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+ + ClFor the forward reaction,
• HCl – proton donor - acid
• H2O – proton acceptor - base
pH and Buffer Solutions
For example,
HCl + H2O ↔ H3O+ + ClFor 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+ + ClHCl - 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+ + ClHNO3 + H2O ↔ H3O+ + NO3HCOOH + H2O ↔ H3O+ + HCOOCH3COOH + H2O ↔ H3O+ + CH3COO-
pH and Buffer Solutions
• Diprotic – donate two protons
e.g, H2SO4, H2CO3
H2SO4 + H2O ↔ H3O+ + HSO4HSO4- + H2O ↔ H3O+ + SO42H2CO3 + H2O ↔ H3O+ + HCO3HCO3- + H2O ↔ H3O+ + CO32-
pH and Buffer Solutions
• Triprotic – donate three protons
e.g, H3PO4
H3PO4 + H2O ↔ H3O+ + H2PO4H2PO4- + H2O ↔ H3O+ + HPO42HPO42- + H2O ↔ H3O+ + PO43-
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
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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 = 14
pH = 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 ↔ HPO42- + 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]