Lecture #11-Buffers and Titrations The Common Ion Effect

Lecture #11-Buffers and Titrations The Common Ion Effect

The Common Ion Effect Shift in position of an equilibrium caused by the addition of an ion taking part in the reaction HA(aq) + H2O(l) A - (aq) + H3O + (aq) Adding a salt containing the anion, NaA, that is the conjugate base of the acid (the common ion) shifts the position of equilibrium to the left. This causes the ph to be higher (lower [H 3 O + ]) than the ph of the acid solution

The Common Ion Effect CH3COOH(aq) + H2O(l) CH3COO - (aq) + H3O + (aq) The Effect of Added Acetate Ion on the Dissociation of Acetic Acid [CH 3 COOH] initial [CH 3 COO - ] added % Dissociation* ph 0.10 0.00 0.10 0.050 0.10 0.10 1.3 0.036 0.018 2.89 4.44 4.74 0.10 0.15 0.012 4.92 * % Dissociation = [CH 3 COOH] dissoc [CH 3 COOH] initial x 100

The Common Ion Effect

Henderson Hasselbach Equation Weak acid equilibrium: HA(aq) + H 2 O(l) H 3 O + (aq) + A (aq) Ka x [HA] = [H3O + ][A-] log [H3O + ]+ log [A-] = log Ka+ log [HA] -log [H3O + ] = -log Ka- log [HA]+ log [A-] ph = pka- log [HA]+ log [A-] Note: If [base] = [acid], ph =pka

5) Calculate the ph of a solution containing 0.22 M acetic acid and 0.13 M sodium acetate. (Ka of acetic acid = 1.8 10 5 ) ph = pk a + log [A-] [HA] = -log(1.8 x 10-5 ) + log (0.13) (0.22) = -log(1.8 x 10-5 ) + log 0.59 = 4.74 + 0.22 = 4.96

ph Buffers

ph Buffers Buffers are solutions that resist changes in ph when an acid or gases is added. Buffers solutions act by neutralizing the added acid or base. Many buffers are made by mixing a solution of a weak acid with a solution of soluble salt containing its conjugate base. There are limits to what buffers can do.

Making a Buffer Solution from a Weak Acid

How a Buffer Works A buffer works by applying Le Chatlier s Principle to the weak acid equilibrium. HA (aq) + H2O (l) <- - - - - - - - - - - - > A - (aq) + H3O + (aq) Weak acid molecules react with added base to neutralize it and produce more Conjugate base anions react with added acid to neutralize it and produce more HA + OH H2O + A A + H3O + HA + H2O

6) Which of the following mixtures will result in a buffer solution? a) HF and NaF b) HF and HCl c) HF and KOH d) NH3 and NaOH e) CH3NH2 and [CH3NH3+][Cl-] f) CH3NH2 and HCl

Buffer Range Buffers are most effective when the ratio of acid/base is 1:1 Buffers can be effective when the ratio of acid/ base is 0.1 < 10 Buffer Range : pk a ± 1 Buffer Capacity Buffer Capacity : The quantity of acid or gas that a buffer can neutralize while maintaining ph within the desired range. (Proportional to component concentrations) Buffers are most effective when the acid and base concentrations are high.

Buffer Range and Capacity The greater the initial concentrations of HA and A, the less the change in ph as acid or base is added to the buffer solution.

7) Determine the number of moles of sodium acetate that must be added to 250.0 ml of 0.16 M acetic acid in order to prepare a ph 4.70 buffer (Ka = 1.76 10 5 ). 4.70 = 4.75 + log [A-] [HA] 0.89 = [A-] [0.16] -0.05 = log [A-] [0.16] [A-] = 0.14 M 10-0.05 = [A-] [0.16] (0.14 mol/l) x (.250 L) = 0.035 mol

Indicators and Acid-Base Titration

Acid-Base Titrations A solution of known concentration is slowly added to a solution of unknown concentration ( or vis a vis) until the reaction is complete. At this point stoichiometric calculations can determine the concentration of the unknown solution. An indicator may be added to determine the endpoint of the titration. Indicators are organically derived dyes which change color at specific ph s.

ph Indicators

Titrations Curves A plot of ph vs amount of added titrant. The inflection point of the curve is the equivalence point of the titration. Prior to the equivalence point, the solution in the flask is in excess, so the ph is closest to its ph. The ph of the equivalence point depends on the ph of the salt solution. Beyond the equivalence point, the added solution is in excess, so the ph approaches its ph.

14 12 10 8 6 Titration of a Strong Base with a Strong Acid H3O + (aq) + OH - (aq) 2 H2O(l) HCl(aq) + NaOH(aq) H2O(l) + NaCl(aq) Equivalence point ph=7.00 A solution of a neutral salt!! 4 2 5 10 15 20 25 30 35 40 ml of titrant

Titration of a Strong Base with a Strong Acid HCl(aq) + NaOH(aq) H2O(l) + NaCl(aq) At equivalence point, use stoichiometry: Moles acid added = moles base in sample NaCl = neutral salt; ph = 7.00

14 Titration of a Weak Base with a Strong Acid HCl (aq) + NH3(aq) NH4Cl (aq) 12 10 8 6 4 2 5 10 15 20 25 30 35 40

14 Titration of a Weak Base with a Strong Acid HCl (aq) + NH3(aq) NH4Cl (aq) 12 10 8 6 4 Midpoint ph=9.25 [NH3] = [NH4 + ] Buffering region Equivalence point ph=5.?? [NH4 + ][Cl-] (aq) A solution of an acidic salt!! 2 Only strong acid present 5 10 15 20 25 30 35 40

Titration of a Weak Base with a Strong Acid HCl(aq) + NH3(aq) H2O(l) + NH4Cl(aq) The initial ph is the ph of the weak base. The first half of the titration is a buffering region. At equivalence point: [NH 4+ ][Cl - ]= acidic salt; ph < 7.00 Midpoint in titration: [HA] = [A ]

Calculating ph for The Titration of a Weak Base with a Strong Acid 1) The initial ph is the ph of the weak base solution. Calculate ph as in a weak base equilibrium. 2) Before equivalence point, solution becomes a buffer. Using reaction stoichiometry, calculate [HA] and [A-]; Calculate ph using Henderson-Hasselbalch equation. 3) At the equivalence point, Calculate ph as in a weak acid equilibrium.

14 12 10 8 6 Titration of a Strong Acid with a Strong Base Only strong base present Equivalence point ph=7.00 A solution of a neutral salt!! 4 2 Only strong acid present H3O + (aq) + OH - (aq) 2 H2O(l) 5 10 15 20 25 30 35 40

Titration of a Strong Acid with a Strong Base HCl(aq) + NaOH(aq) H2O(l) + NaCl(aq) At equivalence point: Na + Cl - = neutral salt; ph = 7.00

Titration of a Weak Acid with a Strong Base 14 CH3COOH (aq) + NaOH(aq) CH3COO - Na + (aq) + H2O(l) 12 10 8 6 4 2 5 10 15 20 25 30 35 40

Titration of a Weak Acid with a Strong Base 14 12 CH3COOH (aq) + NaOH(aq) CH3COO - Na + (aq) + H2O(l) Only strong base present 10 8 6 4 2 Midpoint ph=4.75 [CH3COOH] = [CH3COO - ] Equivalence point ph=8.?? [CH3COO-][Na+] (aq) A solution of a basic salt!! Buffering region 5 10 15 20 25 30 35 40

Titration of a Weak Acid with a Strong Base CH3COOH(aq) + NaOH(aq) H2O(l) + CH3COONa(aq) The initial ph is the ph of the weak acid. The first half of the titration is a buffering region. At equivalence point: [CH 3 COO-][Na+] = basic salt; ph > 7.00 Midpoint in titration: [HA] = [A ]

Calculating ph for The Titration of a Weak Acid with a Strong Base 1) The initial ph is the ph of the weak acid solution. Calculate ph as in a weak acid equilibrium. 2) Before equivalence point, solution becomes a buffer. Using reaction stoichiometry, calculate [HA] and [A-]; Calculate ph using Henderson-Hasselbalch equation. 3) At the equivalence point, Calculate ph as in a weak base equilibrium.

Various Weak Acid-Strong Base Titrations

Titration of a Weak Acid with a Weak Base 14 CH3COOH (aq) + NH3(aq) CH3COO - NH4 + (aq) 12 10 8 6 Equivalence point [CH3COO-][NH4+] (aq) 4 2 5 10 15 20 25 30 35 40

8) A 25.0 ml sample of 0.100 M acetic acid (Ka = 1.76 x 10-5, pka = 4.754) is titrated with 0.125 M NaOH. Sketch the graph of the titration curve (ph vs ml of titrant) by determining the ph of the solution after 0.00, 10.0, 20.0 and 30.0 ml of the base have been added. CH3COOH (aq) + NaOH(aq) CH3COO - Na + (aq) + H2O(l) Start with the stoichiometry: moles of acetic acid = moles of NaOH ph M1V1 = M2V2 (0.0250 L)(0.100 M) = (? L)(0.125 M) 2.50 x 10-3 mol = (? L)(0.125 mol/l)] 20? = (0.020 L) = volume of NaOH added to equivalence point Volume of titrant (NaOH) added

8) A 25.0 ml sample of 0.100 M acetic acid (Ka = 1.76 x 10-5, pka = 4.754) is titrated with 0.125 M NaOH. Sketch the graph of the titration curve (ph vs ml of titrant) by determining the ph of the solution after 0.00, 5.00,10.0, 15.0, 20.0 and 30.0 ml of the base have been added. CH3COOH (aq) + NaOH(aq) CH3COO - Na + (aq) + H2O(l) ph 14 12 10 8 6 4 2 1 2 3 4 5 6 1 25.0 ml of 0.100 M acetic acid 2 } 3 25.0-45.0 ml of a buffer solution 4 5 6 45.0 ml containing mostly sodium acetate 55.0 ml containing sodium acetate + NaOH 5 10 15 20 25 30 35 40 Volume of titrant (NaOH) added

8) A 25.0 ml sample of 0.100 M acetic acid (Ka = 1.76 x 10-5, pka = 4.754) is titrated with 0.125 M NaOH. Sketch the graph of the titration curve (ph vs ml of titrant) by determining the ph of the solution after 0.00, 5.00,10.0, 15.0, 20.0 and 30.0 ml of the base have been added. CH3COOH (aq) + NaOH(aq) CH3COO - Na + (aq) + H2O(l) 6 1 What is the ph of 0.100 M acetic acid? ph 2.876 1 2 3 4 5 R I C CH3COOH CH3COO - H3O + 0.100 0.00 0.00 - x + x + x 20 Volume of titrant (NaOH) added E 0.100- x x x Ka = x 2 0.100-x [H+] = x = 1.33 x 10-3 ph = 2.876 = 1.76 x 10-5 x 2 = 1.76 x 10-6

ph 8) A 25.0 ml sample of 0.100 M acetic acid (Ka = 1.76 x 10-5, pka = 4.754) is titrated with 0.125 M NaOH. Sketch the graph of the titration curve (ph vs ml of titrant) by determining the ph of the solution after 0.00, 5.00,10.0, 15.0, 20.0 and 30.0 ml of the base have been added. CH3COOH (aq) + NaOH(aq) CH3COO - Na + (aq) + H2O(l) 8.750 5 6 5 What is the ph of 0.100 M sodium acetate?? But it isn t 0.100 M sodium acetate!! [CH3COO - Na + ] = (2.50 x 10-3 mol)/(0.0450 L) 2.876 2 3 4 = 0.0556 M Kb = 5.68 x 10-10 5 10 15 20 25 30 35 40 Volume of titrant (NaOH) added R I C CH3COO - CH3COOH OH - 0.0556 0.00 0.00 - x + x + x Kb = x 2 = 5.68 x 10-10 0.0556-x E 0.0556- x x x x 2 = 3.16 x 10-11 ph = 8.750 [OH-] = x = 5.62 x 10-6 M [H+] = 1.78 x 10-9 M

8) A 25.0 ml sample of 0.100 M acetic acid (Ka = 1.76 x 10-5, pka = 4.754) is titrated with 0.125 M NaOH. Sketch the graph of the titration curve (ph vs ml of titrant) by determining the ph of the solution after 0.00, 5.00,10.0, 15.0, 20.0 and 30.0 ml of the base have been added. CH3COOH (aq) + NaOH(aq) CH3COO - Na + (aq) + H2O(l) ph 6 2 3 4 8.750 5.231 4.754 4.276 2.876 2 3 4 Points 2, 3, and 4 are in the buffer region and are governed by the Henderson-Hasselbach equation: 5 10 15 20 25 30 35 40 Volume of titrant (NaOH) added [1] 2 ph = 4.754 + log = 4.276 [3] [1] 3 ph = 4.754 + log = 4.754 [1] [3] 4 ph = 4.754 + log = 5.231 [1]

ph 8) A 25.0 ml sample of 0.100 M acetic acid (Ka = 1.76 x 10-5, pka = 4.754) is titrated with 0.125 M NaOH. Sketch the graph of the titration curve (ph vs ml of titrant) by determining the ph of the solution after 0.00, 5.00,10.0, 15.0, 20.0 and 30.0 ml of the base have been added. CH3COOH (aq) + NaOH(aq) CH3COO - Na + (aq) + H2O(l) 12.356 8.750 5.231 4.754 4.276 2.876 5 10 15 20 25 30 35 40 Volume of titrant (NaOH) added 6 6 Point 6 is far beyond the equivalence point and the buffering region. 20.0 ml of the titrant were used to neutralize the acetic acid. Only the remaining 10.0 ml of the strong base influences the ph. [Na + OH - ] = (0.125 mol/l)(0.0100 L)/(0.0550 L) = 0.0227 M poh = 1.644 ph = 12.356