Strong Acid-Strong Base Titrations



 

Learning Objective

  • Calculate the concentration of an unknown strong acid given the amount of base necessary to titrate it.

Key Points

    • An acid-base titration is used to determine the unknown concentration of an acid or base by neutralizing it with an acid or base of known concentration.
    • Neutralization is the reaction between an acid and a base, producing a salt and a neutralized base.
    • A strong acid yields a weak conjugate base (A), so a strong acid is also described as an acid whose conjugate base is a much weaker base than water.

Terms

  • strong baseA strong base is a basic chemical compound that is able to deprotonate very weak acids in an acid-base reaction. Common examples of strong bases are the hydroxides of alkali metals and alkaline earth metals, such as NaOH and Ca(OH)2. Very strong bases are even able to deprotonate very weakly acidic C–H groups in the absence of water.
  • strong acidA strong acid is one that completely ionizes (dissociates) in water; in other words, one mole of a strong acid (HA) dissolves in water yielding one mole of H+ and one mole of the conjugate base, A−.
  • titrationThe determination of the concentration of some substance in a solution by slowly adding measured amounts of some other substance (normally using a burette) until a reaction is shown to be complete—for instance, by the color change of an indicator.
  • stoichiometryThe calculation of relative quantities or reactants and products in chemical reactions.

An acid-base titration is used to determine the unknown concentration of an acid or base by neutralizing it with an acid or base of known concentration. Using the stoichiometry of the reaction, the unknown concentration can be determined. It makes use of the neutralization reaction that occurs between acids and bases and the knowledge of how acids and bases will react if their formulas are known.

Stages of a Strong Acid-Strong Base Titration

A strong acid-strong base titration is performed using a phenolphthalein indicator. Phenolphtalein is chosen because it changes color in a pH range between 8.3 – 10. It will appear pink in basic solutions and clear in acidic solutions. In the case of a strong acid-strong base titration, this pH transition would take place within a fraction of a drop of actual neutralization, since the strength of the base is high.

The addition of reactants is done from a burette. The reactant of unknown concentration is deposited into an Erlenmeyer flask and is called the analyte. The other reactant of known concentration remains in a burette to be delivered during the reaction. It is known as the titrant. The indicator—phenolphthalein, in this case—has been added to the analyte in the Erlenmeyer flask.

TitrationTitration of an acid-base system using phenolphthalein as an indicator.

Neutralization is the reaction between an acid and a base, producing a salt and neutralized base. For example, hydrochloric acid and sodium hydroxide form sodium chloride and water:

[latex]HCl (aq) + NaOH (aq) \rightarrow H_2O (l) + NaCl (aq)[/latex]

Neutralization is the basis of titration. A pH indicator shows the equivalence point—the point at which the equivalent number of moles of a base have been added to an acid. It is often wrongly assumed that neutralization should result in a solution with pH 7.0; this is only the case in a strong acid and strong base titration.

Example:

What is the unknown concentration of a 25.00 mL HCl sample that requires 40.00 mL of 0.450 M NaOH to reach the equivalence point in a titration?

[latex]HCl (aq) + NaOH (aq) \rightarrow H_2O (l) + NaCl (aq)[/latex]

Step 1: First calculate the number of moles of NaOH added during the titration.

[latex]0.450 \frac{moles}{L} NaOH \times0.0400 L = 0.018\ moles\ NaOH[/latex]

Step 2: Use stoichiometry to figure out the moles of HCl in the analyte.

The mole ratio between HCl and NaOH in the balanced equation is 1:1.

[latex]0.018 \ moles \ NaOH \times \frac{1\ mole \ HCl}{1\ mole\ NaOH} = 0.018 \ moles \ HCl[/latex]

Step 3: Calculate the molar concentration of HCL in the 25.00 mL sample.

Molarity of HCl = [latex]\frac {0.018 \ moles \ HCl}{0.025 \ L \ HCl} = 0.72 \ Molar \ HCl[/latex]