalline compound into a 150 mlbeaker and record the mass.
  4. Add 4 mL of distilled water to a 10 mL graduated cylinder and transfer to the beaker containingthe crystals. Swirl the beaker until the crystals are completely dissolved.
  5. Set a clean and dry 150 mL beaker under the ion exchange column and quantitatively transfer thesolution containing the green crystalline compound to the resin column. Allow the solution torun through the column to just above the top of the resin and collect it in the beaker.
  6. Rinse the 50 mL beaker that originally contained the green crystalline solution with 4 mL ofdistilled water and pour it onto the resin column. Run this solution through the column to the topof the resin bed and collect the eluate in the 150 mL beaker.
  7. Repeat the procedure of rinsing your solution with 4 mL of distilled water and running the solution through theion exchange resin one more time.
  8. Prepare to do a pH titration by putting a stirring bar in the green crystalline solution and settingthe 150 mL beaker containing the solution on top of a magnetic stirrer. Open the program marked “Titration” on the desktop.
  9. Obtain approximately 70 mL of 0.100 M NaOH in a clean dry 100 mL beaker and after rinsingthe buret three times with 5 mL portions of the solution, fill the buret to the zero mark with the0.100 M NaOH solution. Expel air bubbles from the buret.
  10. Calibrate your pH probe as indicated by the instructor.
  11. Put the pH electrode into the 150 mL beaker containing the green solution and add enoughdistilled water to cover the end of the electrode in solution.
  12. Titrate the solution with 0.100 M NaOH. Add 2 mL increments of the NaOH solution until you get to 5mL (near the first equivalence point). Take readings of the mL of NaOH solution and pH aftereach incremental addition.
  13. Between 5 mL and 11 mL make approximately 0.5 mL additions of the NaOH solution.
  14. Between 11 mL and 15 mL make 2 mL additions of the NaOH solution which will get you closeto the second endpoint.
  15. Around the second equivalence point (between 15 mL and 21 mL) make 0.5 mL additions of the NaOHsolution. End the titration with two 2 mL additions.
  16. Make a graph of pH vs. volume of 0.100 M NaOH added. Draw a smooth curve through the points. Determine the equivalence points V1 and V2 from the graph.
  17. Calculate the % K and % Fe in the green crystalline compound.

DATA AND CALCULATIONS:

Include the following in the data/calculations portion of your lab write up:

Concentration of NaOH solution, mass of green crystalline sample, volume in liters of NaOH solution required for first equivalence point, moles of NaOH required for the first equivalence point, moles of K+1 in sample, mass of K+1 in sample, percent K+1 in sample, volume in liters of NaOH solution required for the second equivalence point, moles of NaOH required for the second equivalence point, moles of Fe+3 in sample, mass of Fe+3 in sample, percent of Fe+3 in sample.

SHOW ALL RELEVANT CALCULATIONS THAT TOOK YOU FROM EACH ITEM ABOVE TO THE NEXT!

CONCLUSIONS:

  1. Why is it not necessary to keep track of the amount of distilled water added to the solution of thegreen crystalline compound in step 9 of the procedure?
  2. Describe, using illustrations or explanation, how the ion exchange column operates.
  3. Why is the number of moles of NaOH used to titrate the sample of the green crystalline compound to the first equivalence point equal to the number of moles of potassium in the sample?