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Volumetric analysis

Volumetric analysis To understand volumetric analysis, we must understand the types of reaction that happen in it. Types of reactions used in volumetric analysis : I- Ionic combination reactions:- - The reaction goes to completion due to formation of slightly ionizable or slightly

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Volumetric analysis

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  1. Volumetric analysis To understand volumetric analysis, we must understand the types of reaction that happen in it. Types of reactions used in volumetric analysis : I- Ionic combination reactions:- - The reaction goes to completion due to formation of slightly ionizable or slightly insoluble products. a- Neutralization reaction : In which acid reacts with base to form slightly ionized water. H+ + OH- H2O b- Formation of precipitate : Ag+ + Cl- AgCl  c- Formation of slightly ionizable complex : Ag+ + 2 CN- [Ag(CN)2]- Ca+2 + H2Y-2 [EDTA] 2H+ + CaY-2 [Ca-EDTA complex] • II- Electron transfer reactions : • In which electron transfer from one reactant to another. It is called • (oxidation -reduction reactions) Ce+4 + Fe+2 Ce+3 + Fe+3 i.e. Fe+2 Fe+3 + e oxidation (loss of es.) Ce+4 + e Ce+3 reduction (gain of es)

  2. Acid-Base Acid- base theories 1- Arrhenius theory :- Acid : Is the substance which ionize to give H+ eg. HCl Base : Is the substance which ionize to give OH- eg NaOH 2- Bronsted - Lowry theory :- Acid : Is the substance which donate proton. Base : Is the substance which accept proton. Every acid has a conjugate base and the base has conjugate acid. The stronger the acid the weaker its conjugate base and vice versa. Eg. HCl + H2O Cl- + H3O+ Acid base conj.base conj.acid Eg. NH3 + H2O NH4+ + OH- base acid conj.acid conj.base N.B. Water behave as acid or base because it is neutral.

  3. 3- Lewis theory :- Acid : Is substance which accept lone pair of electrons eg. BF3, AlCl3. Base : Is substance which donate lone pair of electrons eg NH3, amines. Acid-base titration in aq. medium Solns. are classified into :- Electrolytes: Which desociate (ionize) and conduct electricity. orNon electrolytes : Which doesn't ionize and doesn't conduct electricity. Dissociation of water H2O H+ + OH- Dissociation const. Kw = [ H+] [OH-] / [H2O] - Since H2O is weakly dissociated , therfore H­2O is considered unity. therfore Kw = [H+] [OH-] = 10 -14 at 25oc Kw : it is called ionic product of water. At 25oc [H+] =[OH-] = 10-7 If [H+] = [OH-] , therfore soln. is neutral If [H+] > 10-7 eg 10-6, 10-5 , therfore soln. is acidic If [H+] < 10-7 eg 10-8, 10-9 , therfore soln. is alkaline.

  4. Hydrogen exponent : pH pH = -log [H+] i.e. If [H+] = 10-7 pH = - log 10-7 = 7 In acidic side i.e. If [H+] = 10-6 pH = - log 10-6 = 6 In basic side i.e. If [H+] = 10-8 pH = - log 10-8 = 8 i.e. as pH value inc. [H+] conc. decrease. Therefore acid soln has pH less than 7 , alkaline soln. has pH more than 7 and neutral soln. has pH = p OH = 7 pH of acid and bases :- 1- pH of strong acids :- Since strong acids are strongly ionized. Therfore pH = pCa where Ca ( conc. of acid) i.e. 0.1 N HCl pH = - log 0.1= - log 10-1 = 1 2- pH of strong bases :- Since strong bases are completely ionized. Therfore p OH = p Cb where Cb (conc.of base) pH = p Kw – p OH i.e. pH = p Kw – p Cb. i.e. 0.1 N NaOH pH = 14 _ 1 = 13 3- pH of weak acids : pH = 1/2 pCa + 1/2 pKa 4- pH of weak bases :- pH =pKw - 1/2 pCb - 1/2 pKb

  5. 5- pH of salts :- a- Salt of strong acid and strong base eg. NaCl Always neutral i.e. pH = 7 b- Salt of strong acid and weak base eg.NH4Cl Always pH is in the acidic side , calculated from eq. pH = 1/2 pKw - 1/2 pKb + 1/2pCs where Kb (dissociation constant of weak base) Cs (conc. of salt) c- Salt of weak acid and strong base eg. Na Ac Always pH is in the alkaline side, calculated from eq.: pH = 1/2 pKw + 1/2 pKa - 1/2pCs where Ka (dissociation constant of weak acid) Cs (conc. of salt) d- Salt of weak acid and weak base eg. NH4Ac pH is calculated from eq. : pH = 1/2 pKw + 1/2 pKa - 1/2pKb Buffer solutions -Def : They are solns which resist changes in pH upon addition of small amount of acid or base. -They consist of weak acid and its salt or weak base and its salt Type 1- weak acid and its salt eg. HAc and Na Ac pH of this buffer is calculated from the eq.: pH = pKa + log salt / acid

  6. Type 2- weak base and its salt eg. NH4OH and NH4Cl pH of this buffer is calculated from the eq.: pH = pKw - pKb - log salt/base log salt/acid or log salt/base is called buffer ratio if [salt] = [acid] therefore pH =pKa Examples 1- Calculate the pH of a buffer soln. containing 0.1 M acetic acid and 0.1 M sodium acetate pKa =4.76 soln. pH = pKa + log salt / acid pH = 4.76 + log 0.1 / 0.1 = 4.76 Neutralization indicators Color indicators: - Substance which change their color with change in pH are used as neutralization indicators. eg. phenol phthalein"Ph.Ph" (one color indicator), methyl orange"M.O" (2 color indicator) eg. Ph.Ph. =8-10 M.O. =3.3-4.4 M.R. = 4-6 N.B. the indicator is chosen according to pH of the product.

  7. Neutralization titration curves For neutralization reaction titration . The titration curve is plot of pH versus the mls of titrant. Types of neutralization curves: 1- Strong acid -strong base titration :- eg. HCl and NaOH we have sample of 100 ml HCl and titrate against NaOH. • Before the titration pH is due to • the sample i.e HCl (strong acid) • therfore pH = pCa • b- At the equivalent point : HCl +NaOH NaCl + H2O pH is due to NaCl i.e. pH= 7 (salt of strong acid and strong base) c- After the equivalent point : pH is due to excess titrant i.e. NaOH (strong base) pH = pKw-pCb N.B. The pH rises slowly till 99.9 % of acid is titrated by adding 0.1 ml NaOH pH rises from 4 to 7 then another 0.1 ml after end point pH rises from 7 to 10 i.e. at e.p. pH rises from 4 to 10. So we can use M.O.(3.3 - 4.4)- M.R.(4 - 6)- Ph.Ph.(8 - 10) indicators

  8. 2- Weak acid - Strong base titration :- • eg. CH3COOH (pKa = 4.74) and NaOH • We have sample of 100 ml 0.1 N HAc and tirate against 0.1 N NaOH • Before the titration pH is due to • the sample i.e. HAc (weak acid) • therfore pH = 1/2pKa + 1/2 pCa • = 2.37 + 0.5 = 2.87 • b- During titration eg. after adding 40 ml NaOH HAc + NaOH NaAc + H2O HAc will be present ( not completely neutralized) and NaAc i.e. weak acid and its salt, therfore pH is calculated from eq. of acidic buffer pH = pKa + log salt/ acid = 4.74 + log 40/60 = 4.53 c- At the equivalence point (end point) all HAc is neutralized and only NaAc is present , therfore pH is calculated from: pH = 1/2 pKw + 1/2 pKa -1/2 pCs (salt of weak acid and strong base) = 7 + 2.37 -1/2 (- log 100 x 0.1 /200) = 8.72 d- After end point the pH is calculated from excess titrant i.e. NaOH. N.B. The end point is at the alkaline side and abrupt change in the curve is from pH 7 to pH 11. Therfore M.O and M.R. indicator are not suitable. Therfore use Ph.Ph. ind or any ind of pH range on the alkaline side.

  9. 3- Strong acid - weak base titration :- eg. NH4OH (pKb = 4.74)and HCl We have sample 100 ml 0.1 N NH4OH and titrate against 0.1 NHCl a- Before titration pH is due to the sample NH4OH pH is calculated from eq. of weak base. pH = pKw - 1/2 pKb -1/2 pCb = 14 – 2.37 - 0.5 =11.13 b-During titration eg. After adding 90 ml HCl NH4 OH + HCl NH4Cl + H2O NH4OH will be present(not completely neutralized) and NH4Cl i.e. weak baseand its salt.therfore pH is calculated from eq. of basic buffer pH = pKw -pKb - log salt/ base = 14 - 4.74 - log 90/10 = 8.31 c- At the equivalence point (end point) all NH4OH is neutralized and NH4Cl is only present , therfore pH is calculated from : pH = 1/2 pKw- 1/2 pKb + 1/2 pCs( salt of strong acid and weak base) = 7 - 2.37 + 1/2 (- log 100 x 0.1 /200) = 5.13 d-After the end point the pH is calculated from excess titrant i.e. HCl N.B. The end point is at the acidic side = 5.13 .The abrupt change is from pH 6 to 4 , Therfore Ph.Ph. indicator cannot be used. Use M.O. or M.R. or any indicator of pH range at acidic side.

  10. Applications 1- Direct titration methods :- Direct titration is useful for :- A- Strong acid B- Strong base C- Weak acid or base if Ka and Kb not less than 10-7 Determination of acids :- 1- Strong acids can be titrated against strong alkali using Ph.Ph. or M.O. On titrating weak acid only Ph.Ph. is suitable. 2- Acids like benzoic acid , salicylic acid which are not insoluble in water are dissolved in neutral ethanol then add water and titrate against NaOH using Ph.Ph. as indicator. 3- Boric acid: weak acid is a monobasic acid i.e. release 1 H+ , can be titrated against NaOH only after potentiation by adding any poly hydroxy compound eg. glycerol using Ph.Ph. as indicator. Determination of bases :- 1- Strong base can be titrated against strong acids using M.O. or Ph.Ph . For weak bases we use M.O. indicator or any indicator of pH range on the acidic side.

  11. 2- Displacement titration It is used for easily hydrolysable salts :- A- Salt of strong base and weak acid eg. borax, Na2CO3 B- Salt of weak base and strong acid eg. FeCl3, Al2(SO4)3 N.B.Always titrate the strong part of the salt. eg.1. KCN We titrate KOH by standard acid eg. HCl eg.2.Borax Na2B4O7 Borax hydrolyze in water to give: Na2B4O7 + 7 H2O 4H3BO3 + 2 NaOH v.weak acid tit.  2 HCl using M.O titrate  HCl using M.O indicator.(let reading = x) eg.3.Na2CO3 sodium carbonate. Na2CO3HCl NaHCO3 HCl NaCl + H2O +CO2 pH = 8.3 3.8 Ph.Ph. M.O.

  12. Na2CO3 can be determined by titration against HCl by 2 methods : a- Using M.O. as indicator it will give total CO3-2 b- Using Ph.Ph. as indicator it will give 1/2 CO3-2 and considered as half neutralization step. But care that the 1st step to NaHCO3 takes place on 2 separate steps; Na2CO3 + 2HCl 2NaCl + CO2 +H2O Na2CO3 +CO2 + H2O 2 NaHCO3 2Na2CO3 +2HCl 2NaHCO3 + 2 NaCl i.e. For 1/2 neutralization, we must prevent the escape of CO2 by : 1- cooling 2- dilution 3-stirring 4- dipping the nozzle of burrete under the surface of the soln.

  13. 3- Indirect or back (residual titration) - In it we add Known excess of standard to the sample and titrate the excess unreacted standard. Conc. of sample A = Known excess standard - b When do we use back titration ? 1- When sample is volatile .eg. NH3, formic acid. 2- When sample is insoluble eg. ZnO, CaO, CaCO3, BaCO3 3- When reaction require heat of standard solution. 4- When reaction proceed only in presence of excess reagent eg. with lactic acid. Determination of inorganic ammonium salts : NH4Cl + NaOH NaCl + NH3 + H2O Add known excess NaOH , then boil to remove NH3 and titrate excess NaOH using HCl and M.R. as indicator. Determination of nitrogen in organic compounds : (byKjeldahl's method) : Organic cpd Conc.H2SO4 Δ NH3 (NH4)2SO4 -Nitrogen of organic cpd is reduced to NH3 by digestion with conc. H2SO4 in the presence of K2SO4 or Na2SO4 (to raise the boiling point of acid) and CuSO4 or HgO as a catalyst.

  14. -The organic cpd. is oxidised to CO2 and the acid is reduced to SO2 and nitrogen to NH3 which is fixed with excess acid as (NH4)2SO4 then add known excess NaOH and titrate excess unreacted alkali against HCl as before. 4- Other indirect Titrations : Determination of esters :- Esters are hydrolysed by reflux with known excess of NaOH, cool and the excess unreacted NaOH is titrated against standard HCl using Ph.Ph. indicator. Determination of ammonium salt and amino acids (Formol titration) Another method for determination of ammonium salts is formol titration . When formaldehyde is added to the sample , hexamethylene tetramine (hexamine) which is neutral is formed and equivalent amount of acid which can be titrated against NaOH. 4NH4Cl + 6HCHO (CH2)6N4 + 4HCl + 6H2O 4(NH4)2SO4 + 6HCHO (CH2)6N4 + 2H2SO4 + 6H2O N.B. HCHO must be neutralized from any formic acid due to aerial oxidation .

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