CH302_022411 - Today Titration determining something about...

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Principles of Chemistry II © Vanden Bout Today Titration determining something about an unknown by reacting it with a known solution Polyprotic Acids
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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
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Principles of Chemistry II © Vanden Bout Neutralize first 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
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Principles of Chemistry II © Vanden Bout Neutralize first 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
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Principles of Chemistry II © Vanden Bout Neutralize first 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
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Principles of Chemistry II © Vanden Bout Neutralize first 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
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Principles of Chemistry II © Vanden Bout Neutralize first 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 .........
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