Teaching Bufferingby Comparing Observed and Expected Robert J. Kosinski C. Kaighn Morlok Clemson University Clemson, South Carolina
Why Buffering? • Biologically important, but a difficult topic for students to understand. • A simple example of homeostasis and how homeostatic mechanisms can be overwhelmed. • Lends itself to problem-solving and scientific deduction from data.
Why Observed and Expected? • Science tests observations against expected results if some hypothesis is true (e.g., null hypothesis of no treatment differences). • Formulation of the expected outcome is a key test to see if a student understands an experiment.
What Is a Buffer? • Minimizes changes in pH. • Absorbs hydrogen ions when acid is added and gives up hydrogen ions when base is added. • In other words, if there is no buffering and I add 100 H+ ions, I might see almost all of them in the solution. • If there is buffering, I might see only one H+. The rest have been absorbed.
The Two Components Buffers consist of a basic component (absorbs H+, buffers against acids) and an acid component (releases H+, buffers against bases).
The Phosphate Buffer System • Acid component = monobasic phosphate (H2PO4-) H2PO4- + OH- HPO42- + H2O • Basic component = dibasic phosphate (HPO42-) HPO42- + H+ H2PO4- + H2O
The Spreadsheet • Data entry on sheet 1. • Model of the phosphate buffer system on sheet 2. • Download the spreadsheet (and lab exercise and PowerPoint) from http://biology.clemson.edu/bpc/bp/Lab/110/buffering.html (p. 5).
Exercise A • “Mystery solutions” of either distilled water, H2PO4- alone, HPO42- alone, or equal concentrations of both. • Student team titrates one solution with either acid or base. • Team types data into a data entry spreadsheet. • Whole class determines what each mystery solution is from type of buffering that is shown.
Exercise B • Base doesn’t destroy H2PO4-, it converts H2PO4- to HPO42-. • Acid converts HPO42- to H2PO4-. • As you add acid and base, the acid and basic components of the buffer are constantly converting back and forth.
Three Buffering Curves • 0.01 M H2PO4- and 0.01 M HPO42- • 0.02 M H2PO4-, half-titrated with NaOH • 0.02 M HPO42-, half-titrated with HCl
Exercise B • Base converts H2PO4- to HPO42-; buffer with base added to it will be exhausted sooner when titrated with base but later when titrated with acid. • Acid converts HPO42- to H2PO4-; buffer with acid added to it will be exhausted sooner when titrated with acid but later when titrated with base. • Under these conditions, unmodified buffer should run out when 5 mL of titrant has been added.
Exercise B • More ambitious than Exercise A. • Every team titrates balanced buffer with both acid and base. • Team then adds 0, 2, 4, 6, or 8 mL of either acid or base to remaining 100 mL. • Exchanges bottles with another team. • Team titrates this with both acid and base. • Determines what was added and how much, and states this to the class. • Team that modified the bottle tells them whether they were right or wrong.
Acid or Base, and How Much? 2 mL base/50 mL
Acid or Base, and How Much? 3 mL acid/50 mL
Acid or Base, and How Much? 1 mL base/50 mL
Acid or Base, and How Much? 1 mL acid/50 mL
Exercise B Disadvantages • More complicated to explain. • Every team does four titrations instead of one (two for unmodified buffer and two for modified). • They have to add the secret amount of acid or base and exchange bottles. • Sometimes results are ambiguous. • Team might be embarrassed if they’re wrong about what was added.
Exercise B Advantages • Teaches a more complex lesson about interconversion of buffer components. • Demands more understanding. • Challenge of making a public statement of conclusions increases engagement. • If they’re right, they’re rewarded a little public recognition.
Download from… http://biology.clemson.edu/bpc/bp/Lab/110/buffering.html (p. 5).