Chapter 5&6

3,564 views

Published on

analytical Chemistry just the basic one. consists of compleximetric titration and the principle titration

0 Comments
4 Likes
Statistics
Notes
  • Be the first to comment

No Downloads
Views
Total views
3,564
On SlideShare
0
From Embeds
0
Number of Embeds
3
Actions
Shares
0
Downloads
41
Comments
0
Likes
4
Embeds 0
No embeds

No notes for slide
  • ???
  • ???
  • pembolehubah
  • Guna K a
  • MgY 2- = Mg complex EDTA
  • K f =formation constant
  • Chapter 5&6

    1. 1. PRINCIPLES OF NEUTRALIZATION TITRATIONS Buffer solution Calculating pH in titrations of strong acids and strong bases Calculating pH in weak acid (or base) titrationsCOMPLEX-FORMATION TITRATIONS Indicators for EDTA titrations Some applications in human life
    2. 2.  Standard Solutions - strong acids or strong bases because they will react completely. › Acids: hydrochloric (HCl), perchloric (HClO 4), and sulfuric (H2SO4) – also very hazardous. › Bases: sodium hydroxide (NaOH), potassium hydroxide (KOH) Variables: temperature, ionic strength of medium and presence of organic solvents or colloidal particles
    3. 3.  Acid/Base Indicators: a weak organic acid or weak Indicators organic base whose undissociated form differs in color from its conjugate form (In would be indicator). HIn + H2O  In- + H3O+ or In + H2O  HIn+ + OH- (acid color) (base color) (base color) (acid color) Ka = [H3O+][In-] [HIn] [H3O+] = Ka[HIn] [In-]
    4. 4. › HIn pure acid color: [HIn]/[In-] ≥ 10› HIn pure base color: [HIn]/[In-] ≤ 0.1 ~The ratios change from indicator to indicator~› Substitute the ratios into the rearranged K a: [H3O+] = 10 Ka (acid color) [H3O+] = 0.1Ka (base color)› To obtain the indicator pH range,  acid color pH = -log (10 Ka) = pKa + 1  base color pH = -log (0.1 Ka) = pKa – 1
    5. 5. Commonly Used Indicators Indicator pH Range Acid Base Thymol Blue 1.2-2.8 red yellow Thymol blue 8.0-9.6 yellow blue Methyl yellow 2.9-4.0 red yellow Methyl orange 3.1-4.4 red orange Bromcresol green 4.0-5.6 yellow blue Methyl red 4.4-6.2 red yellow Bromcresol purple 5.2-6.8 yellow purple Bromothymol Blue 6.2-7.8 yellow blue Phenol red 6.4-8.0 yellow red Cresol purple 7.6-9.2 yellow purple Phenolphthalein 8.0-10.0 colorless red Thymolphthalein 9.4-10.6 colorless blue Alizarin yellow GG 10.0-12.0 colorless yellow
    6. 6.  A titration curve is constructed by plotting pH of the solution during titration as ordinates and the amount of acid or base added. These curves are useful to indicate equivalence point graphically. graphically The nature of titration curve depends on the ionization constants of acid and base employed in titration i.e., their strength. The principles of acid–base equilibria are important for the construction and interpretation of titration curves in neutralization titrations.
    7. 7.  2 general types of titration curves are encountered in titrimetric methods. First type called a sigmoidal curve: important observations curve are confined to a small region (± 0.1 to ± 0.5 mL) surrounding the equivalence point. Second type called a linear- segment curve, measurement curve are made on both side and away from the equivalence point.
    8. 8.  A buffer is a mixture of a weak acid and its conjugate base or a weak base and its conjugate acid that resists change in pH. A solution containing a weak acid, HA and its conjugate base, A-, may be acidic, base or neutral depending upon the position of two competitive equilibria: HA + H2O  H3O+ + A- @ A- + H2O  OH- + HA Ka = [H3O+][A-] Kb = [OH-][HA] [HA] [A-] If the first equilibrium lies father to the right than the second, the solution is acidic. If the second equilibrium is more favorable, the solution is basic.
    9. 9.  In principle, the calculations work but there are uncertainties in numerical values of dissociation constants & simplifications used in calculations. How to Prepare/Get: › Making up a solution of approximately the desired pH and then adjust by adding acid or conjugate base until the required pH is indicated by a pH meter › Empirically derived recipes are available in chemical handbooks and reference works › Biological supply houses offer a variety of such buffers
    10. 10. The effect of dilution The pH of a buffer solution is remains essentially independent of dilution until the concentrations of the species are decreased to the point so that we cannot assume that the differences between the hydronium and hydroxide ion concentrations is negligible when calculating the concentration of the species.The effect of Added Acids and Bases buffers are resistant to pH change after addition of small amounts of strong acids or bases
    11. 11. There are 4 distinctly different types of calculations to derive atitration curve for a weak acid @ weak base.1.At the beginning: pH is calculated from the concentration of beginningthat solute and its dissociation constant.2.After various increments of titrant has been added: pH is addedcalculated by the analytical concentrations of the conjugate baseor acid and the residual concentrations of the weak acid or base3.At the equivalence point: the pH is calculated from the pointconcentration of the conjugate of the weak acid or base ~ a salt4.Beyond the equivalence point: pH is determined by the pointconcentration of the excess titrant
    12. 12. Determine the pH for the titration of 50 mL of 0.10 M acetic acid afteradding 0.00, 10.00, 50.00, and 50.10 mL of 0.10 M sodium hydroxide HOAc + H2O  H3O+ + OAc- OAc- + H2O  HOAc + OH- Ka = 1.75 x 10 -5Initial pH: pH Ka = [H3O+]2 / cHOAc [H3O+] = 1.32 x 10-3 pH = -log(1.32 x 10-3) = 2.88pH after titrant has been added (5.00 mL NaOH): NaOH *the buffer solution now has NaOAc & HOAc* cHOAc = mol original acid – mol base added total volume
    13. 13. cHOAc = (50 x 0.10) – (10.00 x 0.10) = 0.067M 60 cNaOAc = mol base added total volume cNaOAc = (10.00 x 0.10) = 0.017M 60*we can then substitute these concentrations into the dissociation- constant expression for acetic acid* [H3O+] = Ka x [HOAc] [NaOAc] [H3O+] = 1.75 x 10-5 x [0.067] [0.017] [H3O+] = 7.00 x 10-5 pH = -log(7.00 x 10-5) = 4.16
    14. 14.  Equivalence Point (50.00 mL NaOH): *all the acetic acid has been converted to sodium acetate* OAc- + H2O  HOAc + OH- [HOAc] = [OH-] In the present sample, the NaOAc concentration is 0.05M. Thus : [OAc-]= 0.05 M *we can substitute this in to the base-dissociation constant (Kb) for OAc-* Kb = [OH-][HOAc] = Kw [OAc-] Ka [OH-]2 = 1.00 x 10-14 0.05 1.75 x 10-5
    15. 15.  Beyond the Equivalence Point (50.10 mL NaOH):* the excess base and acetate ion are sources of the hydroxide ion, but the acetate ion concentration is so small it is negligible* [OH-] = cNaOH = mol base added – original mol acid total volume [OH-] = (50.10 x 0.100) – (50.00 x 0.100) 100.10 [OH-] = 1.00 x 10-4 pH = 14.00 – (-log(1.00 x 10-4)) = 10.00
    16. 16.  The Effect of Concentration: the change in pH in the Concentration equivalence-point region becomes smaller with lower analyte and reagent concentrations (0.1 M NaOH versus 0.001 M NaOH) The Effect of Reaction Completeness: pH change in the Completeness equivalence-point region becomes smaller as the acid become weaker (the reaction between the acid and the base becomes less complete) Choosing an Indicator: the color change must occur in Indicator the equivalence-point region
    17. 17. What is the pH of a solution that is 0.4 M in formic acid and 1 Min sodium formate? HCOOH + H2O  H3O+ + HCOO- Ka = 1.80 x 10-4 HCOO- + H2O  HCOOH + OH- Kb = Kw/Ka = 5.56 x 10-11  [HCOO-] ≈ cHCOONa = 1 M [HCOOH] ≈ cHCOOH = 0.4 M [H3O+] = (1.80 x 10-4) x (0.400) = 7.20 x 10-5 (1.00) pH = -log(7.20 x 10-5) = 4.14
    18. 18. Calculate the pH of a solution that is 0.2 M in NH3 and 0.3 M inNH4Cl. NH4+ + H2O  NH3 + H3O+ Ka = 5.70 x 10-10 NH3 + H2O  NH4+ + OH- Kb = Kw/Ka = 1.75 x 10-5  [NH4+] ≈ cNH4Cl = 0.3 M [NH3] ≈ cNH3 = 0.2 M [H3O+] = (5.70 x 10-10) x (0.3) = 8.55 x 10-10 (0.2) pH = - log (8.55 x 10-10) = 9.07
    19. 19. o Buffer Capacity - the number of moles of strong acid or strong base that causes one liter of the buffer to change pH by one unit.o Calculate the pH change that takes place when a 100 mL portion of 0.05 M NaOH is added to a 400 mL buffer consisting of 0.2 M NH3 and 0.3 M NH4Cl (see example for “Buffers Formed from a Weak Base and its Conjugate Acid”) An addition of a base converts NH4+ to NH3: NH4+ + OH-  NH3 + H2O The concentration of the NH3 and NH4Cl change: cNH3 = original mol base + mol base added total volume cNH3 = (400 x 0.2) + (100 x 0.05) = 0.170 M 500
    20. 20. o Calculate the pH change that takes place when a 100 mL portion of 0.05 M NaOH is added to a 400 mL buffer consisting of 0.2 M NH3 and 0.3 M NH4Cl (see example for “Buffers Formed from a Weak Base and its Conjugate Acid”) cNH4Cl = original mol acid – mol base added total volume cNH4CL = (400 x 0.30) - (100 x 0.05) = 0.230 M 500 [H3O+] = (5.70 x 10-10) x (0.230) =7.71 x 10-10 (0.170) pH = -log (7.71 x 10-10) = 9.11 ∆ pH = 9.11 – 9.07 = 0.04
    21. 21. › Pre-equivalence: calculate the concentration of the acid Pre-equivalence from is starting concentration and the amount of base that has been added, the concentration of the acid is equal to the concentration of the hydroxide ion and you can calculate pH from the concentration.› Equivalence: the hydronium and hydroxide ions are Equivalence present in equal concentrations› Post-equivalence: the concentration of the excess base Post-equivalence is calculated and the hydroxide ion concentration is assumed to be equal to or a multiple of the analytical concentration, the pH can be calculated from the pOH
    22. 22. Do the calculations needed to generate the hypothetical titrationcurve for the titration of 50 mL of 0.05 M HCl with 0.10 M NaOH › Initial Point: the solution is 0.05 M in H3O+, so Point pH = -log(0.05) = 1.30 › Pre-equivalence Point (after addition of 10 mL reagent) cHCl = mmol remaining (original mmol HCl – mmol NaOH added) total volume (mL) = (50 mL x 0.05 M) – (10 mL x 0.10 M) 50.0 mL + 10.00 mL = 2.5 x 10-2 M pH = -log(2.5 x 10-2) = 1.602
    23. 23. › Equivalence Point – neither HCl nor NaOH is in excess. So the concentration of OH- and H3O+ is equal. [OH-] = [H3O+], pH = 7› Post-equivalence Point (after addition of 25.10 mL reagent) cHCl = mmol NaOH added – original mmol HCl total volume solution = (25.10 mL x 0.10 M) – (50 mL x 0.05 M) 50.0 mL + 25.10 mL = 1.33 x 10-4 M pOH = -log(1.33 x 10-4) = 3.88 pH = 14 – pOH = 10.12
    24. 24. › Pre-equivalence: calculate the concentration of the base from is starting concentration and the amount of acid that has been added, the concentration of the base is equal to the concentration of the hydronium ion and you can calculate pOH from the concentration, and then the pH› Equivalence: the hydronium and hydroxide ions are present Equivalence in equal concentrations, so the pH is 7› Post-equivalence: the concentration of the excess acid is calculated and the hydronium ion concentration is the same as the concentration of the acid, and the pH can be calculated
    25. 25.  Metal ions are Lewis acids - accepting electrons pairs from electron-donating ligands that are Lewis bases. Monodentate ligand: binds to a metal ion through only one ligand atom. Multidentate ligand: attaches to a metal ion through more ligand than one ligand atom, also known as chelating ligand. ligand• Chelate effect - the ability of multidentate ligands to form more stable metal complexes than those formed by similar monodentate ligands
    26. 26.  A titration based on complex formation is called a complexometric titration. titration Following are structures of analytically useful chelating agents:
    27. 27.  EDTA is an abbreviation for ethylenediaminetetraacetic acid, acid a compound that most widely used complexometric titrant. EDTA has 6 potential sites for bonding a metal ion: the 4 carbonyl groups and the 2 amino groups, each of the latter with an unshared pair of electron. Thus, EDTA is a hexadentate ligand.
    28. 28. • EDTA form strong 1 : 1 complexes with many metal ions; the coordination is through the 4 O atoms and 2 N atoms
    29. 29.  The most common technique to detect the end point in EDTA titrations is to use a metal ion indicator. are compounds whose color changes when they bind to a metal ion. Useful indicators must bind metal less strongly than EDTA does. Masking agent: In a direct titration, analyte is titrated with standard EDTA. The analyte is buffered to a pH at which the conditional formation constant for the metal-EDTA complex is large and the color of the free indicator is distinctly different from that of the metal- indicator complex.
    30. 30. There are several methods for the determinationof cations with EDTA.Direct titrationBack titrationDisplacement titration
    31. 31.  The solution containing the metal ion to be determined is buffered to the desired pH and titrated directly with the standard EDTA solution. It may be necessary to prevent precipitation of the hydroxide of the metal (or a basic salt) by the addition of some auxiliary complexing agent, such as tartrate or citrate or agent triethanolamine. At the equivalence point the magnitude of the concentration of the metal ion being determined decreases immediately. This is generally determined by the change in colour of a metal indicator or by amperometric, spectrophotometric, or potentiometric methods.
    32. 32.  Back-titration procedures are used when no suitable indicator is available, when the reaction between analyte and EDTA is slow, or when the analyte forms a precipitate at the pH required for its titration. In such cases, an excess of standard EDTA solution is added until the reaction is judged complete. The excess of the EDTA is back-titrated with a standard metal ion solution; a solution of zinc chloride or sulphate or of magnesium chloride or sulphate is often used for this purpose. The end point is detected with the aid of the metal indicator which responds to the zinc or magnesium ions introduced in the back-titration.
    33. 33.  Displacement titrations may be used for metal ions that do not react (or react unsatisfactorily) with a metal indicator, that are more stable than those of other metals such as magnesium and zinc. An unmeasured excess of a solution containing the Mg @ Zn complex of EDTA is introduced into analyte solution. If the metal cation M2+ forms a more stable complex than Mg or Zn, the following displacement reaction occurs : The amount of magnesium ion set free is equivalent to the cation present and can be titrated with a standard solution of EDTA and a suitable metal indicator.
    34. 34.  Complexometric is widely used in the medical industry because of the microliter-size sample involved. The method is efficient in research related to the biological cell. › Ability to titrate the amount of ions available in a living cell. › Ability to introduce ions into a cell in case of deficiencies.
    35. 35.  Complexometric titration is an efficient method for determining the level of hardness of water. Caused by accumulation of mineral ions, pH of water is increased. Softening of hard water is done by altering the pH of the water reducing the concentration of the metal ions present.

    ×