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EDTA – An Introduction CHM 103 Sinex

EDTA – An Introduction CHM 103 Sinex. EDTA is ethylenediaminetetraacetic aicd For more information on EDTA – see the MOTM for March 2004.MOTM for March

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Page 1: EDTA – An Introduction CHM 103 Sinex. EDTA is ethylenediaminetetraacetic aicd For more information on EDTA – see the MOTM for March 2004.MOTM for March

EDTA – An Introduction

CHM 103Sinex

Page 2: EDTA – An Introduction CHM 103 Sinex. EDTA is ethylenediaminetetraacetic aicd For more information on EDTA – see the MOTM for March 2004.MOTM for March

EDTA is ethylenediaminetetraacetic aicd

For more information on EDTA – see the MOTM for March 2004.

Six binding sites4 H on carboxylic acid groups (H’s are removed, so –COO-)2 lone pairs of electrons on nitrogensEDTA Chime structure

Page 3: EDTA – An Introduction CHM 103 Sinex. EDTA is ethylenediaminetetraacetic aicd For more information on EDTA – see the MOTM for March 2004.MOTM for March

Fe+3 + EDTA-4 FeEDTA-1

1:1 metal ion to EDTA

metal ion with octahedral geometry

Page 4: EDTA – An Introduction CHM 103 Sinex. EDTA is ethylenediaminetetraacetic aicd For more information on EDTA – see the MOTM for March 2004.MOTM for March

chemistry• Ca+2 with EDTA – 1:1• at equivalence point:

moles Ca = moles EDTA

• so to find moles: moles = M x VL

• standard calcium solution from CaCO3

CaCO3(s) + 2HCl(aq) CaCl2(aq) + CO2(g) + H2O

• indicator – MgIn- HIn-2 wine red sky blue

After Ca+2 reacts with EDTA, EDTA complexes Mg+2 from indicator.

Page 5: EDTA – An Introduction CHM 103 Sinex. EDTA is ethylenediaminetetraacetic aicd For more information on EDTA – see the MOTM for March 2004.MOTM for March

indicator

Tracking the metal ion - EDTA titration

Page 6: EDTA – An Introduction CHM 103 Sinex. EDTA is ethylenediaminetetraacetic aicd For more information on EDTA – see the MOTM for March 2004.MOTM for March

If a student masses out 0.5251 g calcium carbonate, dissolves it in HCl, and then places it into a 500.0 mL volumetric flask, what is the molarity of the calcium ion?

0.5251 g x 1 mole/100g = 5.251 x 10-3 moles Ca+2

5.251 x 10-3 moles Ca+2/0.5000L = 0.1050 M

answer

If 25.00 mL of the Ca+2 solution above requires 18.93 mL of EDTA to titrate, what is the molarity of the EDTA?

MCa x VCa = MEDTA x VEDTA hint