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Published byΑλάστωρ Μπουκουβαλαίοι Modified over 6 years ago
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Simplify Titration Curves that Buffer With Visual Analysis
Amy Zitzelberger Hazel Park High School Hazel Park, MI Jamie Benigna The Roeper School Birmingham, MI
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Graph originally from: MIT Open Classroom
HA A- H+ HA HA Point A. Before titration begins: Ka = x2 / [HA] where x = [H+]
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HA OH- A- Na+ H2O Early Buffering Region. Before ½ equivalence point when buffering occurs: pH = pKa + log [A-] / [HA] where [A-] = OH- used so far in moles, and [HA] is remaining moles.
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HA A- OH- H2O A- Na+ Na+ Early Buffering Region. Before ½ equivalence point when buffering occurs: pH = pKa + log [A-] / [HA] where [A-] = OH- used so far in moles, and [HA] is remaining moles.
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Point C: ½ equivalence pt: pH = pKa [A-] = [HA], and log(1) = 0
OH- A- Na+ OH- Na+ A- Na+ H2O Point C: ½ equivalence pt: pH = pKa [A-] = [HA], and log(1) = 0
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HA A- OH- A- Na+ OH- Na+ A- Na+ OH- A- Na+ Later Buffering Region. After ½ equivalence point when buffering occurs: pH = pKa + log [A-] / [HA] where [A-] = OH- used so far in moles, and [HA] is remaining moles.
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HA A- OH- A- Na+ OH- Na+ A- Na+ A- OH- OH- A- Na+ Na+ Later Buffering Region. After ½ equivalence point when buffering occurs: pH = pKa + log [A-] / [HA] where [A-] = OH- used so far in moles, and [HA] is remaining moles.
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HA Na+ A- OH- OH- A- H2O OH- HA A- Na+ OH- Na+ A- Na+ A- OH- OH- A- Na+ Na+ Point D. Equivalence point: Kb = x2 / [A-] where x = [OH-], and [A-] = moles OH- used over total volume.
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HA Na+ OH- Na+ A- OH- OH- A- A- Na+ OH- Na+ A- Na+ A- OH- OH- A- Na+ Na+ Beyond the Equivalence Pt. pOH = ̶ log[OH-] where the OH- is only the moles of excess OH- divided by the total volume.
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HA Na+ OH- Na+ A- OH- OH- A- A- Na+ OH- Na+ A- Na+ A- OH- OH- OH- A- Na+ Na+ Na+ Beyond the Equivalence Pt. pOH = ̶ log[OH-] where the OH- is only the moles of excess OH- divided by the total volume.
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To Summarize E A: Before the Equiv. Pt. B: Buffering Region C: ½ way to Equiv. Pt. D: Equivalence Point D E: After Equiv. Pt. B C A
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A For example: 1) A solution of 0.20 M nitrous acid. To Summarize E
A: Before the Equiv. Pt. B: Buffering Region C: ½ way to Equiv. Pt. D: Equivalence Point D E: After Equiv. Pt. A B C A For example: 1) A solution of M nitrous acid.
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To Summarize E A: Before the Equiv. Pt. B: Buffering Region C: ½ way to Equiv. Pt. D: Equivalence Point D E: After Equiv. Pt. E B C A 25.0 mL of 0.10 M nitrous acid with 26.5 mL of 0.10 M NaOH added to it.
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D 25.0 mL of 0.30 M KOH is added to 25.0 mL of 0.30 M HF. To Summarize
A: Before the Equiv. Pt. B: Buffering Region C: ½ way to Equiv. Pt. D: Equivalence Point D E: After Equiv. Pt. D B C A 25.0 mL of 0.30 M KOH is added to 25.0 mL of 0.30 M HF.
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To Summarize E A: Before the Equiv. Pt. B: Buffering Region C: ½ way to Equiv. Pt. D: Equivalence Point D E: After Equiv. Pt. C B C A A titration that has 25.0 mL of 1.00 M acetic acid with 25.0 mL of 0.50 M NaOH.
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To Summarize E A: Before the Equiv. Pt. B: Buffering Region C: ½ way to Equiv. Pt. D: Equivalence Point D E: After Equiv. Pt. B B C A When 30.0 mL of 0.20M propanoic acid has been titrated with 20.0 mL of 0.20 M NaOH.
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