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Titration Equilibria for AP Chemistry

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By — McGraw-Hill Professional
Updated on Feb 9, 2011

Practice problems for these concepts can be found at:

An acid–base titration is a laboratory procedure commonly used to determine the concentration of an unknown solution. A base solution of known concentration is added to an acid solution of unknown concentration (or vice versa) until an acid–base indicator visually signals that the end point of the titration has been reached. The equivalence point is the point at which a stoichiometric amount of the base has been added to the acid. Both chemists and chemistry students hope that the equivalence point and the end point are close together.

If the acid being titrated is a weak acid, then there are equilibria which will be established and accounted for in the calculations. Typically, a plot of pH of the weak acid solution being titrated versus the volume of the strong base added (the titrant) starts at a low pH and gradually rises until close to the equivalence point, where the curve rises dramatically. After the equivalence point region, the curve returns to a gradual increase. This is shown in Figure 15.3.

In many cases, one may know the initial concentration of the weak acid, but may be interested in the pH changes during the titration. To study the changes one can divide the titration curve into four distinctive areas in which the pH is calculated.

  1. Calculating the initial pH of the weak acid solution is accomplished by treating it as a simple weak acid solution of known concentration and Ka.
  2. As base is added, a mixture of weak acid and conjugate base is formed. This is a buffer solution and can be treated as one in the calculations. Determine the moles of acid consumed from the moles of titrant added—that will be the moles of conjugate base formed. Then calculate the molar concentration of weak acid and conjugate base, taking into consideration the volume of titrant added. Finally, apply your buffer equations.
  3. At the equivalence point, all the weak acid has been converted to its conjugate base. The conjugate base will react with water, so treat it as a weak base solution and calculate the [OH] using Kb. Finally, calculate the pH of the solution.
  4. After the equivalence point, you have primarily the excess strong base that will determine the pH.

Titration Equilibria

Let's consider a typical titration problem. A 100.0 mL sample of 0.150 M nitrous acid (pKa = 3.35) was titrated with 0.300 M sodium hydroxide. Determine the pH of the solution after the following quantities of base have been added to the acid solution:

  1. 0.00 mL
  2. 25.00 mL
  3. 49.50 mL
  4. 50.00 mL
  5. 55.00 mL
  6. 55.00 mL
  1. 0.00 mL. Since no base has been added, only HNO2 is present. HNO2 is a weak acid, so this can only be a Ka problem.
  2. Quadratic needed: x2 + 4.47 × 10–5x – 6.70 × 10–5 = 0

    (extra sig. figs.)

      x = [H+ ] = 8.0 × 10–3M pH = 2.10
  3. 25.00 mL. Since both an acid and a base are present (and they are not conjugates), this must be a stoichiometry problem. Stoichiometry requires a balanced chemical equation and moles.
  4. Na+ + NO2 could be written as NaNO2, but the separated ions are more useful.

    Based on the stoichiometry of the problem, and on the moles of acid and base, NaOH is the limiting reagent.

    The stoichiometry part of the problem is finished.

    The solution is no longer HNO2> and NaOH, but HNO2 and NO2 (a conjugate acid–base pair).

    Since a CA/CB pair is present, this is now a buffer problem, and the Henderson–Hasselbalch equation may be used.

      pH + pKa + log (CB/CA) = 3.35 – log (0.00750/0.00750) = 3.35

    Note the simplification in the CB/CA concentrations. Both moles are divided by exactly the same volume (since they are in the same solution), so the identical volumes cancel.

  5. 49.50 mL. Since both an acid and a base are present (and they are not conjugates), this must be a stoichiometry problem again. Stoichiometry requires a balanced chemical equation and moles.
  6. Based on the stoichiometry of the problem, and on the moles of acid and base, NaOH is the limiting reagent.

    The stoichiometry part of the problem is finished.

    The solution is no longer HNO2 and NaOH, but HNO2 and NO2 (a conjugate acid–base pair).

    Since a CA/CB pair is present, this is now a buffer problem, and the Henderson–Hasselbalch equation may be used.

      pH = pKa + log (CB/CA) = 3.35 + log (0.0148/0.0002) = 5.2
  7. 50.00 mL. Since both an acid and a base are present (and they are not conjugates), this must be a stoichiometry problem. Stoichiometry requires a balanced chemical equation and moles.
  8. Based on the stoichiometry of the problem, and on the moles of acid and base, both are limiting reagents.

    The stoichiometry part of the problem is finished.

    The solution is no longer HNO2 and NaOH, but an NO2 solution (a conjugate base of a weak acid).

    Since the CB of a weak acid is present, this is a Kb problem.

  9. 55.00 mL. Since both an acid and a base are present (and they are not conjugates), this must be a stoichiometry problem. Stoichiometry requires a balanced chemical equation and moles.
  10. Based on the stoichiometry of the problem, and on the moles of acid and base, the acid is now the limiting reagent.

    The strong base will control the pH.

      [OH] = 0.0015 mol/0.155 L = 9.7 × 10–3 M

    The stoichiometry part of the problem is finished.

    Since this is now a solution of a strong base, it is now a simple pOH/pH problem.

        pOH = – log 9.7 × 10–3 = 2.01
      pH = 14.00 – pOH = 14.00 – 2.01 = 11.99
  11. 75.00 mL. Since both an acid and a base are present (and they are not conjugates), this must be a stoichiometry problem. Stoichiometry requires a balanced chemical equation and moles.
  12. Based on the stoichiometry of the problem, and on the moles of acid and base, the acid is now the limiting reagent.

    The strong base will control the pH.

      [OH] = 0.0075 mol/0.175 L = 4.3 × 10–2 M

    The stoichiometry part of the problem is finished.

    Since this is now a solution of a strong base, it is now a simple pOH/pH problem.

      pOH = –log 4.3 × 10–2 = 1.37
      pH = 14.00 – pOH = 14.00 – 1.37 = 12.63

Practice problems for these concepts can be found at:

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