In the Classroom edited by
Applications and Analogies
Ron DeLorenzo
Canadian Tire Money: An Analogy for Use When Discussing Weak Acid–Strong Base Titrations
Middle Georgia College Cochran, GA 31014
Arthur M. Last Department of Chemistry, University College of the Fraser Valley, Abbotsford, BC, Canada V2S 7M9;
[email protected] Past experience has indicated that many of my students respond positively to the use of monetary analogies to assist them in understanding some of the concepts that are generally encountered in a first-year university chemistry course (1, 2). Recently, I resorted to using another of these analogies while discussing the changes in pH that occur during the titration of a weak acid, such as acetic acid, with a strong base. Weak Acid–Strong Base Titration Problem We were discussing the type of problem that is typically found in introductory chemistry textbooks, specifically the titration of 50.0 mL of 0.100 M acetic acid with 0.100 M sodium hydroxide, and some students were having difficulty in calculating the amount (i.e., number of moles) of each species present at the half-equivalence point. The amount of each species is found prior to determining their respective concentrations and subsequently substituting these concentrations into the Ka expression. At the half-equivalence point of the titration, half of the original 5.00 × 10᎑3 moles of acetic acid initially present has been converted to acetate ions: CH3CO2H(aq) + OH−(aq)
CH3CO2−(aq) + H2O(l)
However, because the acetate ion is the conjugate base of a weak acid, a proportion of these ions will be hydrolyzed back to undissociated acetic acid: CH3CO2−(aq) + H2O(l)
CH3CO2H(aq) + OH−(aq)
If the amount of acetate ion hydrolyzed back to undissociated acetic acid is x moles when equilibrium is established at the half-equivalence point, the amount of acetate ion present will be (2.50 × 10᎑3 − x) moles and the amount of acetic acid will be (2.50 × 10᎑3 + x) moles. It is usual to assume that x
