16.3 Weak Acids Weak Bases Titration

Transcription

1 16.3 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

2 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 (Stoichiometric calculation) Click for simulation Equilb Buffer Hydrolysis Stoic Excess 2 Weak Acids Weak Bases Titration

3 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 0%-, 50%-, 95%-, 100%- and 105% of equivalent point. Analyte : ml 0.400M HOCl: Titrant: M KOH HOCl + KOH g H 2 O + OCl - + K + HOCl = 0.400M 10.00ml = 4.0 mmol HOCl Volume KOH corresponding to 0%-, 50%-, 95%-, 100%- and 105% M KOH = 0ml, Misc. Info.: HOCl: K a = pk a = Weak Acids Weak Bases Titration

4 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 0%-, 50%-, 95%-, 100%- and 105% of equivalent point. Analyte : ml 0.400M HOCl: Titrant: M KOH HOCl + KOH g H 2 O + OCl - + K + HOCl = 0.400M 10.00ml = 4.0 mmol HOCl Volume KOH corresponding to 0%-, 50%-, 95%-, 100%- and 105% M KOH = 0ml, 5.0ml, 9.5ml, 10.0ml, 10.5 ml Misc. Info.: HOCl: K a = pk a = Weak Acids Weak Bases Titration

5 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! K a H3O + + OCl - i 0.4M Ex 0 0 C -x -x +x +x e 0.4M-x Ex +x +x K a = x 2 =.4M x, 0.4M x 0.4M ( ) x = [ H30 + ]= x = M ph = Weak Acids Weak Bases Titration

6 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 HOCl + KOH! H2O + OCl- + K+ s 4mmol 2mmol Ex 0 - R f 2mmol 0 Ex+2 2mmol - c.133m - Ex.133M Notice that the concentration of the acid and its conjugate are equal. In the mass action expression these two terms cancel. ph = pka. Buffer situation-(type-2) A. Long Approach Note that the excess 2mmol of HOCl will dissociate in water to HOCl + H2O " H3O+ + OCl- K a i.133m Ex M C -x -x +x +x e.133-x Ex x.133+x B Simple Approach Buffer solution using Henderson Hasselbalch equation and the srfc table above ph = pk a + log C b C a ph = log.133m.133m ph = 7.5 The calculation is a simple equilibrium analysis- K a = = x( x), 0.133M ±x 0.133M x ( ) x = ,Assumption checks! [ H30 + ]= M ph = Weak Acids Weak Bases Titration

7 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! H2O + OCl- + K+ s 4 mmol 3.8 mmol Ex 0 - R f 0.2 mmol 0 Ex 3.8 mmol - c M - Ex 0.195M Noticed that the acid and conjugate are present in the same solution. This is a common ion effect or buffer type problem. Use the HH equation. Buffer situation-(type-2) A. Long Approach Note that the excess 0.2mmol (or M) of HOCl will dissociate in water HOCl + H2O " H3O+ + OCl- K a B Simple Approach Buffer solution using Henderson Hasselbach equation and srfc table ph = pk a + log C b C a 0.195M ph = log M ph = ph = 8.78 i M Ex M C -x -x +x +x e M - x Ex x 0.195M+x x ( x) K a = = ( x) % ' ( * & ) = x = x,assumption checks! M = + H ,- / = ph 0.195M +x 0.195M M x M 7 Weak Acids Weak Bases Titration

8 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): 100%, VT = 20.0 ml HOCl + KOH! H2O + OCl- + K+ s 4mmol 4mmol Ex 0 - R f 0 0 Ex 4mmol - c - - Ex 0.2M - 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 K b i 0.2M Ex 0 0 C -x -x +x +x e 0.2-x Ex x x K b = K w K a = = x 2 ( 0.2- x) x 0.2M x = x = M % OH ( &' )* M, poh = 3.59 ph = Weak Acids Weak Bases Titration

9 Titration (4iv): Weak Acid (or Weak Base) with Strong Base (or strong Acid) 105% Type IV Calculation Stoichiometry 105% KOH added (TYPE 4 Strong Base) HOCl + KOH! H 2 O + OCl- + K+ s 4mmol 4.2mmol Ex 0 - R f 0 0.2mmol Ex 4mmol - c M Ex.195M - Strong Base calculation (Type-4) Since the excess is KOH, a strong acid, then the ph (or poh in this case) is determine by the following dissociation of KOH: A simple check shows that [OH - ] 2 is negligible. OCl - + H 2 O! OH - + HOCl K b i M Ex 0 0 C -x -x +x +x e M-x Ex x x K b = K w = K a = x x x 2 Therefore you see that - [OH ] Total ( ), x M = x = [OH - ] 2 = M = [OH - ] 1 + [OH - ] 2 = M KOH! OH - + K i M 0 0 C M M M e M M [OH - ] 1 = M poh = 2.01 ph = [OH ] Total = M poh = ph = Note that the 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

10 Titration Curve: Strong Acid / Strong Base Titration of 0.500M HCl with 0.500M NaOH Titration of M HCl with M NaOH. A strong acid-strong base titration curve, showing how the ph increases as M NaOH is added to 20.00mL of M HCl. The equivalence point ph is The steep portion of the curve includes the transition intervals of bromothymol blue, phenolphthalein and bromophenol blue. 10 Weak Acids Weak Bases Titration

11 Titration of ml M HCH 3 CH 2 OOO with M NaOH Weak acid - Strong base Titration Curve Features: WA - SB Titration curve for a weak acid by a strong base: 40.00mL of M CH 3 CH 2 OOOH by M NaOH. When exactly one-half the acid is neutralized, [CH 3 CH 2 COOH] = [CH 3 CH 2 COO - ] and ph = pk a = 8.80 The equivalent point is above 7.00 because the solution contains the weak base CH 3 CH 2 COO -. 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

12 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 E is the next strongest acid followed by acid D, acid C, Acid B and acid A. Acid A is the weakest among the weak acid. The Ka s of each acid is determined by reading the ph half way to the equivalent volume for each acid. B C D E F 12 Weak Acids Weak Bases Titration

13 Titration Curve Features: WB -SA Titration of M NH 3 with M HCl Weak base - Strong acid A weak base-strong acid titration curve, showing how the ph decreases as M HCl is added to 40.00mL of M NH 3. When exactly one-half the base is neutralized, [NH 3 ] = [NH 4+ ] and poh = pk b of NH 3 (4.76) or the ph = 14 - pk b of NH 3 (9.24). Note that this ph value (9.24) is actually the pka of NH 4+, the conjugate of NH 3. Note that the equivalent point is below 7.00 because the solution contains the weak acid NH + 4. Methyl red is a suitable indicator for this titration but phenolphthalein is not because its color changes over a large volume range. 13 Weak Acids Weak Bases Titration

14 Titration Curve: Polyprotic Titration Curve of 0.100M H 2 SO 3 with M NaOH Curve for the titration of a weak polyprotic acid. Titrating 40.00mL of M H 2 SO 3 with M NaOH leads to a curve with two buffer regions and two equivalence points. Because the K a values are separated by several orders of magnitude, in effect the titration curve looks like two weak acidstrong base curves attached. The ph of the first equivalence point is below 7 because the solution contains HSO 3-, which is a stronger acid than it is a base. K a of HSO - 3 = ; K b of HSO - 3 = Weak Acids Weak Bases Titration

15 Titration Curve: Poly-Basic Titration Curve of 0.10 M Na 2 CO 3 with 0.10 M HCl. 1. CO H + g HCO H 2 O 2. HCO H + g H 2 CO 3 + H 2 O 15 Weak Acids Weak Bases Titration

16 Various Titration Curve Example of Various: Titration Curves Analytes Strong Weak Poly Acids Base 16 Weak Acids Weak Bases Titration

17 Summary There are two main type of titration problems. The strategy to solve them are: 1) SA-SB: Strong acid being titrated with a strong base (or vice versa). This is a stoichiometry type of problem. The ph at the equivalent point = a) Weak acid being titrated with a strong base. In this type of problem, the there are four sub-problems that must be solved. a) At 0% titrant, the problem is an equilibrium type. The Ka of the weak acid will determine the extent of ionization and therefore the ph. b) At 1-99% titrant, the problem is a buffer type. Use the Henderson-Hasselbach Equation to solve for the ph c) At equivalence point, this is now a hydrolysis problem. Recall that the conjugate base of the weak acid now reacts with water to produce OH-. The solution will be basic at the equivalent point. d) Pass the equivalence point, this is now a stoichiometry problem. The excess titrant base will dictate the ph of the solution. 2b) Weak base being titrated with a strong acid. This is the same type of problem as (2a) above except in this case a weak base is being neutralized by the strong acid titrant. 17 Weak Acids Weak Bases Titration