Displacement titration as analytical technique
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Displacement titration as analytical technique

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Displacement titration as analytical technique

Displacement titration as analytical technique

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  • 1. Displacement Titration Pabitra Kumar Mani , Assoc. Prof., Ph.D. ACSS, BCKV
  • 2. Volumetric analysis (Titrimetry) (i) Acid-Base and (ii) Displacement titrimetry Titrations of anions of weak acids (Bronsted Bases) with strong acids (Displacement Titrations) On the Bronsted theory the so –called titration of solutions of hydrolysed salts, is merely the titration of a weak Bronsted base(carbonate ion , borate ion, acetate ion) with a strong acid. B+A- + HX ⇋ HA + B+XSalt strong Acid weak Acid Salt B+A- + DOH ⇋ BOH + D + AStrong base weak base CH3COONa + HCl = CH3.COOH + NaCl The weak acetic acid was apparently displaced by the strong HCl acid, and the process was refrred to as displacement titration
  • 3. Salt of weak acid vs agt. a strong acid 100 ml of 0.2 (N) KCN ----- 0.2(N) HCl Initial pH of 0.2 N KCN KCN + H2O = KOH + HCN for HCN, Ka= 7.2 X 10-10, pka=9.2 pH= ½ pKw + ½ pKa + ½log c = 7 + 4.6 + ½log 0.2 =11.45 The pH at e.p is due to 0.1 (N) HCN produced (vol.is diluted so, (N) reduces) KCN+ HCl = HCN + KCl pH = ½ pKa - ½log c = 4.6 + 0.5 = 5.1 When 0.1 ml acid (HCl) is in excess, pH= 4.0 Thus the inflexion region in the titration curve covers the pH range 5-3.7 and the indicator like M.O, BCG, M.R can be used with success for detecting the e.p.
  • 4. Borax (Na2B4O7) can similarly be titrated with HCl Na2B4O7 +2HCl +5H2O = 4 H3BO3+ 2NaCl (0.2N) 0.2 N weak monobasic acid Ka= 6x10-10 The pH at e.p. (pH=5.1) is due to 0.1 M boric acid, i.e., 5.1. Further addition of HCl will cause a sharp decrease of pH and any indicator covering the pH range 3.7-5.1 may be used, Bromo cresol green, methyl orange and methyl red
  • 5. Titration of a carbonate ion with strong acid In case of Na2CO3 can be tritrated in 2 stages according to the equn. Na2CO3 + HCl = NaHCO3 + NaCl........1 NaHCO3 + HCl = H2CO3 + NaCl...........2 At 1st e.p. , pH = ½pK1 + ½pK2 = 8.3 Here indicator PhTh becomes colourless and can be used to detect the e.p. The pH at the 2nd e.p. is due to that of H2CO3 produced in soln. pH = ½pK1 - ½log c = 3.8 c= 0.1 M So, indicator that can be used is M.O; congo red and BPB
  • 6. Salt of Weak base-Strong Base NH4Cl (0.2 N) - NaOH (0.2 N) NH4Cl + NaOH = NH4OH + NaCl The pH at e.p will be the alkaline side, and it can be calculated in the same manner as the pH of a free weak base like NH 4OH At pH due to NH4OH (0.1N) pH = pKw - ½pKb + ½log c = 14-2.4 + 3.8+ ½log (0.1)= 11.1
  • 7. Criteria for using a pH indicator There must be at least 2 units of pH change near the stoichiometric end point for the solution of 0.1 ml of the titrant pH on the either side of the e.p due to ± 0.1 ml addition of the titrant can be obtained from relevant equations and the difference will indicate whether the change is large enough to permit a sharp end point to be determined. If the pH change is satisfactory an indicator should be selected which changes colour at or near the e.p. Generally if pK In-a ±1 value for an indicator forms within the range of pH change at the e.p. the indicator is suitable for the determination of the e.p
  • 8. Other types of pH indicators (i) Mixed indicators: Mixture of 2 or more simple indicators or mixture of an indicator with a suitable dye are often employed for better detection of end point. in pH titration. The purpose of using mixed indicator is to render the colour changes more contrasting or to obatined a sharp colour change in a narrow range of pH. Thus methylene Blue (dye) (red-ox indicator) modifies the yellow to red colour change of Methyl orange (M.O.+ Methylene blue) : green- gray-violet M.O : Yellow to Red Equal parts of neutral red indicator and methylene blue (both 0.1% concn) makes the colour changes more contrasting from green to violet and both these mixed indicators are suitable for titration of very weak base like pyridine agt. a strong acid. The main indicator mixed with a pre or oxillary indicator or two indicators with overlapping pH ranges may help the colour change to take place over a narrower pH change. Thus 1 drop of Sodium Cresol Red with 3 drops of Na-Thymol Blue (both 0.1%) enables a pH of 8.3 to be exactly detected 8.2(pink)—8.3 –8.4 (violet)
  • 9. (ii) Universal pH indicators : Multiple range pH indicators are prepared by mixing a no. of indicators in proper ratio so that final indicator can show different colours over a very long range of pH. Thus a mixture of 0.05 g M.O., 0.15 g M.R, 0.30 g BTB, 0.35 g PhTh in one Litre of 60% ethanol pH Colour 3 4 5 6 Red Orange Red orange Y 7 8 Y-G Greenish blue 9 10 11 B V R-V
  • 10. Precipitation titrations The most important precipitation processes in titrimetric analysis utilise silver nitrate as the reagent (argentimetric processes)
  • 11. Detection of the end point: INDICATORS Indicator forming a coloured compound with a titrant. The Mohr method for determining chloride serves as an example. The chloride is titrated with std silver nitrate soln. A soluble chromate salt is added as the indicator. This produces a yellow soln. When the pptn of the chloride is complete, the first excess of Ag+ reacts with the indicator to precipitate red silver chromate:
  • 12. In actual practice, The indicator concentration is kept at 0.002 to 0.005 M 2CrO42- + 2H+  2HCrO4 Cr2O72- +H2O Titration should be done in alkaline pH,6-9, in acid soln the rn occurs as follows
  • 13. The action of these indicators is due to the fact at the e.p the indicator is adsorbed by the ppt, and during the process of adsorption a change occurs in the indicator which leads to a substances of difft. colour, they have therefore been termed as adsorption indicator The indicator which is a dye, exists in soln as the ionized form, usually an anion, In-. Consider the titration of Cl- with Ag+. Before the equivalence point, Cl- is in excess, and the primary adsorbed layer is Cl-. This repulses the indicator anion, and the more loosely held secondary (counter ) layer of adsorbed ions is cations such as Na+: Beyond the equivalence point, Ag+ is in excess, and the surface of the precipitate becomes positively charged, with the primary layer being Ag +. This will now attract the indicator anion and adsorb it in the counterlayer:
  • 14. Thus the colour is formed of Ag+In- on the ppt surface which is more intense than In- colour in solution. It is interesting to note that Ag+In- comp. as such is not as intensity coloured as the adsorbed colour on the ppt. Ag-eosinate is soluble in water, and has reddish orange colour but adsorbed Ag-eosinate on AgX is intense reddish violet colour. Fajans (1923) first introduced Fluorescein ----( for Cl-) Eosin------------ (Br- substituted) Erythrosin.............(I- substituted) The adsorption indicator should not be strongly adsorbed on the ppt and it must be reversible
  • 15. One important group of colour indicators is derived from 1:10 phenantholine (ortho-phenanthroline) which forms a 3:1 complex with iron(II). The complex known as 'ferroin‘ undergoes a reversible redox reaction accompanied by a distinct colour change
  • 16. Equivalence point is determined by using an indicator (W.Ostald) (pH-indicator) HIn ⇋ H+ + Acidic Indicator(mol.form) InOH In- HIn and In- are differently coloured ionic form ⇋ In+ + OH- InOH and In+ are differently coloured Basic ionic form Indicator(mol.form) ⇋ Tautomeric transformation HIn Hin* ⇋ H+ + In- HIn and In- are differently coloured HIn ⇋ H+ + In[ H ][ In ] kIn-a = + [ HIn] [H ] + = kIn-a pH = - logkIn-a − [ HIn] [ In ] [ HIn] − - log [ In ] − pH = pKIn-a + log [ In ] − [ HIn] Alkaline colour intensity acidic colour intensity
  • 17. [ In ] When [ HIn] = 10, pH= pKIn-a + 1 [ In ] and When [ HIn] = 1/10, pH= pK In-a - 1 − − Operationally, pH= pKIn-a ± 1 [ In ] 10 times concn of HIn i.e., when the color due to [ In ] is dominant − − During titration, if pH at equivalence point, lies in the range pK In-a ± 1, that indicator can be used to detect the e.p. In the particular titration InOH ⇋ In+ + OH- [ In ][ OH ] = [ InOH ] + KIn-b − [ InOH ] ∴ [OH ] = kIn-b [ In ] - [ In ] + -log[OH ] or, pOH = pKIn-b + log10 - [ InOH ] [ In ] pH = 14 - pKIn-b -log10 [ InOH ] + [ In ] + pH = pKIn-a -log10 [ InOH ] + pH + pOH =pKw=14 pH -14 = - pOH HA ⇋ H+ + ApKa+pKb= pKw