🎧 New: AI-Generated Podcasts Turn your study notes into engaging audio conversations. Learn more

Analytical Chemistry II.pdf

Loading...
Loading...
Loading...
Loading...
Loading...
Loading...
Loading...

Document Details

QuieterStrait2370

Uploaded by QuieterStrait2370

Mindanao State University - Iligan Institute of Technology

Tags

analytical chemistry titration chemical analysis

Full Transcript

Analytical Chemistry II OUTLINE ▪ Acid-Base Titration ▪ Precipitation Titration ▪ Complexometric Titration Analytical Chemistry 2 METHODS OF ANALYSIS Titration ▪ a process by which a stoichiometrically equivalent quantity of standard solution is systematic...

Analytical Chemistry II OUTLINE ▪ Acid-Base Titration ▪ Precipitation Titration ▪ Complexometric Titration Analytical Chemistry 2 METHODS OF ANALYSIS Titration ▪ a process by which a stoichiometrically equivalent quantity of standard solution is systematically added to a known quantity of sample Titrant ▪ reagent that is added incrementally usually from a buret Acid-Base Titration 3 METHODS OF ANALYSIS Standard Solution ▪ reagent of exactly known concentration Primary Standard Solution ▪ prepared from an accurately weighed substance of high degree of purity Acid-Base Titration 4 METHODS OF ANALYSIS Secondary Standard Solution ▪ concentration is obtained by standardization using a primary standard solution Equivalence Point ▪ the point in a titration when the amount of added standard reagent is chemically equal to the amount of analyte Acid-Base Titration 5 METHODS OF ANALYSIS End Point ▪ point in a titration when a physical change that is associated with the condition of chemical equivalence occurs Acid-Base Titration 6 METHODS OF ANALYSIS (Skoog, 2014) Acid-Base Titration 7 TITRATION TECHNIQUES Types ▪ Direct Titration ▪ Back-Titration ▪ Displacement Titration Acid-Base Titration 8 TITRATION TECHNIQUES Direct Titration ▪ a process in which the analyte is made to react with the titrant after addition of an indicator Acid-Base Titration 9 TITRATION TECHNIQUES Back Titration ▪ a process in which the excess of a standard reagent used to react with an analyte is determined by titration with a second standard solution ▪ usually employed when a reaction is slow to go to completion, and a sharp end point cannot be obtained ▪ when the analyte forms precipitates at the pH required for its titration Acid-Base Titration 10 TITRATION TECHNIQUES Displacement Titration ▪ used when no indicator for the analyte is available in EDTA complexometric titration, an unmeasured excess of a solution containing the magnesium or zinc complex is introduced into the analyte solution if the analyte forms a more stable complex than that of magnesium or zinc, the following displacement reaction occurs and the liberated cation is then titrated with standard 𝑀𝑔𝑌 2− + 𝑀2+ → 𝑀𝑌 2− + 𝑀𝑔2+ Acid-Base Titration 11 REQUIREMENTS Reactions Used in Titrimetric Analysis ▪ Reaction must proceed according to a definite chemical equation (no side reaction). ▪ Reaction must be complete (in general, the reaction is essentially complete if K > 107) ▪ Equivalence point must be detected by some method e.g. indicator ▪ Reaction must be rapid Acid-Base Titration 12 REQUIREMENTS Primary Standard ▪ High purity ▪ Stability toward air ▪ Composition does not change with variations in relative humidity (absence of hydrate water) ▪ Available and relatively inexpensive ▪ Reasonably soluble in the titration medium ▪ Reasonably large molar mass to minimize relative error in weighing Acid-Base Titration 13 REQUIREMENTS Acid/Base Indicators ▪ have an indicator range (transition range) of approximately 𝑝𝐾𝑖𝑛 ± 1 ▪ a suitable indicator changes color within 1 drop of the titrant in excess of the equivalence point 𝐻𝐼𝑛 + 𝐻2 𝑂 ↔ 𝐼𝑛− + 𝐻3 𝑂+ 𝐻3 𝑂+ 𝐼𝑛− 𝐾𝑎 = 𝐻𝐼𝑛 Acid-Base Titration 14 REQUIREMENTS Table of Acid/Base Indicators (Skoog, 2014) Acid-Base Titration 15 TITRATION CURVE Acid/Base Indicators ▪ pH or pOH vs mL of titrant ▪ useful for determining whether: feasibility of titration appropriate indicator Acid-Base Titration 16 TITRATION CURVE Titration curves for HCl with NaOH Curve A: 50.00 mL of 0.0500 M HCl with 0.1000 M NaOH Curve B: 50.00 mL of 0.000500 M HCl with 0.00100 M NaOH (Skoog, 2014) Acid-Base Titration 17 TITRATION CURVE Titration curves for Acetic acid with NaOH Curve A: 0.1000 M acid with 0.1000 M base Curve B: 0.001000 M acid with 0.001000 M base (Skoog, 2014) Acid-Base Titration 18 ACID-BASE EQUILIBRIA Strong acids/bases ▪ completely dissociated, hence there is no Ka or Kb associated with them Weak acids/bases ▪ partially dissociated ▪ monofunctional: Ka or Kb ▪ polyfunctional: Ka1, Ka2, etc. or Kb1, Kb2, etc. Acid-Base Titration 19 ACID-BASE EQUILIBRIA Bronsted-Lowry Concept ▪ conjugate acid-base pairs: the stronger the acid, the weaker the conjugate base and vice versa; K a x Kb = K w 𝑎1 → 𝑏1 + 𝐻+ 𝑏2 → 𝐻+ + 𝑎2 Acid-Base Titration 20 ACID-BASE EQUILIBRIA 𝑯𝟑 𝑶+ in solutions of weak acids 𝐻3 𝑂+ 𝐴− 𝐻𝐴 + 𝐻2 𝑂 ↔ 𝐻3 𝑂+ + 𝐴− 𝐾𝑎 = 𝐻𝐴 2𝐻2 𝑂 ↔ 𝐻3 𝑂+ + 𝑂𝐻− 𝐾𝑤 = 𝐻3 𝑂+ 𝑂𝐻− MBE: 𝐻3 𝑂+ = 𝐴− + 𝑂𝐻− ≅ 𝐴− 𝐶𝐻𝐴 = 𝐻𝐴 + 𝐴− Acid-Base Titration 21 ACID-BASE EQUILIBRIA 𝑯𝟑 𝑶+ in solutions of weak acids 𝐻3 𝑂+ 2 𝐾𝑎 = 𝐶𝐻𝐴 − 𝐻3 𝑂+ 𝑖𝑓 𝐶𝐻𝐴 > 1000𝐾𝑎 𝐻3 𝑂+ = 𝐾𝑎 𝐶𝐻𝐴 𝑖𝑓 𝐶𝐻𝐴 ≤ 1000𝐾𝑎 𝐻3 𝑂+ 2 + 𝐾𝑎 𝐻3 𝑂+ − 𝐾𝑎 𝐶𝐻𝐴 = 0 Acid-Base Titration 22 ACID-BASE EQUILIBRIA Salts of weak electrolytes ▪ react with water to produce either hydrogen or hydroxide ions 𝑂𝐴𝑐 − + 𝐻2 𝑂 ↔ 𝐻𝑂𝐴𝑐 + 𝑂𝐻 − b1 a2 a1 b2 + 𝑁𝐻4 + 𝐻2 𝑂 ↔ 𝑁𝐻3 + 𝐻3 𝑂+ a1 b2 b1 a2 Acid-Base Titration 23 ALPHA (𝜶) EXPRESSIONS 𝜶 = fraction of species ▪ For a monoprotic weak acid HA: 𝐻3 𝑂+ 𝐴− 𝐻𝐴 + 𝐻2 𝑂 ↔ 𝐻3 𝑂+ + 𝐴− 𝐾𝑎 = 𝐻𝐴 𝐻𝐴 𝐻𝐴 𝐻𝐴 𝐻3 𝑂+ 𝛼0 = 𝛼𝐻𝐴 = = − = = 𝐶𝐻𝐴 𝐻𝐴 + 𝐴 𝐻𝐴 𝐻3 𝑂+ + 𝐾𝑎 𝐻𝐴 + 𝐾𝑎 𝐻3 𝑂+ 𝐴− 𝐴− 𝐴− 𝐾𝑎 𝛼1 = 𝛼𝐴− = = − = + − = + +𝐾 𝐶𝐻𝐴 𝐻𝐴 + 𝐴 𝐻3 𝑂 𝐴 𝐻3 𝑂 𝑎 + 𝐴− 𝐾𝑎 Acid-Base Titration 24 ALPHA (𝜶) EXPRESSIONS 𝜶 = fraction of species ▪ For a tripotic weak acid H3A: 𝐻3 𝑂+ 3 𝛼0 = 𝛼𝐻3 𝐴 = 𝐻3 𝑂+ 3 + 𝐻3 𝑂+ 2 𝐾𝑎1 + 𝐻3 𝑂+ 𝐾𝑎1 𝐾𝑎2 + 𝐾𝑎1 𝐾𝑎2 𝐾𝑎3 𝐻3 𝑂+ 2 𝐾𝑎1 𝛼1 = 𝛼𝐻2 𝐴− = 𝐻3 𝑂+ 3 + 𝐻3 𝑂+ 2 𝐾𝑎1 + 𝐻3 𝑂+ 𝐾𝑎1 𝐾𝑎2 + 𝐾𝑎1 𝐾𝑎2 𝐾𝑎3 Acid-Base Titration 25 ALPHA (𝜶) EXPRESSIONS 𝜶 = fraction of species ▪ For a tripotic weak acid H3A: 𝐻3 𝑂+ 𝐾𝑎1 𝐾𝑎2 𝛼2 = 𝛼𝐻𝐴2− = 𝐻3 𝑂+ 3 + 𝐻3 𝑂+ 2 𝐾𝑎1 + 𝐻3 𝑂+ 𝐾𝑎1 𝐾𝑎2 + 𝐾𝑎1 𝐾𝑎2 𝐾𝑎3 𝐾𝑎1 𝐾𝑎2 𝐾𝑎3 𝛼3 = 𝛼𝐴3− = 𝐻3 𝑂+ 3 + 𝐻3 𝑂+ 2 𝐾𝑎1 + 𝐻3 𝑂+ 𝐾𝑎1 𝐾𝑎2 + 𝐾𝑎1 𝐾𝑎2 𝐾𝑎3 Acid-Base Titration 26 EXAMPLE What is the pH of a solution that is 0.400 M in formic acid and 1.00 M in sodium formate? 𝐻𝐶𝑂𝑂𝐻 + 𝐻2 𝑂 ↔ 𝐻3 𝑂+ + 𝐻𝐶𝑂𝑂− 𝐾𝑎 = 1.80 𝑥 10−4 𝐻𝐶𝑂𝑂− + 𝐻2 𝑂 ↔ 𝐻𝐶𝑂𝑂𝐻 + 𝑂𝐻− 𝐾𝑏 = 5.56 𝑥 10−11 Analytical Chemistry 27 EXAMPLE Because the Ka for formic acid is orders of magnitude larger than the Kb for formate, the solution is acidic, and Ka determines the H3O+ concentration 𝐻3 𝑂+ 𝐻𝐶𝑂𝑂 − 𝐾𝑎 = = 1.80 𝑥 10−4 𝐻𝐶𝑂𝑂𝐻 𝐻𝐶𝑂𝑂 − ≈ 𝑐𝐻𝐶𝑂𝑂− = 1.00 𝑀 𝐻𝐶𝑂𝑂𝐻 ≈ 𝐶𝐻𝐶𝑂𝑂𝐻 = 0.400 𝑀 0.400 𝐻3 𝑂+ = 𝑥1.80 𝑥 10−4 = 7.20 𝑥 10−5 1.00 Analytical Chemistry 28 EXAMPLE Notice that our assumptions that 𝐻3 𝑂+ ≪ 𝑐𝐻𝐶𝑂𝑂𝐻 and that 𝐻3 𝑂+ ≪ 𝑐𝐻𝐶𝑂𝑂− are valid, thus 𝑝𝐻 = −𝑙𝑜𝑔 7.20 𝑥 10−5 = 4.14 Analytical Chemistry 29 EXAMPLE Calculate the hydronium ion concentration in 0.120 M nitrous acid. 𝐻𝑁𝑂2 + 𝐻2 𝑂 ↔ 𝐻3 𝑂+ + 𝑁𝑂2− Ka= 7.1 x 10-4 Analytical Chemistry 30 EXAMPLE 𝐻3 𝑂+ 𝑁𝑂3− 𝐾𝑎 = = 7.1 𝑥 10−4 𝐻𝑁𝑂2 𝐻3 𝑂+ = 𝑁𝑂3− 𝐻𝑁𝑂2 = 0.120 − 𝐻3 𝑂+ 𝐻3 𝑂+ 𝐻3 𝑂+ −4 𝐾𝑎 = = 7.1 𝑥 10 0.120 − 𝐻3 𝑂+ Analytical Chemistry 31 EXAMPLE Assuming 𝐻3 𝑂+ ≪ 0.120 then 𝐻3 𝑂+ 2 𝐾𝑎 = = 7.1 𝑥 10−4 0.120 𝐻3 𝑂+ = 9.2 𝑥 10−3 𝑀 OR Solving for the quadratic equation 𝐻3 𝑂+ = 8.9 𝑥 10−3 𝑀 Analytical Chemistry 32 BUFFERS ▪ resist large changes in pH when an acid or base is added or when the solution is diluted ▪ usually made from a weak acid-conjugate base pair or a weak base-conjugate acid pair ▪ buffer action of buffer consisting of HA and A- Addition of acid: 𝐴− + 𝐻3 𝑂+ ↔ 𝐻𝐴 + 𝐻2 𝑂 Addition of base: 𝐻𝐴 + 𝑂𝐻 − ↔ 𝐴− + 𝐻2 𝑂 the conjugate base reacts with H3O+ added to the buffer; the conjugate acid reacts with OH- added to the buffer solution, thereby resisting a large change in pH Acid-Base Titration 33 BUFFERS pH of buffers consisting of HA and A- ▪ Two competitive equilibria: 𝐻3 𝑂+ 𝐴− 𝐻𝐴 + 𝐻2 𝑂 ↔ 𝐻3 𝑂+ + 𝐴− 𝐾𝑎 = 𝐻𝐴 𝐾𝑤 𝑂𝐻 − 𝐻𝐴 𝐴− + 𝐻2 𝑂 ↔ 𝐻𝐴 + 𝑂𝐻− 𝐾𝑏 = = 𝐾𝑎 𝐴− MBE: 𝐻𝐴 = 𝐶𝐴 − 𝐻3 𝑂+ + 𝑂𝐻− 𝐴− = 𝐶𝑁𝑎𝐴 + 𝐻3 𝑂+ − 𝑂𝐻− Since 𝐻3 𝑂+ and 𝑂𝐻− have very little difference, then both may be neglected in the MBEs. However, if 𝐶𝐻𝐴 ≤ 1000𝐾𝑎 , then either 𝐻3 𝑂+ or 𝑂𝐻 − must be retained in the MBEs. Acid-Base Titration 34 BUFFERS pH of buffers consisting of HA and NaA 𝐻3 𝑂+ 𝐴− 𝐾𝑎 = 𝐻𝐴 Taking the log of both sides of the equation and rearranging gives the Henderson-Hasselbalch Equation 𝐴− 𝑝𝐻 = 𝑝𝐾𝑎 + 𝑙𝑜𝑔 𝐻𝐴 Acid-Base Titration 35 BUFFERS pH of buffers consisting of B and BH+ 𝐵𝐻+ 𝑂𝐻− 𝐵 + 𝐻2 𝑂 ↔ 𝐵𝐻+ + 𝑂𝐻− 𝐾𝑏 = 𝐵 𝐵𝐻+ 𝑝𝑂𝐻 = 𝑝𝐾𝑏 + 𝑙𝑜𝑔 𝐵 Acid-Base Titration 36 BUFFERS Buffer Capacity ▪ the number of moles of a strong acid or a strong base that causes 1.00 L of the buffer to undergo a 1.00- unit change in pH ▪ the higher the concentration of the two buffer components, the greater is the buffer capacity Acid-Base Titration 37 BUFFERS Buffer Capacity ▪ the closer the ratio of [HA] to [A-], the greater is the buffer capacity, maximum when [HA] = [A-] 𝑝𝐻 = 𝑝𝐾𝑎 ▪ the pKa of the acid chosen must lie within ± 1 unit of the desired pH in order for the buffer to have a reasonable capacity Acid-Base Titration 38 pH OF AMPHIPROTIC SALTS 𝐻3 𝑂+ 𝐴2− 𝐻𝐴− + 𝐻2 𝑂 ↔ 𝐴2− + 𝐻3 𝑂+ 𝐾𝑎2 = 𝐻𝐴− 𝐾𝑤 𝐻2 𝐴 𝑂𝐻− 𝐻𝐴− + 𝐻2 𝑂 ↔ 𝐻2 𝐴 + 𝑂𝐻− 𝐾𝑏2 = = 𝐾𝑎1 𝐻𝐴− + 𝐾𝑎2 𝐶𝑁𝑎𝐻𝐴 + 𝐾𝑤 𝐶𝑁𝑎𝐻𝐴 𝐻3 𝑂 = ≅ 𝐾𝑎1 𝐾𝑎2 𝑖𝑓 ≫ 1 𝑎𝑛𝑑 𝐶𝑁𝑎𝐻𝐴 𝐾𝑎1 1+ 𝐾𝑎1 𝐾𝑎2 𝐶𝑁𝑎𝐻𝐴 ≫ 𝐾𝑤 Acid-Base Titration 39 EXAMPLE Calculate the pH change that takes place when a 100- mL portion of 0.0500 M NaOH is added to 400 mL of the buffer solution that is 0.200 M in NH3 and 0.300 M in NH4Cl. The initial pH of the buffer solution is 9.07. 𝑁𝐻4+ + 𝐻2 𝑂 ↔ 𝑁𝐻3 + 𝐻3 𝑂+ 𝐾𝑎 = 5.70 𝑥 10−10 𝑁𝐻3 + 𝐻2 𝑂 ↔ 𝑁𝐻4+ + 𝑂𝐻− 𝐾𝑏 = 1.75 𝑥 10−5 Analytical Chemistry 40 EXAMPLE Adding NaOH converts part of the NH4+ in the buffer to NH3 𝑁𝐻4+ + 𝑂𝐻 − ↔ 𝑁𝐻3 + 𝐻2 𝑂 The analytical concentrations of NH3 and NH4Cl then become 0400 𝑥 0.20 + 100 𝑥 0.0500 85.0 𝑐𝑁𝐻3 = = = 0.170 𝑀 500 500 0400 𝑥 0.30 − 100 𝑥 0.0500 85.0 𝑐𝑁𝐻4𝐶𝑙 = = = 0.230 𝑀 500 500 Analytical Chemistry 41 EXAMPLE When substituted into the acid dissociation-constant expression for NH4+, these values yield 𝐻3 𝑂+ 𝑁𝐻3 𝐾𝑎 = + = 5.70 𝑥 10−10 𝑁𝐻4 𝐻3 𝑂+ 0.170 = 5.70 𝑥 10−10 0.230 0.230 𝐻3 𝑂+ = 5.70 𝑥 10−10 𝑥 = 7.71 𝑥 10−10 𝑀 0.170 𝑝𝐻 = − log 7.71 𝑥 10−10 = 9.11 ∆𝑝𝐻 = 9.11 − 9.07 = 0.04 𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑒 Analytical Chemistry 42 MIXTURE OF BASES Two Separate Titration Method ▪ one titration uses phenolphthalein (phth, pH 8.0 - 9.6) 𝑉𝑝ℎ𝑡ℎ = 𝑉𝐻𝑐𝑙 required to reach eq. pt. ▪ another titration uses bromocresol green (bcg, pH 3.8 - 5.3) or methyl orange (mo, pH 3.1 – 4.4) 𝑉𝑏𝑐𝑔 = 𝑉𝑚𝑜 = 𝑉𝐻𝑐𝑙 required to reach eq. pt. Acid-Base Titration 43 MIXTURE OF BASES Double Indicator Method ▪ one titration using 2 indicators ▪ phth is added at the start of titration 𝑉1 = 𝑉𝐻𝑐𝑙 required to reach eq. pt. ▪ bcg or mo is added when phth changes color 𝑉2 = 𝑎𝑑𝑑𝑖𝑡𝑖𝑜𝑛𝑎𝑙 𝑉𝐻𝑐𝑙 required to reach the 2nd eq. pt. Acid-Base Titration 44 Kjeldahl Method Determination of Organic Nitrogen Sample bound nitrogen Concentrated H2SO4 𝑁𝐻4 + Distilled into excess strong acid 𝑁𝐻3 𝑙𝑖𝑏𝑒𝑟𝑎𝑡𝑒𝑑 Excess acid is back-titrated with standard base Acid-Base Titration 45 Protein Content in Sample % 𝑝𝑟𝑜𝑡𝑒𝑖𝑛 = %N x factor 6.25 for meats 6.38 for dairy products 5.70 for cereals Acid-Base Titration 46 TITRATION CURVE ▪ before equivalence point, determine amount of unreacted analyte ▪ at equivalence point, use Ksp ▪ beyond equivalence point, determine amount of excess titrant Precipitation Titration 47 TITRATION CURVE Effect of Concentration ▪ A: 50.00 mL of 0.05000 M NaCl titrated with 0.1000 M AgNO3 ▪ B: 50.00 mL of 0.00500 M NaCl titrated with 0.01000 M AgNO3 (Skoog, 2014) Precipitation Titration 48 TITRATION CURVE Effect of Concentration ▪ the greater the concentration of analyte and titrant, the ∆𝑝𝐴𝑔 greater , steeper slope at ∆𝑉 equivalence point region (sharper endpoint) (Skoog, 2014) Precipitation Titration 49 TITRATION CURVE Reaction Completeness ▪ The smaller the value of Ksp, the greater K and the more complete is the reaction, the ∆𝑝𝐴𝑔 greater , steeper slope ∆𝑉 at equivalence point region (sharper endpoint) (Skoog, 2014) Precipitation Titration 50 INDICATORS Precipitation Titration Involving Silver ▪ Mohr Method: formation of a colored precipitate ▪ Volhard Method: formation of a colored complex ▪ Fajans’ Method: adsorption of a colored organic compound (weak acid or base); for precipitates passing through the colloidal state Precipitation Titration 51 MOHR METHOD Reaction 𝐴𝑔+ + 𝑋 − ↔ 𝐴𝑔𝑋(𝑠) Endpoint 2𝐴𝑔+ + 𝐶𝑟𝑂42− ↔ 𝐴𝑔2 𝐶𝑟𝑂4(𝑠) (𝑏𝑟𝑖𝑐𝑘 𝑟𝑒𝑑) Application Direct: Cl-, Br-, CN- with Ag+ as titrant Indirect: Ag+ + Cl- (measured excess); back titrate Cl- with Ag+ Limitations pH range of 6-10 Acidic solution: 2𝐶𝑟𝑂42− + 2𝐻 + ↔ 𝐶𝑟2 𝑂72− + 𝐻2 𝑂 Basic solution: 2𝐴𝑔+ + 𝑂𝐻 − ↔ 2𝐴𝑔𝑂𝐻(𝑠) ↔ 𝐴𝑔2 𝑂 + 𝐻2 𝑂 Precipitation Titration 52 VOLHARD METHOD Reaction 𝐴𝑔+ + 𝑆𝐶𝑁 − ↔ 𝐴𝑔𝑆𝐶𝑁(𝑠) Endpoint 𝐹𝑒 3+ + 𝑆𝐶𝑁 − ↔ 𝐹𝑒𝑆𝐶𝑁 2+ (𝑏𝑙𝑜𝑜𝑑𝑦 𝑟𝑒𝑑) Application Direct: Ag+ with SCN- as titrant Indirect: halide ions X + Ag+ (measured excess); back-titrate Ag+ with SCN- In the indirect analysis of Cl-, AgCl must be filtered first before back-titration to prevent low Cl- analysis because AgSCN is less soluble than AgCl: 𝐴𝑔𝐶𝑙 + 𝑆𝐶𝑁 − ↔ 𝐴𝑔𝑆𝐶𝑁(𝑠) + 𝐶𝑙 − Limitations pH range must be acidic Basic solution: 𝐹𝑒 3+ + 𝐻2 𝑂 ↔ 𝐹𝑒 𝑂𝐻 3(𝑠) + 3𝐻 + Precipitation Titration 53 FAJANS METHOD ▪ uses an adsorption indicator, an organic compound that adsorbs onto or desorbs from the surface of the solid in a precipitation titration ▪ the adsorption or desorption occurs near the equivalence point and results not only in a color change but also in the transfer of color from the solution to the solid or vice versa Precipitation Titration 54 DEFINITION Lewis acid ▪ electron pair acceptor (e.g. Ca2+, Mg2+, Zn2+) Lewis base ▪ electron pair donor (ligand); must have at least a lone pair of electrons, (e.g. NH3) Complexometric Titration 55 DEFINITION Chelate ▪ a cyclic complex formed when a cation is bonded by two or more donor groups contained in a single ligand https://en.wikipedia.org/wiki/Chelation Dentate ▪ means having tooth-like projections Complexometric Titration 56 DEFINITION Coordination Number ▪ number of covalent bonds that a cation tends to form with electron donors Complexometric Titration 57 TITRATION CURVE Titration of 60.0 mL of a solution that is 0.020 M in metal M with A: a 0.020 M solution of the tetradentate ligand D to give MD as the product B: a 0.040M solution of the bidentate ligand B to give MB2 C: a 0.080 M solution of the unidentate ligand A to give MA4 The overall formation constant for each (Skoog 2014) product is 1020 Complexometric Titration 58 TITRATION CURVE Multidentate ligands are preferred over unidentate ligands because they: ▪ generally react more completely with cations (sharper end points) ▪ ordinarily react with metal ions in a single step process (sharper end points) (Skoog 2014) Complexometric Titration 59 EDTA Ethylenediaminetetraacetic acid (EDTA) (Skoog 2014) 𝐻4 𝑌 + 𝐻2 𝑂 ↔ 𝐻3 𝑌 − + 𝐻3 𝑂+ 𝐾𝑎1 𝐻3 𝑌 − + 𝐻2 𝑂 ↔ 𝐻2 𝑌 2− + 𝐻3 𝑂+ 𝐾𝑎2 𝐻2 𝑌 2− + 𝐻2 𝑂 ↔ 𝐻𝑌 3− + 𝐻3 𝑂+ 𝐾𝑎3 𝐻𝑌 3− + 𝐻2 𝑂 ↔ 𝑌 4− + 𝐻3 𝑂+ 𝐾𝑎4 Complexometric Titration 60 EDTA 𝑌 4− 𝑌 4− 𝛼4 = = 𝐶𝑦 𝐻4 𝑌 + 𝐻3 𝑌 − + 𝐻2 𝑌 2− + 𝐻𝑌 3− + 𝑌 4− 𝐾𝑎1 𝐾𝑎2 𝐾𝑎3 𝐾𝑎4 𝛼4 = + 4 𝐻 + 𝐻 + 3 𝐾𝑎1 + 𝐻 + 2 𝐾𝑎1 𝐾𝑎2 + 𝐻 + 𝐾𝑎1 𝐾𝑎2 𝐾𝑎3 + 𝐾𝑎1 𝐾𝑎2 𝐾𝑎3 𝐾𝑎4 𝛼4 , the fraction of EDTA in the Y4- form, increases with increasing pH EDTA titrations are done in basic pH to ensure enough Y4-, which is the form needed in titration Complexometric Titration 61 FORMATION CONSTANTS 𝑀𝑌 (𝑛−4) absolute 𝑀𝑛+ + 𝑌 4− ↔ 𝑀𝑌 (𝑛−4) 𝐾𝑓 = 𝑛+ 4− = 𝐾𝑎𝑏𝑠 formation 𝑀 𝑌 constant 𝑀𝑌 (𝑛−4) since 𝑌 4− = 𝛼4 𝐶𝑌 , then 𝐾𝑓 = 𝑀𝑛+ 𝛼4 𝐶𝑌 𝑀𝑌 (𝑛−4) effective ′ 𝑤ℎ𝑒𝑟𝑒 𝛼4 𝐾𝑓 = = 𝐾𝑒𝑓𝑓 = 𝐾𝑓 formation 𝑀𝑛+ 𝛼4 constant Whereas Kf or Kabs is constant, Keff or Kf’ depends on pH Complexometric Titration 62 EDTA TITRATION Indicators ▪ also form chelates with metal ions (e.g. Eriochrome Black T (EBT) and Calmagite → 𝐻2 𝐼𝑛− ) ▪ EBT decomposes slowly with standing ▪ order of stability of complexes 𝐶𝑎𝑌 2− > 𝑀𝑔𝑌 2− > 𝑀𝑔𝐼𝑛− > 𝐶𝑎𝐼𝑛− Complexometric Titration 63 EDTA TITRATION CURVE Effect of pH Influence of pH on the titration of 0.0100 M Ca2+ with 0.0100 M EDTA As pH increases, 𝛼 increases and ∆𝑝𝐶𝑎 increases (steeper slope) ∆𝑉 (Skoog 2014) at the equivalence point Complexometric Titration 64 EDTA TITRATION CURVE Effect of completeness of reaction Titration curves for 50.0 mL of 0.0100 M solutions of various cations at pH 6.0. (Skoog 2014) As Kf increases, the reaction becomes more complete ∆𝑝𝑀 increases (steeper slope) ∆𝑉 Complexometric Titration 65 EDTA TITRATION CURVE Minimum pH needed for the titration of various cations with EDTA (Skoog 2014) Complexometric Titration 66 AUXILIARY COMPLEXING AGENTS ▪ are added to prevent precipitation of the hydrous oxides (e.g. NH3) 𝑍𝑛2+ + 𝑌 4− ↔ 𝑍𝑛𝑌 2− 𝑍𝑛𝑌 2− 𝑍𝑛𝑌 2− 𝐾𝑓 = 2+ 4− = 𝑍𝑛 𝑌 𝛼𝑍𝑛 𝐶𝑍𝑛 𝛼𝑌 𝐶𝑌 𝑍𝑛𝑌 2− 𝛼𝑍𝑛 𝛼𝑌 𝐾𝑓 = = 𝐾𝑒𝑓𝑓 = 𝐾 𝐶𝑍𝑛 𝐶𝑌 𝑍𝑛2+ 𝛼𝑍𝑛 = 𝑍𝑛2+ + 𝑍𝑛 𝑁𝐻3 2+ + 𝑍𝑛 𝑁𝐻3 2+ 2 + 𝑍𝑛 𝑁𝐻3 2+ 3 + 𝑍𝑛 𝑁𝐻3 2+ 4 Complexometric Titration 67 AUXILIARY COMPLEXING AGENTS Effect of completeness of reaction As NH3 concentration ∆𝑝𝑀 increases decreases ∆𝑉 (less steep slope) at the equivalence point region (Skoog 2014) Complexometric Titration 68 LIEBIG TITRATION 𝐵𝑎𝑠𝑖𝑠: 2𝐶𝑁 − + 𝐴𝑔+ ↔ 𝐴𝑔(𝐶𝑁)− 2 𝐾~1021 Endpoint: appearance of turbidity due to the precipitation of silver cyanide + − 𝐴𝑔 + 𝐴𝑔(𝐶𝑁)2 ↔ 2𝐴𝑔𝐶𝑁(𝑠) + − 𝑜𝑟 𝐴𝑔 + 𝐴𝑔(𝐶𝑁)2 ↔ 𝐴𝑔 𝐴𝑔 𝐶𝑁 2 (𝑠) Deniges modification: iodide ion is added since AgI is less soluble than silver cyanide NH3 is added Ag(NH3)2+ forms which retards the precipitation of AgI until very near the equiv. pt. Complexometric Titration 69 EXAMPLE Determine the pH of the resulting solution of 50.00 mL of 0.0500 M HCl titrated with various amounts of 0.1000 M NaOH at 25°C. (a) 10mL; (b)25.10 mL Analytical Chemistry 70 EXAMPLE Before any base is added, the solution is 0.0500 M in 𝐻3 𝑂+ , and 𝑝𝐻 = −𝑙𝑜𝑔 𝐻3 𝑂+ = − log 0.0500 = 1.30 𝑛𝑜. 𝑚𝑚𝑜𝑙 𝐻𝐶𝑙 𝑟𝑒𝑚𝑎𝑖𝑛𝑖𝑛𝑔 𝑎𝑓𝑡𝑒𝑟 𝑎𝑑𝑑𝑖𝑜𝑛 𝑜𝑓 𝑁𝑎𝑂𝐻 𝑐𝐻𝐶𝑙 = 𝑡𝑜𝑡𝑎𝑙 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑛𝑜. 𝑚𝑚𝑜𝑙 𝐻𝐶𝑙 − 𝑚𝑚𝑜𝑙 𝑁𝑎𝑂𝐻 𝑎𝑑𝑑𝑒𝑑 = 𝑡𝑜𝑡𝑎𝑙 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 50.00 𝑚𝐿 𝑥 0.0500 𝑀 − 10.00 𝑚𝐿 𝑥 0.1000 𝑀 = 50 + 10 𝑚𝐿 = 2.50 𝑥 10−2 𝑀 Analytical Chemistry 71 EXAMPLE 𝑝𝐻 = −𝑙𝑜𝑔 𝐻3 𝑂+ = − log 2.50 𝑥 10−2 𝑀 = 1.60 𝑛𝑜. 𝑚𝑚𝑜𝑙 𝑁𝑎𝑂𝐻 𝑎𝑑𝑑𝑒𝑑 − 𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑛𝑜. 𝑜𝑓 𝑚𝑚𝑜𝑙𝑒𝑠 𝐻𝐶𝑙 𝑐𝑁𝑎𝑂𝐻 = 𝑡𝑜𝑡𝑎𝑙 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 25.10 𝑚𝐿 𝑥 0.100 𝑀 − 50.00 𝑚𝐿 𝑥 0.050 𝑀 = 50 + 25.10 𝑚𝐿 = 1.33 𝑥 10−4 𝑀 𝑝𝑂𝐻 = −𝑙𝑜𝑔 𝑂𝐻 − = − log 1.33 𝑥 10−4 𝑀 = 3.88 𝑝𝐻 = 14.00 − 𝑝𝑂𝐻 = 10.12 Analytical Chemistry 72 REFERENCES Skoog, Douglas A, Donald M West, and Stanley R Crouch. 2014. Fundamentals Of Analytical Chemistry 9E. Australia: Cengage Learning®. Valera, Florenda. n.d. "Intro And Review Of Basic Concepts In Anal Chem". Presentation, University of the Philippines - Diliman. Analytical Chemistry 73

Use Quizgecko on...
Browser
Browser