Chapter 16 Flashcards

16.1-16.3 Common Ion Effect, Buffer Solutions

Common Ion Effect

  • The presence of a common ion suppresses the ionization of a weak acid or weak base.
  • This phenomenon can be understood through Le Chatelier’s Principle.
  • For example, consider what happens to the H+H^+ concentration when CH<em>3COONaCH<em>3COONa is added to a solution of CH</em>3COOHCH</em>3COOH.

Example Calculation: Common Ion Effect

  • Calculate the pH of a 0.30 M CH<em>3COOHCH<em>3COOH solution. Also, calculate the pH of a solution containing 0.30 M CH</em>3COOHCH</em>3COOH and 0.20 M CH<em>3COONaCH<em>3COONa. The K</em>aK</em>a of CH3COOHCH_3COOH is 1.8×10−51.8 \times 10^{-5}.

Henderson-Hasselbalch Equation

  • The Henderson-Hasselbalch equation simplifies solutions involving the common ion effect.

  • It assumes that the [H+][H^+] is small compared to the initial concentration of the acid/conjugate base.

    pH=pKa+log[A−][HA]pH = pK_a + log \frac{[A^-]}{[HA]}

Example Calculation : Henderson-Hasselbalch Equation

  • What is the pH of a solution containing 0.5 M NH<em>3NH<em>3 and 0.2 M NH</em>4ClNH</em>4Cl? The K<em>bK<em>b for NH</em>3=1.8×10−5NH</em>3 = 1.8 \times 10^{-5}. To solve for pH, first find pOH with the Henderson-Hasselbalch equation, then convert
    pOH=pK<em>b+log[NH</em>4+][NH3]pOH= pK<em>b + log\frac{[NH</em>4^+]}{[NH_3]}

Buffer Solutions

  • Buffers are chemical systems that resist pH changes by neutralizing added acid or base.
  • A buffer contains significant amounts of both a weak acid and its conjugate base (or a weak base and its conjugate acid).
    • The weak acid neutralizes added base.
    • The conjugate base neutralizes added acid.
  • Examples:
    • CH<em>3COONa/CH</em>3COOHCH<em>3COONa/CH</em>3COOH
    • KH<em>2PO</em>4/K<em>2HPO</em>4KH<em>2PO</em>4/K<em>2HPO</em>4
    • HCN/KCNHCN/KCN

Action of a Buffer

  • Buffers work through equilibrium. When acid is added, it shifts towards the production of HA. When base is added, it shifts towards the production of A−A^-.

Calculating pH Changes in Buffers

  • A 2.0L buffer solution contains 0.1 mol of HC<em>2H</em>3O<em>2HC<em>2H</em>3O<em>2 and 0.1 mol of KC</em>2H<em>3O</em>2KC</em>2H<em>3O</em>2. The value of K<em>aK<em>a for HC</em>2H<em>3O</em>2HC</em>2H<em>3O</em>2 is 1.8×10−51.8 \times 10^{-5}. Calculate the pH of the buffer:

    • First calculate the pH of the buffer before adding KOH. This can be done using the Henderson-Hasselbalch equation:
      pH=pKa+log[A−][HA]pH = pK_a + log \frac{[A^-]}{[HA]}
  • 0. 010 moles of solid KOH are added to the buffer. Calculate the new pH of the buffer:

    • Then, calculate the new pH of the buffer after adding KOH. Because you are adding a strong base to a weak acid, this will change the concentration. Treat it like a limiting reagent, so subtract the 0.01 moles from the moles of HC<em>2H</em>3O<em>2HC<em>2H</em>3O<em>2 and add it to the moles of KC</em>2H<em>3O</em>2KC</em>2H<em>3O</em>2, then do the Henderson-Hasselbalch equation again with the new values.
  • For comparison, calculate the pH of a 0.01 M solution of just KOH in pure water:

    • First, since [OH−]=0.01M[OH^-] = 0.01M, pOH=−log(0.01)=2pOH=-log(0.01)=2, so pH=14−2=12pH= 14-2=12

Buffer Effectiveness

  • Buffer Capacity: The amount of acid or base it can effectively neutralize.
    • Most effective when the concentrations of acid and conjugate base are equal.
    • Most effective when the concentrations of acid and conjugate base are high.
  • Example: A 1.0L buffer solution is 1.0 M in HF and 0.050 M in NaF. Which action will destroy the buffer?
    • A) adding 0.05 mol of HCl
    • B) adding 0.05 mol of NaF
    • C) adding 0.050 mol of NaOH
      • The correct answer is C) adding 0.050 mol of NaOH because NaOH will neutralize HF, turning it into F−F^-. Because F−F^- is already at 0.050 M, adding 0.050 mol of NaOH will overwhelm the buffering system.
    • D) None of the above

Buffer Range

  • Buffer Range: The pH range over which a particular acid and conjugate base can be effective.
  • The effective range for a buffering system is generally one pH unit on either side of the pKapK_a.

Preparing a Buffer

  • Which acid would you choose to combine with its sodium salt to make a solution buffered at pH 4.25?
    • 1. HClO<em>2HClO<em>2 pK</em>a=1.95pK</em>a = 1.95
    • 2. HNO<em>2HNO<em>2 pK</em>a=3.34pK</em>a = 3.34
    • 3. HCHO<em>2HCHO<em>2 pK</em>a=3.74pK</em>a = 3.74
    • 4. HClOHClO pKa=7.54pK_a = 7.54
      • The best choice would be #3 since it is closest to 4.25
  • For the best choice, calculate the ratio of the conjugate base to the acid required to attain the desired pH.