Chromatography

• the process of separating and analyzing mixtures

• involves a mobile phase and a stationary phase

Types include:

• Gas chromatography• High performance liquid chromatography

Gas chromatography

• Separation and analysis of a gas

High performance liquid chromatography

• separation and analysis of a liquid• Types include: High performance

partition- immiscible stationary phase• High performance adsorption- finely

divided silica and alumina is the stationary phase

• Ion exchange

Ion Exchange

• makes use of ion exchange resins for the stationary phase.

• Resins: may be cation or anion, essentially trades ions with those in a solution (mobile phase)

Discussion Questions

2. The significance of the steps in the ion-exchange procedure (from preparation to analysis) a) Soaking the resin in water for an hour before introducing to the I/E column. Why can’t a dry resin be used?- Increase the moisture content of the resin- Increase strength of cross linakages

b) Keeping the water level above the top of the resin- air pockets - bar the solution from reacting with the

resin- causing errors in the reaction- Make sure all resin can and will react

with the solution

c) Addition of H2SO4 to the column - Regeneration-> ions are eliminated -The resin is returned to its protonated form: (RSO3)nM + H2SO4 2(RSO3-)H+ + (Mn+)→(SO4-2)

d) Rate flow of 15 drops per minute - makes sure a complete reaction occurs- through this rate hydrogen ions can

completely exchange with the copper(II) ions

- If rate is faster, copper may not bind well with the resin -> copper may be flush down as eluate

e) Washing the column until the eluate has the pH equal to the pH of distilled water- Copper <-> Hydrogen ions- Hydrogen ions flush down as eluate- pH decreases because of this- When pH equals 5 (dH2O), hydrogen ions

are no longer flushing down as eluate- Hydrogen ions are displaced completely

by copper

3. Complete balanced equation of the I/E column

2RSO3-H+ + Cu2+ (RSO3)2Cu + 2H+

4. Factors that affect ion-exchange - Resin (size, type, etc.)- Rate of drops- Retention- how long it takes for the

mobile phase to pass through the stationary phase

5. M Cu2+ in the unknown sample

Trial M Cu2+

1 0.0446

2 0.0446

3 0.06695

Average 0.05205– when 3 is rejected, 0.0446

Calculations

• Aliquot Factor:10mL 2500 ppm Cu 50mL100mL

10mL 10mL

€

10010x5010

= 50

M NaOH (Standardization)

1KHP:1NaOH

Average M NaOH= 0.0892€

(0.4126gKHP)x1molKHP204.2g

x1NaOH1KHP

x1

(0.0225L)= 0.0898M

Ppm Cu2+

Average ppm Cu2+= 2834.1516 ppm Cu2+

€

€

0.0892MNaOHx(0.0002L)x1H +

1NaOHx1Cu2+

2H + x63.546g1Cu2+ x

1000mg1g

x1

0.010Lx50

SW

• An aliquot solution of 20.00mL Cu2+ was taken from a solution of 100.0mL 3000.ppm Cu2+. That solution was then pass through a column with resin. Eluate was then titrated with NaOH (0.8620M). It was found out that 0.5000mL of NaOH was used until the endpoint was achieved. A. How much ppm (experimental) of Cu2+? B. What is the percent error?  

Cause and Effect• 1. Effect of air spaces on the ppm.

(Liquid level fell below resin)• 2. Effect of pH water not being equal to

pH of eluate to the ppm. (pH of eluate is higher than pH water)

• 3. Effect of using the wrong resin on a solution on ppm.

• 4. Effect of rate flow being more than 15 drops per minute on the ppm.

• 5. Effect of not washing the resin with sulfuric acid on the ppm.