BUFFER SOLUTIONS What is a buffer solution? Definition A buffer solution is one which resists changes in pH when small quantities of an acid or an alkali are added to it.

Examples 1 Acetic acid & sodium acetate ( CH3COOH + CH3COONa) 2 Formic acid & potassium formate (HCOOH + HCOOK) 3 Benzoic acid and & sodium benzoate ( C6H5COOH + C6H5COONa) 4 Phosphoric acid & Sodium phosphate ( H3PO4 + Na3PO4) A basic buffer is a mixture of weak base and its salt with a strong acid. 5 ammonium hydroxide & ammonium chloride ( NH4OH + NH4Cl )

100 cm3 0.1M CH3COOH and 0.1M CH3COO- Na+ It tends to resist changes in pH when small amounts of acid or base are added, and their pH is not affected by dilution. It is an ACIDIC BUFFER solution! (i.e. weak acid + conjugate base) CH3COOH + H2O CH3COO- + H3O+ 100 cm3 0.1M CH3COOH and 0.1M CH3COO- Na+ When acid is added, the additional H3O+ would combine with the conjugate base i.e. [H3O+] does not change much, pH remains almost unchanged. When base is added, some H3O+ ions are reacted. Some acid molecules dissociate to produce H3O+ i.e. [H3O+] does not change much, pH remains almost unchanged.

NH3 + H2O NH4+ + OH- 100 cm3 0.1M NH3 and 0.1M NH4Cl It is an BASIC BUFFER solution! (i.e. weak base + conjugate acid) NH3 + H2O NH4+ + OH- When acid is added, some OH- are neutralized 100 cm3 0.1M NH3 and 0.1M NH4Cl i.e. [OH-] does not change much, pH remains almost unchanged. When base is added, additional OH- ions would combine with NH4+ i.e. [OH-] does not change much, pH remains almost unchanged.

pH of buffer is NOT affected by dilution.  WHY? conjugate base acid pKa = [A-(aq)] [HA(aq)] - log pH constant at constant temp. pH of the buffer could be adjusted by varying the ratio of HA and A-. During dilution, both [A-] and [HA] are diluted to the same extent. Thus the ratio [A-]/[HA] is unchanged. Therefore, pH of buffer remains unchanged in dilution!

Acid-Base Eqm (4): Buffer Solutions Acid-Base Eqm (EXP.4): Buffer Solutions 0.01 M HCl Q.: What is the new pH after addition of 0.01M HCl ? (assume no volume change) 100 cm3 H2O [H3O+] = 0.01 100/1000 = 0.10 mol  pH = - log(0.10) = 1  pH would decrease from 7 to 1 after adding 0.01 M HCl !

100 cm3 0.1M CH3COOH and 0.1M CH3COO- Na+ 0.01 M HCl Q.: What is the new pH after addition of 0.01 M HCl ? (assume no volume change) x (0.1+x) 0.1 – x = 1.7610-5 x = 1.7610-5 pH = 4.75 Before addition of HCl: 100 cm3 0.1M CH3COOH and 0.1M CH3COO- Na+ After addition of HCl: extra [H3O+] = 0.01M [CH3COO-] remained = 0.1 – 0.01 = 0.09M (0.09+y) y 0.11 – y = 1.7610-5 y = 2.1510-5 pH = -log (y) = 4.67  pH would decrease from 4.75 to 4.67 after adding 0.01 M HCl!

tends to resist the changes in pH when small amounts of acid is added Note the difference … tends to resist the changes in pH when small amounts of acid is added 100cm3 H2O 100cm3 0.1M CH3COOH and 0.1M CH3COO-Na+ Initial pH 7 4.75 pH after adding 0.01 mol HCl 1 4.67 i.e. it is a “BUFFER” solution!

Q. 1:. Calculate the pH value of the resultant solution for 10cm3 of 0 Q.1: Calculate the pH value of the resultant solution for 10cm3 of 0.20M CH3COOH are mixed with 10cm3 of 0.40M CH3COONa. (Ka = 1.7510-5 mol dm-3) After mixing, new [CH3COOH] = 0.10M new [CH3COONa] = 0.20M x(0.20+x) 0.10 – x 1.7510-5 = pH of acidic buffer could be adjusted by [acid] and [salt]. x = 8.75 10-6  pH = 5.06 An acid-salt mixture (acid and salt have comparable conc.)  ACIDIC BUFFER!

Q. 2:. Calculate the pH value of the resultant solution for 10cm3 of 1 Q.2: Calculate the pH value of the resultant solution for 10cm3 of 1.00M NH3 are mixed with 10cm3 of 1.00M NH4Cl. (Kb = 1.7810-5 mol dm-3) After mixing, new [NH3] = 0.50M new [NH4+] = 0.50M x(0.50+x) 0.50 – x 1.7810-5 = pH of basic buffer could be adjusted by [base] and [salt]. x = 1.78 10-5  pOH = 4.75  pH = 9.25 A base-salt mixture (base and salt have comparable conc.)  BASIC BUFFER!