Chapter 9 Complexation and Precipitation Titrations

Forming Complexes Most metals ions react with electron-pair donors to form coordination compounds or complexes. The donor species (ligand) must have at least one pair of unshared electrons available for bond formation. Cu(H 2 O) 4 +2, Cu(NH 3 ) 4 +2, Cu(NH 2 CH 2 COO) 2 CuCl 4 -2

Ligands A ligands is a neutral molecule or ion having a lone pair that can be used to form a bond to a metal ion. Chelating agents: unidentate, bidentate, tridentate, tetradentate, pentadentate, hexadentate

Titration of a Single Anion Calculate the pAg of the solution during the titration of 50ml of 0.05M NaCl with 0.1M AgNO 3 after the addition of the following volumes of reagent: (a) 0ml (b) 24.5 ml (c) 25 ml (d) 25.5ml K sp =1.82×10 -10

A: 50 ml of 0.05M NaCl with 0.1 M AgNO 3 B: 50 ml of 0.005M NaCl with 0.01 M AgNO 3

Signaling the End Point Mohr method Formation of a colored precipitate The Mohr method must be carried out at a pH of 7-10 because chromate ion is the conjugate base of the weak chromic acid. white red

Ethylenediaminetetraacetic Acid The EDTA molecule has six potential sites for bonding a metal ion: the four carboxyl groups and two amino groups.

Acidic Properties of EDTA H4YH4Y

H 3 Y - K 1 =1.02×10 -2 H 2 Y -2 K 2 =2.14×10 -3

HY -3 K 3 =6.92×10 -7 Y -4 K 4 =5.5×10 -11

The Nature of EDTA Complexes with Metal Ions The reagent combines with metal ions in a 1:1 ratio regardless of the charge on the cation.

Equilibrium Calculations Involving EDTA EDTA titrations are always performed in solutions that are buffered to a known pH to avoid interferences by other cations or to ensure satisfactory indicator behavior.

Conditional Formation Constant

Calculation of the Cation Concentration in EDTA Solutions Calculate  4 Calculate conditional formation constants K ’ 0.015M-xx Calculate equilibrium [Ni +2 ] at pH=3 and pH=8

pH 44 4 × × × × × × × × × × ×10 -4 Value for  4 for EDTA at selected pH value

Indicator for EDTA Titrations Ertichrome Black T K 1 =5× K 2 =2.8× red blue blue orange red blue

At the end point: MgIn + EDTA  MgEDTA + In - (red) (colourless) (colourless) (Blue) Before Titration: Mg 2+ + In -  MgIn (colourless) (blue) (red) During Titration: Before the end point Mg 2+ + EDTA  MgEDTA (free Mg 2+ ions) (Solution red due to MgIn complex) Compounds changing colour when binding to metal ion. K f for Metal-In - < K f for Metal-EDTA.

Requirements for Indicator Metal-indicator complex must be less stable than the metal-EDTA complex. Binding between metal and indicator must not be too weak. It has to avoid EDTA replacing at the beginning of the titration. In general, the metal-indicator complex should be 10 to 100 times less stable than the metal- titrant complex.

Titration of Ca +2 and Mg +2 The K’ of EDTA of Ca +2 and Mg +2 are too close to differentiate between them in an EDTA titration. Generally, they will titrate together. This titration is used to determine total hardness of water.

Titration of Ca +2 EB-T cannot be used to indicate the direct titration of Ca +2 in the absence of Mg +2 with EDTA. The indicator forms too weak a complex with Ca +2 to give a sharp end point.

Resolution A small measured amount of Mg +2 is added to the Ca +2 solution. Ca +2 gives more stable K’ than Mg +2. A correction is made for the amount of EDTA used for titration of the Mg +2.

At the end point: MgIn + EDTA  MgEDTA + In - (red) (colourless) (colourless) (Blue) Before Titration: Ca +2 +EDTA+Mg 2+ +In -  MgIn+CaEDTA During Titration: Before the end point Ca 2+ + EDTA  CaEDTA K f for Mg +2 -EDTA < K f for Ca +2 -EDTA.

EDTA Titration Curve Region 1 Excess M n+ left after each addition of EDTA. Conc. of free metal equal to conc. of unreacted M n+. Region 2 Equivalence point:[M n+ ] = [EDTA] Some free M n+ generated by MY n-4  M n+ + EDTA Region 3 Excess EDTA. Virtually all metal in MY n-4 form.