Unit 8.4 Notes

Acid-Base Reactions and Buffers

Determining pH with Strong Acid and Strong Base Mixtures

When mixing strong acids and strong bases in unequal amounts, the dominant factor in determining the resulting solution's pH is the excess reagent after neutralization. The overall reaction between hydrogen ions (H+) and hydroxide ions (OH-) leads to the formation of water (H2O):

H+ (aq) + OH- (aq) → H2O (l)

Steps to Determine the pH:

1. Calculate the Moles: Find the number of moles of both acid and base present in the solution.

2. Identify Limiting Reagent: Determine which reactant is in excess by subtracting the smaller number of moles from the larger number.

3. Calculate Molarity of Excess Reagent: Divide the moles of the excess reagent by the total volume of the combined solution.

4. Determine pH: Use the pH square (previously covered) to find the pH value based on the concentration of the excess reagent.

Example Calculation:

What is the pH of a solution composed of:

• 500.0 mL of 0.250 M sodium hydroxide (NaOH)

• 400.0 mL of 0.200 M hydrochloric acid (HCl)?

Determining pH with Strong Acid/Base and Weak Acid/Base Mixtures

In reactions where weak acids are mixed with strong bases (and vice versa), different outcomes can occur:

1. Excess Weak Acid/Base (Buffer Formation): If the weak acid or base is in excess, a buffer solution is formed, containing high concentrations of both the conjugate acid and base. Buffers maintain stable pH levels despite the addition of strong acids or bases. The pH of the buffer can be calculated using the Henderson-Hasselbalch equation.

2. Excess Strong Acid or Base: If the strong acid or base is in excess, the resulting pH is determined by the concentration of the remaining strong acid or base, following the same calculations as in strong acid/strong base reactions.

3. Equimolar Weak Acid & Strong Base or Weak Base & Strong Acid (Equivalence Point): If equivalent moles of weak acid and strong base (or weak base and strong acid) are mixed, the pH at this equivalence point depends on the properties of the conjugate acid or base formed. The pH can be determined as follows:

   • Calculate the initial molarity of the conjugate acid/base using the moles and the total volume.

   • Determine the K value using: Kb = Kw/Ka (for conjugate bases) or Ka = Kw/Kb (for conjugate acids).

   • Use an ICE (Initial, Change, Equilibrium) table to establish equilibrium concentrations.

   • Solve for equilibrium concentrations of OH- or H3O+ with the respective reactions:

     • Weak Acid Reaction: A- (aq) + H2O (l) ⇌ HA (aq) + OH- (aq)

     • Weak Base Reaction: HB+ (aq) + H2O (l) ⇌ B (aq) + H3O+ (aq)

Weak Acid / Weak Base Reactions

When combining weak acids and weak bases, the resulting solution will react to achieve equilibrium:

HA (aq) + B (aq) ⇌ A- (aq) + HB+ (aq)

Example Problems

• Net Ionic Equations: Write the net ionic equations for reactions involving nitrous acid (HNO2) with the following:

  • A) Ammonia (NH3)

  • B) Potassium hydroxide (KOH)

  • C) Water (H2O)

• Identifying Acid/Base Situations: For various mixtures of acids and bases:

  1. 200.0 mL of 0.10 M HClO2 with 100.0 mL of 0.20 M Mg(OH)2: Determine strong/weak characteristics and identify excess.

  2. 50.0 mL of 0.50 M HBr with 100.0 mL of 0.25 M LiOH: Analyze the resultant acid/base profile.

  3. 10.0 mL of 1.0 M HI with 10.0 mL of 2.0 M NH3: Classify the strength of the acids and bases.

• Example Calculation of pH: For the scenario of mixing 100.0 mL of 0.100 M acetic acid (HCH3COO) with 100.0 mL of 0.100 M potassium hydroxide (KOH), and given the Ka of acetic acid as 1.8 x 10^-5.

• Example Calculation of pH using Lactic Acid: For a solution containing 90.0 mL of 0.345 M sodium hydroxide (NaOH) mixed with 50.0 mL of 0.123 M lactic acid (HC3H5O3), and considering the Ka of lactic acid as 1.38 x 10^-4.

• pH Calculation of Hydrazonic Acid Reaction: For a solution with 100.0 mL of 0.900 M hydrazoic acid (HN3) and 50.0 mL of 0.300 M sodium hydroxide (NaOH) considering the dissociation reaction.