CH302_022411 - Principles of Chemistry II © Vanden Bout...

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Unformatted text preview: Principles of Chemistry II © Vanden Bout Today Titration determining something about an unknown by reacting it with a known solution Polyprotic Acids Principles of Chemistry II © Vanden Bout Titration Why do a titration. You have a solution with an unknown property Unknown Concentration? Unknown Ka (Kb)? Both. Slowly neutralize the solution by adding a strong base (acid) monitor the pH with each addition Principles of Chemistry II © Vanden Bout Neutralize frst Then look at the neutralization From last class equilibrium imagine a 100 mL solution with 0.1 M Acetic Acid (initial .01 moles) we add 10 mL oF 0.1M NaOH in each titration step (add 0.001 moles) HA OH- Initial (moles) AFter Neutralization Equilibrium pH Volume (L) 0.010 0.000 0.010 0.10 2.87 0.11 3.79 A- HA OH- A- 0.001 0.00 0.00 0.010 0.000 0.009 0.000 0.000 0.001 Principles of Chemistry II © Vanden Bout Neutralize frst Then look at the neutralization From last class equilibrium imagine a 100 mL solution with 0.1 M Acetic Acid (initial .01 moles) we add 10 mL oF 0.1M NaOH in each titration step (add 0.001 moles) HA OH- Initial (moles) AFter Neutralization Equilibrium pH Volume (L) 0.010 0.000 0.010 0.10 2.87 0.11 3.79 0.009 0.12 4.15 A- HA OH- A- 0.001 0.001 0.00 0.00 0.00 0.010 0.000 0.009 0.008 0.000 0.000 0.000 0.001 0.002 Principles of Chemistry II © Vanden Bout Neutralize frst Then look at the neutralization From last class equilibrium imagine a 100 mL solution with 0.1 M Acetic Acid (initial .01 moles) we add 10 mL oF 0.1M NaOH in each titration step (add 0.001 moles) HA OH- Initial (moles) AFter Neutralization Equilibrium pH Volume (L) 0.010 0.000 0.010 0.10 ......... 2.87 0.11 3.79 0.009 0.12 4.15 A- HA OH- A- 0.001 0.001 0.00 0.00 0.00 0.010 0.000 0.009 0.008 0.000 0.000 0.000 0.001 0.002 0.006 0.001 0.005 0.005 0.000 0.005 0.15 4.75 Principles of Chemistry II © Vanden Bout Neutralize frst Then look at the neutralization From last class equilibrium imagine a 100 mL solution with 0.1 M Acetic Acid (initial .01 moles) we add 10 mL oF 0.1M NaOH in each titration step (add 0.001 moles) HA OH- Initial (moles) AFter Neutralization Equilibrium pH Volume (L) 0.010 0.000 0.010 0.10 ......... 2.87 0.11 3.79 0.009 0.12 4.15 A- HA OH- A- 0.001 0.001 0.00 0.00 0.00 0.010 0.000 0.009 0.008 0.000 0.000 0.000 0.001 0.002 0.006 0.001 0.005 0.005 0.000 0.005 0.15 ......... 0.001 0.001 0.009 0.000 0.000 0.010 0.15 4.75 8.78 Principles of Chemistry II © Vanden Bout Neutralize frst Then look at the neutralization From last class equilibrium imagine a 100 mL solution with 0.1 M Acetic Acid (initial .01 moles) we add 10 mL oF 0.1M NaOH in each titration step (add 0.001 moles) HA OH- Initial (moles)...
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This note was uploaded on 03/27/2011 for the course CHEM 302 taught by Professor Mccord during the Spring '10 term at University of Texas.

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CH302_022411 - Principles of Chemistry II © Vanden Bout...

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