Acid/Base Titration Dr. Hisham Ezzat Abdellatef Prof. of Analytical Chemistry Dr. Hisham Ezzat Abdellatef Prof. of Analytical Chemistry Clinical Pharmacy Code: 205

Electrolyte and the theory of electrolytic Dissociation Electrolytes conduct the electric current mineral acid, caustic alkalies and salts non–electrolytes non–conducting solutions cane sugar, glycerin and ethyl acetate. Electrolytes conduct the electric current mineral acid, caustic alkalies and salts non–electrolytes non–conducting solutions cane sugar, glycerin and ethyl acetate.

Acid–Base Titration Pure water is a bad conductor of electricity acid as HCl H + Cl - base as KOH K + OH - salt as Na 2 SO 4 Na + SO 4 2- Molecule Dissociation ions Pure water is a bad conductor of electricity acid as HCl H + Cl - base as KOH K + OH - salt as Na 2 SO 4 Na + SO 4 2- Molecule Dissociation ions Dissolved in water

Strong and weak electrolytes NaCl ⇌ Na + + Cl – K 2 SO 4 ⇌ 2K + + SO 4 2– Na 2 HPO 4 ⇌ 2Na + + H + + PO 4 3 Arrhenius therefore introduced a quantity "a", called "the degree of dissociation“ "a“ = 1 "strong electrolyte“ “a” very far from unity.= weak electrolyte NaCl ⇌ Na + + Cl – K 2 SO 4 ⇌ 2K + + SO 4 2– Na 2 HPO 4 ⇌ 2Na + + H + + PO 4 3 Arrhenius therefore introduced a quantity "a", called "the degree of dissociation“ "a“ = 1 "strong electrolyte“ “a” very far from unity.= weak electrolyte

Law of mass action "The rate of a chemical reaction is proportional to the active masses of the reacting substances." in diluted solution = concentration "The rate of a chemical reaction is proportional to the active masses of the reacting substances." in diluted solution = concentration

Law of mass action V f = [A].[B]. K f V b = [C]. [D]. K b K f [A]. [B] = K b [C]. [D] V f = [A].[B]. K f V b = [C]. [D]. K b K f [A]. [B] = K b [C]. [D]

The dissociation of water H 2 O ⇌ H + + OH – [H + ]. [OH]  = K w ………..(2) "The ionic product of water" H 2 O ⇌ H + + OH – [H + ]. [OH]  = K w ………..(2) "The ionic product of water"

The dissociation of water 25 o C; the value of K w [H + ]. [OH  ] = 10  14 [H + ] 2 = K w = 1 x 10  14 ………(3) 25 o C; the value of K w [H + ]. [OH  ] = 10  14 [H + ] 2 = K w = 1 x 10  14 ………(3)

Hydrogen ion exponent (pH): [H + ] 10 –6, 10 –5 1 x 10  7 10 –8, 10 –9 Acid base Neutral [H + ] 10 –6, 10 –5 1 x 10  7 10 –8, 10 –9 Acid base Neutral

Procedure for Titration pH is defined as equal to the logarithm of the hydrogen ion concentration with a negative sign. pH = –log [H + ] pH is defined as equal to the logarithm of the hydrogen ion concentration with a negative sign. pH = –log [H + ]

The pH scale acid neutral base acid neutral base Stomach juice: pH = 1.0 – 3.0Human blood: pH = 7.3 – 7.5 Lemon juice: pH = 2.2 – 2.4Seawater: pH = 7.8 – 8.3 Vinegar: pH = 2.4 – 3.4Ammonia: pH = 10.5 – 11.5 Carbonated drinks: pH = 2.0 – M Na 2 CO 3 : pH = 11.7 Orange juice: pH = 3.0 – M NaOH: pH = 14.0

Figure : The pH scale and pH values of some common substances

Figure: A pH meter

Figure: Indicator paper being used to measure the pH of a solution

p p pK w = pH + pOH  pH = pK w – pOH or pH = 14 – pOH or pOH = 14 – pH pK w = pH + pOH  pH = pK w – pOH or pH = 14 – pOH or pOH = 14 – pH

Acids and bases: Arrhenius theory Acid [H + ] when dissolved in water Bases, [OH – ] Arrhenius theory Acid [H + ] when dissolved in water Bases, [OH – ]

Acid and base

pH calculations 1. Solution of strong acids and strong bases [H + ] or [OH – ] =concentration 1. Solution of strong acids and strong bases [H + ] or [OH – ] =concentration

Example 1 Calculate the pH value of a solution of a completely ionised 1.0 N solution of acid; or base. ? Solution: [H + ] = 1M pH = –log 1 = 0 (zero) similarly, in a completely ionised 1.0 N solution of base [OH – ] = 1 M pOH = –log 1 = 0 (zero) Calculate the pH value of a solution of a completely ionised 1.0 N solution of acid; or base. ? Solution: [H + ] = 1M pH = –log 1 = 0 (zero) similarly, in a completely ionised 1.0 N solution of base [OH – ] = 1 M pOH = –log 1 = 0 (zero)

Example 2 Calculate the [H + ] and pH of N hydrochloric acid? solution [H + ] = N pH =– log (9.0 X 10 –3 ) = – log 9.0 – log 10 –3 pH = – = 2.05 Calculate the [H + ] and pH of N hydrochloric acid? solution [H + ] = N pH =– log (9.0 X 10 –3 ) = – log 9.0 – log 10 –3 pH = – = 2.05

Example 3 Calculate the pH values of a solution of sodium hydroxide whose [OH – ] is 1.05 x 10 –3 ? solution pOH = – (log log 10 –3 ) pOH = – (0.02 –3 ) = 2.98 pH = 14 – 2.98 = Calculate the pH values of a solution of sodium hydroxide whose [OH – ] is 1.05 x 10 –3 ? solution pOH = – (log log 10 –3 ) pOH = – (0.02 –3 ) = 2.98 pH = 14 – 2.98 = 11.02

Example 4 Calculate the hydrogen ion concentration of a solution of pH 5.3? solution pH = – log [H + ] 5.3 = –log [H + ] [H + ] = 5.01 x 10 –6 M Calculate the hydrogen ion concentration of a solution of pH 5.3? solution pH = – log [H + ] 5.3 = –log [H + ] [H + ] = 5.01 x 10 –6 M

Example 5 Calculate the hydroxyl ion concentration of a solution of pH ? solution pOH = 14 – = 3.25 [OH – ] = the antilog of –3.25 [OH – ] = 5.62 x 10 –4 M Calculate the hydroxyl ion concentration of a solution of pH ? solution pOH = 14 – = 3.25 [OH – ] = the antilog of –3.25 [OH – ] = 5.62 x 10 –4 M

pH calculations 2. Solution of weak acids and bases A) Calculation of pH of solution of weak acids pH = ½ (pK a + pC a ) 2. Solution of weak acids and bases A) Calculation of pH of solution of weak acids pH = ½ (pK a + pC a )

B) Calculation of pH of solution of weak bases: pH = pK w – ½ (pK b + pC b ) B) Calculation of pH of solution of weak bases: pH = pK w – ½ (pK b + pC b )

pH of salts A) Salts of strong acids or bases KCl  K + + Cl - K + + OH - ⇌ KOH Cl - + H + ⇌ HCl The equilibrium between the hydrogen and hydroxyl ions in water H 2 O ⇌ H + + OH – A) Salts of strong acids or bases KCl  K + + Cl - K + + OH - ⇌ KOH Cl - + H + ⇌ HCl The equilibrium between the hydrogen and hydroxyl ions in water H 2 O ⇌ H + + OH –

B) Salts of weak acids or bases (hydrolysis) Salts of weak acids (or bases) react with water to give basic (or acidic) solutions Hydrolysis reverse of neutralization B) Salts of weak acids or bases (hydrolysis) Salts of weak acids (or bases) react with water to give basic (or acidic) solutions Hydrolysis reverse of neutralization

i) Salts of weak acids and strong bases: H 2 O + CH 3 COO  ⇌ CH 3 COOH + OH  pH = ½ (pK w – pC a + pK a ) i) Salts of weak acids and strong bases: H 2 O + CH 3 COO  ⇌ CH 3 COOH + OH  pH = ½ (pK w – pC a + pK a )

ii) Salts of weak bases and strong acids: NH H 2 O ⇌ NH 4 OH + H + pH = ½ (pK w + pC s – pK b ) ii) Salts of weak bases and strong acids: NH H 2 O ⇌ NH 4 OH + H + pH = ½ (pK w + pC s – pK b )

iii) Salts of weak acids and weak bases: pH = ½ pK w + ½ pK a – ½pK b iii) Salts of weak acids and weak bases: pH = ½ pK w + ½ pK a – ½pK b

Buffer solutions Resist changes in pH caused by addition of small amounts of acid or base; or upon dilution.

Types of buffers