Volumetric analysis To understand volumetric analysis we must

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 b- Formation of precipitate : Ag+ + Cl H 2 O Ag. Cl c- Formation of slightly ionizable complex : Ag+ + 2 CN Ca+2 + H 2 Y 2 [EDTA] [Ag(CN)2] 2 H+ + Ca. Y 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 Ce+4 + e Fe+3 + e Ce+3 oxidation (loss of es. ) reduction (gain of es)

Acid-Base Acid- base theories 1 - Arrhenius theory : Acid : Is the substance which ionize to give H+ Base : Is the substance which ionize to give OH eg. HCl eg Na. OH 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 + Acid H 2 O base Eg. NH 3 + base H 2 O acid N. B. Cl + conj. base NH 4+ + conj. acid H 3 O+ conj. acid OH conj. base Water behave as acid or base because it is neutral.

3 - Lewis theory : Acid : Is substance which accept lone pair of electrons eg. BF 3, Al. Cl 3. Base : Is substance which donate lone pair of electrons eg NH 3, amines. Acid-base titration in aq. medium Solns. are classified into : Electrolytes : Which desociate (ionize) and conduct electricity. or Non electrolytes : Which doesn't ionize and doesn't conduct electricity. Dissociation of water H 2 O H+ + OH Dissociation const. Kw = [ H+] [OH ] / [H 2 O] Since H 2 O is weakly dissociated , therfore H 2 O is considered unity. therfore Kw = [H+] [OH ] = 10 14 at 25 oc Kw : it is called ionic product of water. At 25 oc [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.

Hydrogen exponent : p. H = log [H+] i. e. If [H+] = 10 7 p. H = log 10 7 = 7 In acidic side i. e. If [H+] = 10 6 p. H = log 10 6 = 6 In basic side i. e. If [H+] = 10 8 p. H = log 10 8 = 8 i. e. as p. H value inc. [H+] conc. decrease. Therefore acid soln has p. H less than 7 , alkaline soln. has p. H more than 7 and neutral soln. has p. H = p OH = 7 p. H of acid and bases : 1 - p. H of strong acids : Since strong acids are strongly ionized. Therfore p. H = p. Ca where Ca ( conc. of acid) i. e. 0. 1 N HCl p. H = log 0. 1= log 10 1 = 1 2 - p. H of strong bases : Since strong bases are completely ionized. Therfore p OH = p Cb where Cb (conc. of base) p. H = p Kw – p OH i. e. p. H = p Kw – p Cb. i. e. 0. 1 N Na. OH p. H = 14 _ 1 = 13 3 - p. H of weak acids : p. H = 1/2 p. Ca + 1/2 p. Ka 4 - p. H of weak bases : p. H =p. Kw - 1/2 p. Cb - 1/2 p. Kb

5 - p. H of salts : a- Salt of strong acid and strong base eg. Na. Cl Always neutral i. e. p. H = 7 b- Salt of strong acid and weak base eg. NH 4 Cl Always p. H is in the acidic side , calculated from eq. p. H = 1/2 p. Kw - 1/2 p. Kb + 1/2 p. Cs where Kb (dissociation constant of weak base) Cs (conc. of salt) c- Salt of weak acid and strong base eg. Na Ac Always p. H is in the alkaline side, calculated from eq. : p. H = 1/2 p. Kw + 1/2 p. Ka - 1/2 p. Cs where Ka (dissociation constant of weak acid) Cs (conc. of salt) d- Salt of weak acid and weak base eg. NH 4 Ac p. H is calculated from eq. : p. H = 1/2 p. Kw + 1/2 p. Ka - 1/2 p. Kb Buffer solutions Def : They are solns which resist changes in p. H 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 p. H of this buffer is calculated from the eq. : p. H = p. Ka + log salt / acid

Type 2 - weak base and its salt eg. NH 4 OH and NH 4 Cl p. H of this buffer is calculated from the eq. : p. H = p. Kw - p. Kb - log salt/base log salt/acid or log salt/base is called buffer ratio if [salt] = [acid] therefore p. H =p. Ka Examples 1 Calculate the p. H of a buffer soln. containing 0. 1 M acetic acid and 0. 1 M sodium acetate p. Ka =4. 76 soln. p. H = p. Ka + log salt / acid p. H = 4. 76 + log 0. 1 / 0. 1 = 4. 76 Neutralization indicators Color indicators: Substance which change their color with change in p. H are used as neutralization indicators. eg. phenol phthalein"Ph. Ph" (one color indicator), methyl orange"M. O" (2 color indicator) eg. Ph. =8 10 M. O. =3. 3 4. 4 M. R. = 4 6 N. B. the indicator is chosen according to p. H of the product.

Neutralization titration curves For neutralization reaction titration. The titration curve is plot of p. H versus the mls of titrant. Types of neutralization curves: 1 - Strong acid -strong base titration : eg. HCl and Na. OH we have sample of 100 ml HCl and titrate against Na. OH. a. Before the titration p. H is due to the sample i. e HCl (strong acid) therfore p. H = p. Ca b At the equivalent point : HCl +Na. OH Na. Cl + H 2 O p. H is due to Na. Cl i. e. p. H= 7 (salt of strong acid and strong base) c After the equivalent point : p. H is due to excess titrant i. e. Na. OH (strong base) p. H = p. Kw p. Cb N. B. The p. H rises slowly till 99. 9 % of acid is titrated by adding 0. 1 ml Na. OH p. H rises from 4 to 7 then another 0. 1 ml after end point p. H rises from 7 to 10 i. e. at e. p. p. H rises from 4 to 10. So we can use M. O. (3. 3 4. 4) M. R. (4 6) Ph. (8 10) indicators