pH of Weak Acids

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pH of Weak Acids

• AP Chemistry
• Unit 9
• Chapter 14

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Strengths of Acids and Bases

• Strength refers to how much an acid or base
  ionizes in a solution.

STRONG WEAK
Ionize completely (100) Example HCl? H Cl- NaOH ? Na OH- Ionize partially (usually lt10) Example HF ? H F- NH3 H2O ? NH4 OH-
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Relative Dissociation of Acids
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Concentration vs- StrengthHA ? H A-
Ka H A- , therefore larger Ka values
indicate stronger acids. HA
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• What is the pH of a 0.25M HF solution?
• The Ka at 25oC 7.1 x 10-4
• HF is a weak acid, so we cant directly calculate
  the pH from the molarity. We need to set up a
  table as we did for other equilibrium problems
• HF(aq) ? H(aq) F-(aq)
• Initial 0.25M 0 M 0 M
• Equilibrium 0.25-x x x
• Ka HF- 7.1 x
  10-4
• HF
• 7.1 x 10-4 x2
• (0.25-x)

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• 7.1 x 10-4 x2
• (0.25-x)
• and x2 7.1 x 10-4x -1.775 x 10-4 0
• Solving using the quadratic equation
• x 1.30 x 10-2 or 1.37x10-2.
• The negative solution is physically impossible
  because x was set to be the H,
• so x 1.30 x 10-2.
• Substituting in this x into the original formula
  for Ka checks, so
• pH-log(1.30 x 10-2) 1.89

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• An easier method than using the quadratic
  equation is to make the assumption that the
  dissociation of HF is so insignificant that the
  initial concentration is not significantly
  changed at equilibrium.
• Therefore, we assume that (0.25 x) 0.25
• So, 7.1 x 10-4 x2
• 0.25
• 1.78 x 10-4 x2
• therefore x 1.33 x 10-2
• pH - log (1.33 x 10-2) 1.88
• Considering we need to round to 2 significant
  figures, our assumption is valid!

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The 5 Rule

• We can ignore the (-x) in the denominator if the
  acid dissociates less than 5
• Otherwise, we must use the quadratic equation.
• Ionization A-
• HAinitial
• Usually a very low Ka value means we can ignore
  the (-x)

X 100
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