CHEM 1011 pH and Buffer Solutions. Brønsted-Lowry Theory Acid-proton donor Base-proton acceptor.

CHEM 1011

pH and Buffer Solutions

pH and Buffer Solutions

Brønsted-Lowry Theory

• Acid-proton donor

• Base-proton acceptor

pH and Buffer Solutions

For example,

HCl + H2O ↔ H3O+ + Cl-

For the forward reaction,

• HCl – proton donor - acid

• H2O – proton acceptor - base

pH and Buffer Solutions

For example,

HCl + H2O ↔ H3O+ + Cl-

For the reverse reaction,

• H3O+ – proton donor - acid

• Cl- – proton acceptor - base

pH and Buffer Solutions

According to Brønsted-Lowry Theory,

every acid-base reaction creates

conjugate acid-base pair.

pH and Buffer Solutions

For example,

HCl + H2O ↔ H3O+ + Cl-

HCl - acid, Cl- - conjugate base

H2O - base, H3O+ - conjugate acid

pH and Buffer Solutions

Acids can be monoprotic, diprotic, or triprotic.

pH and Buffer Solutions

• Monoprotic – donate one proton

e.g., HCl, HNO3, HCOOH, CH3COOH

HCl + H2O ↔ H3O+ + Cl-

HNO3 + H2O ↔ H3O+ + NO3-

HCOOH + H2O ↔ H3O+ + HCOO-

CH3COOH + H2O ↔ H3O+ + CH3COO-

pH and Buffer Solutions

• Diprotic – donate two protons

e.g, H2SO4, H2CO3

H2SO4 + H2O ↔ H3O+ + HSO4-

HSO4- + H2O ↔ H3O+ + SO4

2-

H2CO3 + H2O ↔ H3O+ + HCO3-

HCO3- + H2O ↔ H3O+ + CO3

2-

pH and Buffer Solutions

• Triprotic – donate three protons

e.g, H3PO4

H3PO4 + H2O ↔ H3O+ + H2PO4-

H2PO4- + H2O ↔ H3O+ + HPO4

2-

HPO42- + H2O ↔ H3O+ + PO4

3-

pH and Buffer Solutions

Water can self ionize.

H2O + H2O ↔ H3O+ + OH-

[H3O+] = 1 x 10-7 moles/L

[OH-] = 1 x 10-7 moles/L

Kw = [H3O+] [OH-] = [1 x 10-7] [1 x 10-7] = 1 x 10-14

pH and Buffer Solutions

• Kw same for all aqueous solutions.• Can calculate [OH-] for solution of strong

acid.

• e.g., 0.05 M HNO3

[H3O+] = 5 x 10-2 moles/L

[OH-] = Kw / [H3O+][OH-] = 1 x 10-14 / 5 x 10-2 = 2 x 10-13 mol/L

pH and Buffer Solutions

• Can also calculate [H3O+] of solution of strong base.

• e.g., 0.04 M NaOH

[OH-] = 4 x 10-2 mol/L

[H3O+] = Kw / [OH-]

[H3O+] = 1 x 10-14 / 4 x 10-2 = 2.5 x 10-13

pH and Buffer Solutions

• pH scale goes from 0 to 14

• pH < 7 acidic

• pH = 7 neutral

• pH > 7 basic

pH and Buffer Solutions

• pH = - log [H3O+]

• e.g., pH of 0.05 M HNO3 (a strong acid)

pH = - log [5 x 10-2] = 1.12

pH and Buffer Solutions• The pOH of the solution of a strong base

can also be calculated and used to determine pH.

• e.g., 0.04 M NaOH

pOH = -log[OH-] = -log[4 x 10-2] = 1.10

pH + pOH = pKw = 14pH = 14 - pOH = 14 - 1.10 = 12.9

pH and Buffer Solutions

• You will use pH paper to measure the pHs of your solutions. Although a pH meter would yield more accurate results, it is unnecessary.

pH and Buffer Solutions

Buffer solutions

• Resist changes of pH following the addition of acid or base.

• Combinations of weak acids and their conjugate bases.

pH and Buffer Solutions

• Buffered aspirin

• Blood buffers prevent acidosis and alkalosis.

H2CO3 + H2O ↔ HCO3- + H3O+

H2PO4- + H2O ↔ HPO4

2- + H3O+

pH and Buffer Solutions

• The Henderson-Hasselbalch equation can be used to calculate the pH of a buffer system.

pH = pKa + log [A-] / [HA]

• pH = pKa when [A-] = [HA]

• Buffer activity continues as long as

log [A-] / [HA] ratio kept constant

Conclusions and Calculations

• Answer Post-Lab questions 1 and 3 based upon your observations (omit 2).

Conclusions and Calculations

• For Post-Lab question 4, the HCl reacts with the CH3COO-, and the NaOH reacts with the H3O+

• An acetic acid-acetate buffer system would be represented by

CH3COOH + H2O ↔ CH3COO- + H3O+

Conclusions and Calculations• For an acetic acid-acetate buffer system

pH = pKa + log [CH3COO-] / [CH3COOH]

pKa = [CH3COO-][H3O+] / [CH3COOH] = 4.75

pH = 4.75 + log [CH3COO-] / [CH3COOH]

• Use this relationship when answering Pre-Lab question 5 and Post-Lab questions 5 a. and b.

Conclusions and Calculations• For Post-Lab questions 5 c. and d.,

H2CO3 + H2O ↔ HCO3- + H3O+

pH = pKa + log [HCO3-] / [H2CO3]

pKa = [HCO3-][H3O+] / [H2CO3] = 6.37

pH = 6.37 + log [HCO3-] / [H2CO3]