Homework problems added to syllabus: Friday 9/7 Ch. 8, # 52, 56, 60 Monday 9/10 Ch. 8, # 62, 66

Buffer Solutions

Buffer Solutions Buffer solutions aid in maintaining a consistent pH range for a reaction.

The buffer acts as a sponge for excess acid or base.

Buffer Solutions Changes in reaction pH can be caused by:

Buffer Solutions Changes in reaction pH can be caused by: Acidic or basic product

A + B HC + D

A + B HC + D HC(aq) + H2O H3O+(aq) + C-(aq)

A + B HC + D HC(aq) + H2O H3O+(aq) + C-(aq) [HC] [H3O+]

Buffer Solutions Changes in reaction pH can be caused by: Acidic or basic product Acidic or basic reactant

A + NH3(aq) + H2O(l) A NH3(aq) + NH4+(aq) + OH-(aq)

A + NH3(aq) + H2O(l) A NH3(aq) + NH4+(aq) + OH-(aq) [NH3]

A + NH3(aq) + H2O(l) A NH3(aq) + NH4+(aq) + OH-(aq) [NH3] [OH-]

A buffer consists of a weak acid and its conjugate base or a weak base and its conjugate acid.

A buffer consists of a weak acid and its conjugate base or a weak base and its conjugate acid. A conjugate base or conjugate acid does not necessarily have to come from solution of an acid or base.

Weak acid HA(aq) + H2O(aq) H3O+(aq) + A-(aq)

Weak acid HA(aq) + H2O(aq) H3O+(aq) + A-(aq) [H3O+][A-] Ka = [HA]

Weak acid HA(aq) + H2O(aq) H3O+(aq) + A-(aq) [H3O+][A-] Ka = [HA] [HA] [H3O+] = Ka [A-]

Weak acid HA(aq) + H2O(aq) H3O+(aq) + A-(aq) [HA] [H3O+] = Ka [A-] pH = -log10[H3O+]

Weak acid HA(aq) + H2O(aq) H3O+(aq) + A-(aq) [HA] [H3O+] = Ka [A-] pH = -log10[H3O+] [HA] pH  [A-]

Weak acid HA(aq) + H2O(aq) H3O+(aq) + A-(aq) [HA] [H3O+] = Ka [A-] If [HA] and [A-] are large (  1 M ), the ratio will change slowly if dilute (0.1 M or less) strong acid is added.

CH3COOH 1.0 M NaCH3COO 0.5 M

CH3COOH 1.0 M NaCH3COO 0.5 M CH3COOH(aq) + H2O(l) H3O+(aq) + CH3COO-(aq)

CH3COOH 1.0 M NaCH3COO 0.5 M CH3COOH(aq) + H2O(l) H3O+(aq) + CH3COO-(aq) CH3COOH(aq) + H2O(l) + NaCH3COO (aq) Na+(aq) + H3O+(aq) + 2 CH3COO-(aq)

CH3COOH 1.0 M NaCH3COO 0.5 M CH3COOH(aq) + H2O(l) + NaCH3COO (aq) Na+(aq) + H3O+(aq) + 2 CH3COO-(aq)

CH3COOH 1.0 M Na CH3COO 0.5 M CH3COOH(aq) + H2O(l) + NaCH3COO (aq) Na+(aq) + H3O+(aq) + 2 CH3COO-(aq)

CH3COOH 1.0 M NaCH3COO 0.5 M CH3COOH(aq) + H2O(l) + NaCH3COO (aq) Na+(aq) + H3O+(aq) + 2 CH3COO-(aq) [CH3COOH] [H3O+] [CH3COO-] Start 1.0 0 0.5

CH3COOH 1.0 M NaCH3COO 0.5 M CH3COOH(aq) + H2O(l) + NaCH3COO (aq) Na+(aq) + H3O+(aq) + 2 CH3COO-(aq) [CH3COOH] [H3O+] [CH3COO-] Start 1.0 0 0.5 Change -y +y +y

CH3COOH 1.0 M Na CH3COO 0.5 M CH3COOH(aq) + H2O(l) + NaCH3COO (aq) Na+(aq) + H3O+(aq) + 2 CH3COO-(aq) [CH3COOH] [H3O+] [CH3COO-] Start 1.0 0 0.5 Change -y +y +y Equilibrium 1.0-y y 0.5+y

[CH3COOH] [H3O+] [CH3COO-] Start 1.0 0 0.5 Change -y +y +y Equilibrium 1.0-y y 0.5+y [H3O+][CH3COO-] Ka = [CH3COOH]

[CH3COOH] [H3O+] [CH3COO-] Start 1.0 0 0.5 Change -y +y +y Equilibrium 1.0-y y 0.5+y [H3O+][CH3COO-] (y)(0.5+y) Ka = = (1-y) [CH3COOH] Ka = 1.8 x 10-5

[CH3COOH] [H3O+] [CH3COO-] Start 1.0 0 0.5 Change -y +y +y Equilibrium 1.0-y y 0.5+y (y)(0.5+y) 1.8 x 10-5 = (1-y) Assume y << 1

[CH3COOH] [H3O+] [CH3COO-] Start 1.0 0 0.5 Change -y +y +y Equilibrium 1.0-y y 0.5+y (y)(0.5) 1.8 x 10-5 = (1) Assume y << 1

[CH3COOH] [H3O+] [CH3COO-] Start 1.0 0 0.5 Change -y +y +y Equilibrium 1.0-y y 0.5+y y = 2( 1.8 x 10-5) = 3.6 x 10-5

[CH3COOH] [H3O+] [CH3COO-] Start 1.0 0 0.5 Change -y +y +y Equilibrium 1.0-y y 0.5+y y = 2( 1.8 x 10-5) = 3.6 x 10-5 pH = -log10(3.6x10-5) = 4.44

[CH3COOH] [H3O+] [CH3COO-] Start 1.0 0 0.5 Change -y +y +y Equilibrium 1.0-y y 0.5+y y = 2( 1.8 x 10-5) = 3.6 x 10-5 pH = -log10(3.6x10-5) = 4.44 buffer pH = 2.37 for 1 M CH3COOH

pH = -log10(3.6x10-5) = 4.44 buffer pH = 2.37 for 1 M CH3COOH [H3O+][CH3COO-] Ka = [CH3COOH] CH3COOH(aq) + H2O(l) H3O+(aq) + CH3COO-(aq)

pH = -log10(3.6x10-5) = 4.44 buffer pH = 2.37 for 1 M CH3COOH [H3O+][CH3COO-] Ka = [CH3COOH] [CH3COOH] [H3O+] = Ka [CH3COO-]

Prepare buffer solution of CH3COOH and CH3COO- and include 0.1 M HCl.

Prepare buffer solution of CH3COOH and CH3COO- and include 0.1 M HCl. HCl(aq) + H2O(aq) H3O+ + Cl-(aq)

Prepare buffer solution of CH3COOH and CH3COO- and include 0.1 M HCl. HCl(aq) + H2O(aq) H3O+ + Cl-(aq) HCl is a strong acid 100% dissociation

Prepare buffer solution of CH3COOH and CH3COO- and include 0.1 M HCl. HCl(aq) + H2O(aq) H3O+ + Cl-(aq) HCl is a strong acid 100% dissociation H3O+(aq)+ CH3COO-(aq) CH3COOH(aq) + H2O(l)

Prepare buffer solution of CH3COOH and CH3COO- and include 0.1 M HCl. HCl(aq) + H2O(aq) H3O+ + Cl-(aq) HCl is a strong acid 100% dissociation H3O+(aq)+ CH3COO-(aq) CH3COOH(aq) + H2O(l) Ka of CH3COOH favors products

Assume all HCl reacts with CH3COO- to form an extra 0.1 M CH3COOH and reduces CH3COO- by 0.1 M.

[CH3COOH] [H3O+] [CH3COO-] Start 1.1 0 0.4 Change -y +y +y Equilibrium 1.1-y y 0.4+y [CH3COOH] [H3O+] = Ka [CH3COO-]

[CH3COOH] [H3O+] [CH3COO-] Start 1.1 0 0.4 Change -y +y +y Equilibrium 1.1-y y 0.4+y [CH3COOH] [H3O+] = Ka [CH3COO-] (1.1-y) [H3O+] = Ka = (0.4+y)

[CH3COOH] [H3O+] [CH3COO-] Start 1.1 0 0.4 Change -y +y +y Equilibrium 1.1-y y 0.4+y [CH3COOH] [H3O+] = Ka [CH3COO-] (1.1-y) 1.1 [H3O+] = Ka = Ka (0.4+y) 0.4

[CH3COOH] [H3O+] [CH3COO-] Start 1.1 0 0.4 Change -y +y +y Equilibrium 1.1-y y 0.4+y [CH3COOH] [H3O+] = Ka [CH3COO-] 1.1 [H3O+] = Ka = (1.8 x 10-5)(2.75)= 0.4

[CH3COOH] [H3O+] [CH3COO-] Start 1.1 0 0.4 Change -y +y +y Equilibrium 1.1-y y 0.4+y [CH3COOH] [H3O+] = Ka [CH3COO-] 1.1 [H3O+] = Ka = (1.8 x 10-5)(2.75)= 4.95 x 10-5 0.4

[CH3COOH] [H3O+] [CH3COO-] Start 1.1 0 0.4 Change -y +y +y Equilibrium 1.1-y y 0.4+y [CH3COOH] [H3O+] = Ka [CH3COO-] [H3O+] = 4.95 x 10-5 pH = 4.31

pH 1.0 M CH3COOH 2.37 pH 1.0 M CH3COOH/ 0.5 M CH3COO- 4.44 0.5 M CH3COO-/ 0.1 M HCl 4.31

pH 1.0 M CH3COOH 2.37 pH 1.0 M CH3COOH/ 0.5 M CH3COO- 4.44 0.5 M CH3COO-/ 0.1 M HCl 4.31 The pH decreases due to the increase in CH3COOH concentration.

pH 1.0 M CH3COOH 2.37 pH 1.0 M CH3COOH/ 0.5 M CH3COO- 4.44 0.5 M CH3COO-/ 0.1 M HCl 4.31 pH 0.1 M HCl 1.00

The action of the buffer involves the reaction of the strong acid with the conjugate base of a weak acid.

The action of the buffer involves the reaction of the strong acid with the conjugate base of a weak acid. This generates a weak acid of roughly the equivalent of the added strong acid.

This increased amount of weak acid decreases the pH of the solution, but it is a much smaller decrease than would have occurred with the addition of strong acid only.

CH3COOH(aq) + H2O(l) + NaCH3COO (aq) Na+(aq) + H3O+(aq) + 2 CH3COO-(aq)

Prepare same buffer with 0.5 M HCl

[CH3COOH] [H3O+] [CH3COO-] Start 1.5 0 0.0 Change -y +y +y Equilibrium 1.5-y y +y [CH3COOH] [H3O+] = Ka [CH3COO-]

[CH3COOH] [H3O+] [CH3COO-] Start 1.5 0 0.0 Change -y +y +y Equilibrium 1.5-y y +y [CH3COOH] [H3O+] = Ka [CH3COO-] y2 = Ka[CH3COOH]

[CH3COOH] [H3O+] [CH3COO-] Start 1.5 0 0.0 Change -y +y +y Equilibrium 1.5-y y +y [CH3COOH] [H3O+] = Ka [CH3COO-] y2 = Ka[CH3COOH] = (1.8 x 10-5)(1.5)

[CH3COOH] [H3O+] [CH3COO-] Start 1.5 0 0.0 Change -y +y +y Equilibrium 1.5-y y +y [CH3COOH] [H3O+] = Ka [CH3COO-] y2 = Ka[CH3COOH] = (1.8 x 10-5)(1.5) = 2.7 x 10-5

[CH3COOH] [H3O+] [CH3COO-] Start 1.5 0 0.0 Change -y +y +y Equilibrium 1.5-y y +y [CH3COOH] [H3O+] = Ka [CH3COO-] y2 = 2.7 x 10-5 y = 5.20 x 10-3

[CH3COOH] [H3O+] [CH3COO-] Start 1.5 0 0.0 Change -y +y +y Equilibrium 1.5-y y +y [CH3COOH] [H3O+] = Ka [CH3COO-] y2 = 2.7 x 10-5 y = 5.20 x 10-3 pH = 2.28

pH 1.0 M CH3COOH 2.37 pH 1.0 M CH3COOH/ 0.5 M CH3COO- 4.44 0.5 M CH3COO-/ 0.1 M HCl 4.31 pH buffer + 0.5 M HCl 2.28 pH 0.1 M HCl 1.00

[CH3COOH] [H3O+] [CH3COO-] Start 1.0 0 0.5 Change -y +y +y Equilibrium 1.0-y y 0.5+y [H3O+][CH3COO-] Ka = [CH3COOH] Add 0.51 M HCl

[CH3COOH] [H3O+] [CH3COO-] Start 1.5 0.01 0.0 Change -y +y +y Equilibrium 1.5-y 0.01+y y [H3O+][CH3COO-] Ka = [CH3COOH] Add 0.51 M HCl

[CH3COOH] [H3O+] [CH3COO-] Start 1.5 0.01 0.0 Change -y +y +y Equilibrium 1.5-y 0.01+y y (0.01+y)(y) [H3O+][CH3COO-] Ka = = [CH3COOH] (1.5-y)

[CH3COOH] [H3O+] [CH3COO-] Start 1.5 0.01 0.0 Change -y +y +y Equilibrium 1.5-y 0.01+y y (0.01+y)(y) [H3O+][CH3COO-] Ka = = = (1.5-y) [CH3COOH] (0.01y+y2) = (1.5-y)

[CH3COOH] [H3O+] [CH3COO-] Start 1.5 0.01 0.0 Change -y +y +y Equilibrium 1.5-y 0.01+y y (0.01+y)(y) [H3O+][CH3COO-] Ka = = = (1.5-y) [CH3COOH] y2 + .01y - 2.7 x 10-5 = 0

[CH3COOH] [H3O+] [CH3COO-] Start 1.5 0.01 0.0 Change -y +y +y Equilibrium 1.5-y 0.01+y y (0.01+y)(y) [H3O+][CH3COO-] Ka = = = (1.5-y) [CH3COOH] y2 + .01y - 2.7 x 10-5 = 0 y = 9.93 x 10-3

[CH3COOH] [H3O+] [CH3COO-] Start 1.5 0.01 0.0 Change -y +y +y Equilibrium 1.5-y 0.01+y y (0.01+y)(y) [H3O+][CH3COO-] Ka = = = (1.5-y) [CH3COOH] y2 + .01y - 2.7 x 10-5 = 0 y = 9.93 x 10-3 pH = 2.00

pH 1.0 M CH3COOH 2.37 pH 1.0 M CH3COOH/ 0.5 M CH3COO- 4.44 0.5 M CH3COO-/ 0.1 M HCl 4.31 pH buffer + 0.5 M HCl 2.28 pH buffer + 0.51 M HCl 2.00 pH 0.01 M HCl 2.00

Additional conjugate base will create a solution which can react with more acid before the pH decreases appreciably.

Additional conjugate base will create a solution which can react with more acid before the pH decreases appreciably. This implies that the weak acid concentration should be raised as well.

Determining correct acid-conjugate base combination and concentration for a desired buffer pH.

Determining correct acid-conjugate base combination and concentration for a desired buffer pH. Ka > 10-7 pH < 7

Determining correct acid-conjugate base combination and concentration for a desired buffer pH. Ka > 10-7 pH < 7 Ka < 10-7 pH > 7

HA(aq) + H2O(aq) H3O+(aq) + A-(aq) Ka = [HA]

HA(aq) + H2O(aq) H3O+(aq) + A-(aq) Ka = [HA] [A- ]o,[HA]o = original concentrations

HA(aq) + H2O(aq) H3O+(aq) + A-(aq) Ka = [HA] [A- ]o,[HA]o = original concentrations [HA]  [HA]o [A-]  [A-]o

HA(aq) + H2O(aq) H3O+(aq) + A-(aq) [H3O+][A-]o Ka =  [HA] [HA]o

HA(aq) + H2O(aq) H3O+(aq) + A-(aq) [H3O+][A-]o Ka =  [HA] [HA]o [HA]o [H3O+]  Ka [A-]o

HA(aq) + H2O(aq) H3O+(aq) + A-(aq) [H3O+][A-]o Ka =  [HA] [HA]o [HA]o [H3O+]  Ka [A-]o [HA]o [H3O+]  Ka -log10 = [A-]o

HA(aq) + H2O(aq) H3O+(aq) + A-(aq) [HA]o [H3O+]  Ka [A-]o [HA]o -log10 [H3O+]  Ka = [A-]o [HA]o pH  pKa -log10 [A-]o

HA(aq) + H2O(aq) H3O+(aq) + A-(aq) [HA]o pH  pKa -log10 [A-]o Henderson-Hasselbach equation

[CH3COOH] [H3O+] [CH3COO-] Start 1.0 0 0.5 Change -y +y +y Equilibrium 1.0-y y 0.5+y [CH3COOH] [H3O+] = Ka [CH3COO-] Henderson-Hasselbach equation pH = -log10(3.6x10-5) = 4.44

[HA]o pH  pKa -log10 [A-]o To prepare a buffer of a given pH, start with an acid with pKa  pH.

[HA]o pH  pKa -log10 [A-]o To prepare a buffer of a given pH, start with an acid with pKa  pH. [HA]o  1 [A-]o

[HA]o pH  pKa -log10 [A-]o To prepare a buffer of a given pH, start with an acid with pKa  pH. [HA]o Log10(1) = 0  1 [A-]o

[HA]o pH  pKa -log10 [A-]o Design a buffer with a pH of 9.20

[NH4+]o pH  pKa -log10 [NH3]o Design a buffer with a pH of 9.20 pKa NH4+ = 9.25

[NH4+]o pH  pKa -log10 [NH3]o [NH4+] log10 = pKa - pH [NH3] Design a buffer with a pH of 9.20 pKa NH4+ = 9.25

[NH4+]o pH  pKa -log10 [NH3]o [NH4+] log10 = 9.25 - 9.20 = 0.05 [NH3] Design a buffer with a pH of 9.20 pKa NH4+ = 9.25

[NH4+]o pH  pKa -log10 [NH3]o [NH4+] log10 = 9.25 - 9.20 = 0.05 [NH3] [NH4+] = 100.05 [NH3]

[NH4+]o pH  pKa -log10 [NH3]o [NH4+] log10 = 9.25 - 9.20 = 0.05 [NH3] [NH4+] = 100.05 = 1.12 [NH3]

[NH4+]o pH  pKa -log10 [NH3]o [NH4+] = 100.05 = 1.12 [NH3] [NH4Cl] = 1.12 M [NH3] = 1.0 M

[NH4+]o pH  pKa -log10 [NH3]o [NH4+] = 100.05 = 1.12 [NH3] [NH4Cl] = 1.12 M [NH4Cl] = 0.112 M [NH3] = 1.0 M [NH3] = 0.10 M

Buffers in biological systems

Buffers in biological systems Human blood pH 7.35 - 7.45

Buffers in biological systems Human blood pH 7.35 - 7.45 Buffer system Carbonic acid - hydrogen carbonate

Buffer system Carbonic acid - hydrogen carbonate H3O+(aq) + HCO3-(aq) H2CO3(aq) + H2O(l) CO2(g) + H2O(l)

H3O+(aq) + HCO3-(aq) H2CO3(aq) + H2O(l) CO2(g) + H2O(l) pKa1 H2CO3 = 6.1 Human blood pH 7.35 - 7.45

H3O+(aq) + HCO3-(aq) H2CO3(aq) + H2O(l) CO2(g) + H2O(l) pKa1 H2CO3 = 6.1 Human blood pH 7.35 - 7.45 [HCO3-] Must significantly differ from 1 [H2CO3]

[HCO3-] calculate [H2CO3]

[HCO3-] calculate [H2CO3] [A-]o pH  pKa +log10 [HA]o Henderson-Hasselbach equation

[HCO3-] calculate [H2CO3] [A-]o pH  pKa +log10 [HA]o [A-]o log10 = pH - pKa [HA]o

pH = 7.4 [HCO3-] calculate [H2CO3] pKa = 6.1 [A-]o log10 = pH - pKa [HA]o

pH = 7.4 [HCO3-] calculate [H2CO3] pKa = 6.1 [A-]o log10 = pH - pKa [HA]o [A-]o log10 = 7.4 - 6.1 = 1.3 [HA]o

pH = 7.4 [HCO3-] calculate [H2CO3] pKa = 6.1 [A-]o log10 = pH - pKa [HA]o [A-]o log10 = 7.4 - 6.1 = 1.3 [HA]o 101.3 = 19.95

pH = 7.4 [HCO3-] calculate [H2CO3] pKa = 6.1 [A-]o log10 = pKb - pH [HA]o [A-]o log10 = 7.9 - 7.4 = 0.4 [HA]o [A-]o  20 [HA]o

pH = 7.4 [HCO3-] calculate [H2CO3] pKa = 6.1 [A-]o log10 = pKb - pH [HA]o [A-]o log10 = 7.9 - 7.4 = 0.4 [HA]o [HCO3-] = 0.024 M [A-]o  20 [H2CO3] = 0.0012 M [HA]o

[HCO3-] = 0.024 M [H2CO3] = 0.0012 M High acid capacity Low base capacity

H3O+(aq) + HCO3-(aq) H2CO3(aq) + H2O(l) CO2(g) + H2O(l) pH < 7.35 acidosis

H3O+(aq) + HCO3-(aq) H2CO3(aq) + H2O(l) CO2(g) + H2O(l) pH < 7.35 acidosis Failure to eliminate CO2

H3O+(aq) + HCO3-(aq) H2CO3(aq) + H2O(l) CO2(g) + H2O(l) pH < 7.35 acidosis Failure to eliminate CO2 Heavy exercise

H3O+(aq) + HCO3-(aq) H2CO3(aq) + H2O(l) CO2(g) + H2O(l) pH > 7.45 alkalosis

H3O+(aq) + HCO3-(aq) H2CO3(aq) + H2O(l) CO2(g) + H2O(l) pH > 7.45 alkalosis Eliminating too much CO2 - hyperventilation

H3O+(aq) + HCO3-(aq) H2CO3(aq) + H2O(l) CO2(g) + H2O(l) pH > 7.45 alkalosis Eliminating too much CO2 - hyperventilation Vomiting - removes acid from stomach