17.3 Weak Acids Weak Bases Titration Titration of Weak Acid with Strong Base Titration of Base Acid with Strong Acid Dr. Fred Omega Garces Chemistry 201 Miramar College 1 Weak Acids Weak Bases Titration March 18 Weak Acid (or Weak Base) with Strong Base (or strong Acid) Experimental technique and the concept is similar to that of the titration of a strong acid with a strong base (or vice versa) with equilibrium concept applied. pH calculation involves 4 different type of calculations. i) The analyte alone (equilibrium calculation) ii) Buffer region (Henderson-Hasselbach eqn) iii) Equivalence point (Hydrolysis) iv) Excess titrant Equilb Buffer Hydrolysis Stoic Excess (Stoichiometric calculation) Click for simulation 2 Weak Acids Weak Bases Titration March 18 Titration (WA-SB): Weak Acid (or Weak Base) with Strong Base (or strong Acid) Consider the titration problem: Titration curve for a weak acid (HOCl) and an strong base (KOH). Generate a titration curve upon addition of KOH @ 0%-, 50%-, 95%-, 100%- and 105% of equivalent point. Analyte : 10.00 ml 0.400M HOCl: Titrant: 0.400 M KOH - + HOCl + KOH g H2O + OCl + K HOCl = 0.400M • 10.00ml = 4.0 mmol HOCl Volume KOH corresponding to 0%-, 50%-, 95%-, 100%- and 105% 0.400 M KOH = 0ml, 5.0ml, 9.5ml, 10.0ml, 10.5 ml -8 Misc. Info.: HOCl: Ka = 3•10 pKa = 7.5 4 Weak Acids Weak Bases Titration March 18 Type i: Weak Acid with Strong Base 0 % Type 1: Calculation EQUILBRIUM 0% KOH added (TYPE 1 Weak acid calc.) 0%, VT = 10.0 ml Weak acid pKa - (Type 1 Calculation) pH of solution is determined by the dissociation of the weak acid. HOCl + H2O ! H3O+ + OCl- Ka i 0.4M Ex 0 0 C -x -x +x +x e 0.4M-x Ex +x +x x2 K = 3⋅10 −8 = , 0.4M −x ≈ 0.4M a .4M −x x = .4(3•10−8 ) + -4 [H30 ]= x = 1.10• 10 M pH = 3.96 € 5 Weak Acids Weak Bases Titration March 18 Titration (4ii): Weak Acid (or Weak Base) with Strong Base (or strong Acid) 50 % Type 2: Calculation BUFFER, Henderson- Hasselbalch Eqn 50% KOH added (TYPE 2 Buffer) 50%, VT = 15.0 ml B Simple Approach HOCl + KOH ! H2O + OCl- + K+ Buffer solution using Henderson Hasselbalch s 4mmol 2mmol Ex 0 - equation and the sRfc table above R -2 -2 +2 +2 - f 2mmol 0 Ex+2 2mmol - Cb c .133M - Ex .133M pH = pK + log a C Notice that the concentration of the acid and its conjugate are equal. a In the mass action expression these two terms cancel. pH = pKa. .133M pH = 7.5 + log .133M Buffer situation-(Type-2) pH = 7.5 A. Long Approach Note that the excess 2mmol of HOCl will dissociate in water to " HOCl + H2O H3O+ + OCl- Ka € i .133M Ex 0 0.133M C -x -x +x +x e .133-x Ex x .133+x The calculation is a simple equilibrium analysis- x(0.133+ x) K = 3⋅10−8 = , 0.133M ±x ≈ 0.133M a (0.133- x) −8 x = 3⋅10 ,Assumption checks! + −8 [H30 ]= 3⋅10 M pH = 7.5 6 Weak Acids Weak Bases Titration March 18 € Titration (4ii): Weak Acid (or Weak Base) with Strong Base (or strong Acid) 95% Type II Calculation BUFFER, Henderson- Hasselbalch Eqn 95% KOH added (TYPE 2 Buffer): 95%, VT = 19.5 ml HOCl + KOH ! H O - + 2 + OCl + K B Simple Approach s 4 mmol 3.8 mmol Ex 0 - Buffer solution using Henderson R -3.8 -3.8 +3.8 +3.8 - Hasselbach equation and sRfc table f 0.2 mmol 0 Ex 3.8 mmol - c 1.03•10-2M - Ex 0.195M Cb pH = pKa + log Noticed that the acid and conjugate are present in the same solution. This Ca is a common ion effect or buffer type problem. Use the HH equation. 0.195M pH = 7.5 + log 1.03⋅10-2 M Buffer situation-(Type-2) pH = 7.5 + 1.28 A. Long Approach pH = 8.78 Note that the excess 0.2mmol (or 1.03•10-2M) of HOCl will dissociate in water " HOCl + H2O H3O+ + OCl- Ka € i 1.03•10-2M Ex 0 0.195M C -x -x +x +x e 1.03•10-2M - x Ex x 0.195M+x −8 x(.195 + x) 0.195M +x ≈ 0.195M Ka = 3⋅10 = -2 -2 (1.03⋅10-2 - x) 1.03⋅10 M − x ≈ 1.03⋅10 M -2 8 1.03⋅10 '% 3⋅10− *( & ) = x 0.195 1.58⋅10-9 = x ,Assumption checks! 1.58⋅10-9 M = +H 0+ . ,- 3 /0 8.80 = pH 7 Weak Acids Weak Bases Titration March 18 € Titration (4iii): Weak Acid (or Weak Base) with Strong Base (or strong Acid) 100% Type III Calculation HYDROLYSIS, conjugate back to original 100% KOH added (TYPE 3 Hydrolysis): K 1⋅10−14 x2 100%, VT = 20.0 ml K = w = = 0.2 - x ≈ 0.2M b K −8 (0.2- x) HOCl + KOH ! H2O + OCl- + K+ a 3⋅10 x2 s 4mmol 4mmol Ex 0 - = 3.33⋅10−7 x = 2.58⋅10−4M R -4 -4 +4 +4 - 0.2 f 0 0 Ex 4mmol - % OH− ( = 2.58⋅10−4M, pOH = 3.59 &' )* c - - Ex 0.2M - pH = 10.41 How is the pH or pOH calculated since there are no H30+ or OH-? Actually the excess HOCl does react with water to form H3O+. € Equivalence point calculation, Hydrolysis (Type-3) Note that all the HOCl acid is neutralize by the base. HOCl cannot dissociate in water since there is no excess. But the conjugate base OCl- can react with water in a hydrolysis reaction according to- OCl- + H2O " OH- + HOCl Kb i 0.2M Ex 0 0 C -x -x +x +x e 0.2-x Ex x x 8 Weak Acids Weak Bases Titration March 18 Titration (4iv): Weak Acid (or Weak Base) with Strong Base (or strong Acid) 105% Type IV Calculation Stoichiometry A simple check shows that [OH-]2 is negligible. 105% KOH added (TYPE 4 Strong Base) OCl- + H2O ! OH- + HOCl K HOCl + KOH ! H2O + OCl- + K+ b 9.76e-3 i 0.195 M Ex 0 s 4mmol 4.2mmol Ex 0 - C -x -x +x +x R -4 -4 +4 +4 - e 0.195 M-x Ex 9.76e-3 +x x K f 0 0.2mmol Ex 4mmol - 1 ⋅10−14 (9.76⋅10-3 + x) x K = w = = , 0.195-x ≈ 0.195 M c - -3 Ex .195M - b K −8 0.195-x 9.76•10 M a 3⋅10 ( ) -3 2 (9.76⋅10 x + x ) −7 = 3.33⋅10 (0.195) Strong Base calculation (Type-4) -3 - - Solve Quadratic a=1, b=9.76⋅10 , c= − 6.49 ⋅10 8 x = [OH ] = 6.64 ⋅10−6M 2 Since the excess is KOH, a strong base, then the pH (or pOH in Therefore you see that - this case) is determine by the following dissociation of KOH: [OH−] = [OH-] + [OH-] = 2.54 ⋅10−4M + 9.76⋅10−3 1 2 Total ! KOH OH- + K [OH−] = 1.00 ⋅10−2 ≈ 9.76⋅10−3M Total i 9.76•10-3M 0 0 pOH = 2.010 pH = 11.990 C -9.76•10-3M +9.76•10-3M +9.76•10-3M e - +9.76•10-3M +9.76•10-3M - −3 [OH ] 1 = 9.76 ⋅ 10 M pOH = 2.01 pH = 11.99 Note that the 0.195 M of OCl- will contribute negligible - € amounts of OH as it back reacts with water in a hydrolysis type reaction. 9 Weak Acids Weak Bases Titration March 18 Titration Curve: Strong Acid / Strong Base Titration of 20.0 mL 0.500M HCl with 0.500M TitrationNaOH of 0.100 M HCl with 0.100 M NaOH. A strong acid-strong base titration curve, showing how the pH increases as 0.5000M NaOH is added to 20.00mL of 0.5000M HCl. The equivalence point pH is 7.00. The steep portion of the curve includes the transition intervals of bromothymol blue, phenolphthalein and bromophenol blue . 10 Weak Acids Weak Bases Titration March 18 Titration Curve Features: WA - SB . Titration of 40.00 mL 0.1000M HCH3CH2OOO with 0.1000 M NaOH Weak acid - Strong base Titration curve for a weak acid by a strong base: 40.00mL of 0.1000M CH3CH2OOOH by 0.1000M NaOH. When exactly one-half the acid is neutralized, [CH3CH2COOH] = - [CH3CH2COO ] and pH = pKa = 8.80 The equivalent point is above 7.00 because the - solution contains the weak base CH3CH2COO . Phenolphthalein is a suitable indicator for this titration but Methyl red is not because its color changes over a large volume range. 11 Weak Acids Weak Bases Titration March 18 Titration Curve Features: Monoprotic Acid Titration curve for a series of acids (A - F) being titrated with a strong base A Acid F is the strongest acid, Acid B E is the next strongest acid followed by acid D, acid C, Acid B C and acid A. Acid A is the D weakest among the weak acid. The Kas of each acid is E F determined by reading the pH half way to the equivalent volume for each acid. 12 Weak Acids Weak Bases Titration March 18 Titration Curve Features: WB -SA Titration of 0.1000 M NH3 with 0.1000 M HCl Weak base - Strong acid A weak base-strong acid titration curve, showing how the pH decreases as 0.1000M HCl is added to 40.00mL of 0.1000M NH3.