Lecture 8 Analytical PDF

Course: Principles of Quantitative Analysis Lecture (8): Principles of volumetric analysis Complexmetric Titration D r. A b b a s M a m d o h A b b a s  Complexmetric Titration “A volumetric titration involves the formation of soluble complex...

Course: Principles of Quantitative Analysis Lecture (8): Principles of volumetric analysis Complexmetric Titration D r. A b b a s M a m d o h A b b a s  Complexmetric Titration “A volumetric titration involves the formation of soluble complex between metal ions (as acceptor) and ligands (as donor) to form coordination compounds (complexes) via coordination bonds” The metal ion is known as Central metal atom (M). The anion or neutral molecule is known as Ligand (L)  Complexmetric Titration  Complex (Coordination Compound) Compound results from the combination of a metal ion as (acceptor, Lewis acid) with a ligand (donor, Lewis base) through with coordinated bonds (donor → acceptor). Mn+ + xL → [MLx]n+  Central metals must have small size, vacant (empty) orbitals and ability increases with increasing oxidation state.  The number of coordination bonds around central metal is known as (Coordination number) Ligands (Complexing agents) must have high basicity and contain at least one atom with lone pairs or contain a negative charge.  Complexmetric Titration  Type of complexing agents (ligands) This classification is according to the no. of sites attached to the metal ion 1. Unidentate (H2O, NH3, CN-, Cl-, I-, Br-) 2. Bidentate Ligand  Complexmetric Titration  Type of complexing agents (ligands) 3. Tridentate, tetradentate, pent…etc. (Chelating Agent) Diethylene triamine Triethylene tetramine EDTA Ethylene Diamine Tetra acetic Acid  Complexmetric Titration  Factors affecting stability of complexes  Effect of central metal ion Ionic size (metal radius) Ionic charge (metal charge) Acidity  Effect of Ligand: Basic character (N > O > S > I- > Br- > Cl- > F-) The degree of chelation Steric effect  Complexmetric (EDTA) Titration Titration involving EDTA is known as complexometric titration. EDTA is a hexadentate ligand, containing 4 oxygen and 2 nitrogen donor. 2 moles of hydrogen ions are formed in complexation. EDTA is not a selective chelating agent. H4Y + H2O H3Y - + H+ (pH= 2.7) H3Y - + H2O H2Y2- + H+ (pH = 6.2) H2Y2- + H2O HY3- + H+ (pH = 10.3) HY3- + H O Y4- + H+ (pH = 12 )  Complexmetric Titration  EDTA titration Why EDTA is the best titrant used in complexometric titration? 1. It forms very stable complexes with all metals. 2. One step reaction (sharp abrupt change area). 3. Form soluble complexes (coprecipitation errors are absent) 4. Stoichiometric 1:1 complexes with all metals. 5. Disodium salt of EDTA (Na2H2Y.2H2O) is primary standard. 6. The end point could be easily achieved using metal ion indicators.  Complexmetric Titration  Titration curve When performed the titration at suitable pH, the titration curve will alter with the value of the stability constant. M2+ + Y4- [MY2-] Ksta = Kf =  Complexmetric Titration  Titration curve For the derivation of the titration curve, consider the titration of 50.0 mL of 0.1 M solutions of Mg2+ (buffered at pH=10) with 0.1 M EDTA solution. We calculate the concentration of Mg2+ (as pMg= -log[Mg] and plotted versus the volume added from ETDA 1. Before the addition of EDTA Mg2+ = 0.1 M pMg= - log (0.1) = 1.0  Complexmetric Titration  Titration curve 2. After the addition of EDTA. The unreacted Mg will be calculated after adding 10.0 mL EDTA (the unreacted 40 ml of Mg), so, the concentration of [Mg2+] = (40 x 0.1)/60 = 0.067 pMg = - log(0.067) = 1.18 Adding 30.0 mL EDTA, [Mg2+] = (20 x 0.1)/80 = 0.025, pMg = - log(0.067) = 1.6 Adding 45.0 mL EDTA, [Mg2+] = (5 x 0.1)/95 = 0.0053, pMg = - log(0.067) = 2.28  Complexmetric Titration  Titration curve 2. At the equivalent point. At this point, 50 mL of EDTA has been added to completely react with Mg solution. We will use the stability constant (5.0 x 108) to calculate the pMg. Mg2+ + Y4- ↔ [MgY 2-] kf = = 5.0 x 108 [MgY2-] = (50 x 0.1/100) and (Mg2+) = (Y4-), so, = 5.0 x 108 [Mg2+] = = 1.0 x 10-5 So, pMg= 5  Complexmetric Titration  Titration curve 3. After equivalent point. 60 mL of EDTA have been added to react with the Mg ions solution. We will use the stability constant to calculate the pMg. Mg2+ + Y4- ↔ [MgY]2- kf = = 5.0 x 108 [MgY2-] = (50x0.1/110) and (Mg2+) ≠ (Y4-), and [(Y4-] = (10x0.1)/110 [Mg2+] = = 1 x 10-8 So, pMg= 8  Complexmetric Titration  Detection of End Point: Use Metal Ion Indicators  Characteristics of Metal Ion Indicators 1. The color reaction must be such that before the endpoint. 2. The indicator must be very sensitive to metal ions. 3. The metal-indicator complex must be less stable than the metal-EDTA 4. High color contrast between free and complexed indicators. 5. The color reaction should be specific or at least selective. 6. The above requirements must be achieved within the pH range at which the titration is performed.  Complexmetric Titration  Detection of End Point: Use Metal Ion Indicators  Theory of the use of metal ion indicators M-In + EDTA → M-EDTA + In  Eriochrome black T pH pH H2In- HIn2- In3- 5.3 - 7.3 10.5 - 12.5 Red Blue Yellow - orange Murexide  Complexmetric Titration  Types of EDTA titration. 1. Direct titration The solution containing the metal ion to be determined is buffered to the desired pH (e.g., to pH = 10 with NH4Cl/NH4OH, i.e., the ammoniacal buffer) and titrated directly with the standard EDTA solution in presence of the indicator. Mn+ + In [M-In] Color 1 Color 2 Mn+ + 10 ml of + Ind [M-In] + EDTA [M-EDTA] + In buffer solution (pH = 10) Colorless Color 1  Complexmetric Titration  Types of EDTA titration. 2. Back-titration M + EDTA(excess) → [M-EDTA] n+ Zn2+ or Mg2+ EDTA (excess) + Zn2+ or Mg2+ → [Zn or Mg – EDTA] Zn2+ or Mg2+ + In → [M-In] EDTA(excess) + Mn+ + In +10 ml of buffer solution pH = 10  Complexmetric Titration  Types of EDTA titration. 3. Replacement or substitution titration Mn+ + MgY2− → MY(n−4) + Mg2+ 4. Indirect Titrations 5. Alkalimetric titration Mn+ + H2Y2− → MY(n−4) + 2H+  Complexmetric Titration  Examples  Determination of Water Hardness  Water hardness is due to the presence of Ca2+ & Mg2+ salts.  EDTA forms complex with Ca2+ & Mg2+  Ca-EDTA complex is more stable than Mg-EDTA complex.  At pH 12 EDTA forms complex with Ca2+ only.  Complexmetric Titration  Examples  Determination of Aluminium  Sample of Al3+ is heated with known excess of standard EDTA at pH 7-8.  The solution is then adjusted to pH=10 using ammonia buffer.  The residual EDTA is titrated against standard Zn2+ using EBT indicator.  The color change from blue to wine red.  Complexmetric Titration  Examples  Analysis of Metal Ion Mixtures (selectivity improvements) To increase the selectivity of EDTA, some procedures could be followed Control of pH of the medium Adjustment of oxidation number of metal ion Using masking and demasking agents Using selective metal indicator. Classical separation by precipitation or extraction …etc.  Complexmetric Titration  Examples  Analysis of Metal Ion Mixtures (Control of pH of the medium)  First group: Trivalent & tetravalent cations Bi3+, Fe3+, Th4+, and Hg2+ titrated at pH 1- 3 using conc. HNO3.  Second group: Divalent metals Co2+, Ni2+, Cu2+, Zn2+, pb2+, and Cd2+ titrated at pH 4-6 using acetate buffer.  Third group: Alkaline earth metal (Ba2+, Sr2+, Ca2+, and Mg2+ titrated at pH=10 using ammonia buffer or 8% NaOH.  Complexmetric Titration  Examples  Analysis of Metal Ion Mixtures (Control of pH of the medium) Analysis a mixture of Bi3+ & pb2+ 1.First titrate Bi3+ at pH = 2 using xylenol orange as indicator. 2.Then increased pH to 5 by adding hexamine and titrating pb2+.  Complexmetric Titration  Examples  Analysis of Metal Ion Mixtures (Adjustment of Oxidation Number of Metal Ion) This procedure removes the interferences between metal ions of the same group of pH. Ascorbic acid is a reducing agent used in: Removal of interference of Fe3+ in first group (pH 1-3) → reduced to Fe2+ Removal of interference of Hg2+ in the first group (pH 1-3) → reduced to Hg (ppt.) Removal of interference of Cu2+ in second group (pH 4-6) → reduced to Cu+. Oxidation of Cr3+ to CrO42- Fe2+ , Hg, Cu+, CrO42- do not react with EDTA  Complexmetric Titration  Examples  Analysis of Metal Ion Mixtures (Masking and Demasking Agents)  Masking agents: A reagent added to prevent reaction of some metal ion with EDTA (block metal ions)  These reagents form complexes with interfering ions which are more stable than complexes formed with indicator & EDTA.  Demasking agents: Reagents that release of a metal ion from a masking agent  Complexmetric Titration  Examples  Analysis of Metal Ion Mixtures (Masking and Demasking Agents) Example (titration of Ca and Mg in presence of Zn or Cd) The masking by CN– can be removed by (mixture of formaldehyde – acetic acid) - on addition of demasking agent to [Zn(CN) 4]2-, Zn2+ is liberated and titrated with EDTA.  Complexmetric Titration 50 mL of water sample is transferred into a conical flask. The pH is fixed at pH = 10 using a suitable buffer solution and some drops of EBT indicator are added. This solution is titrated with 0.01 M EDTA. The blue color appears after adding 30 mL of EDTA solution. Calculate the total hardness in molarity and as ppm CaCO3 ?  Complexmetric Titration 100 mL of water sample is transferred into a conical flask. The pH is fixed at 10 and some drops of EBT are added. This solution is titrated with 0.02 M EDTA and the volume at the equivalent point is 35 mL ( this volume is equivalent to Ca + Mg ). Another 100 mL of the water sample is transferred into the conical flask and the pH is fixed at 12 where Mg(OH) 2 is precipitated. A suitable indicator for Ca is added and the solution is titrated with EDTA solution. The volume of EDTA at the equivalent point is 28 mL ( this equivalent to only Ca ). Calculate both ppm CaCO3 and  Complexmetric Titration 100 mL of a paint sample containing Ni2+ ( aw = 58.7 ) is diluted with a buffer solution pH = 10. Then 50 mL of 0.05 M EDTA is added. The excess EDTA is titrated with 0.01 M Mg2+ solution using EBT indicator. If the volume of Mg 2+ solution at the equivalent point is 5 mL calculate the concentration of Ni 2+ in the paint sample in g Ni / L ?