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  • Lecture 23 10/28/05. 50 mL of 0.02 M KOH with 0.1 M HBr Construct curve from 4 points Initial pH X = 0 mL, Y = ? Equivalence point X = V eq, Y = 7 Before
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Lecture 23 10/28/05

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Page 1: Lecture 23 10/28/05. 50 mL of 0.02 M KOH with 0.1 M HBr Construct curve from 4 points Initial pH X = 0 mL, Y = ? Equivalence point X = V eq, Y = 7 Before

Lecture 2310/28/05

Page 2: Lecture 23 10/28/05. 50 mL of 0.02 M KOH with 0.1 M HBr Construct curve from 4 points Initial pH X = 0 mL, Y = ? Equivalence point X = V eq, Y = 7 Before

50 mL of 0.02 M KOH with 0.1 M HBr Construct curve from 4 points

• Initial pH• X = 0 mL, Y = ?

• Equivalence point• X = Veq, Y = 7

• Before Equivalence point• X = ?, Y = ?

• After Equivalence point• X = ?, Y = ?

Page 3: Lecture 23 10/28/05. 50 mL of 0.02 M KOH with 0.1 M HBr Construct curve from 4 points Initial pH X = 0 mL, Y = ? Equivalence point X = V eq, Y = 7 Before

Titration of a weak acid with strong base

• 50 mL of 0.02 M MES• [2-(n-morpholino)ethanesulfonic acid]

• pKa = 6.15

• with 0.100 M NaOH.

Page 4: Lecture 23 10/28/05. 50 mL of 0.02 M KOH with 0.1 M HBr Construct curve from 4 points Initial pH X = 0 mL, Y = ? Equivalence point X = V eq, Y = 7 Before
Page 5: Lecture 23 10/28/05. 50 mL of 0.02 M KOH with 0.1 M HBr Construct curve from 4 points Initial pH X = 0 mL, Y = ? Equivalence point X = V eq, Y = 7 Before
Page 6: Lecture 23 10/28/05. 50 mL of 0.02 M KOH with 0.1 M HBr Construct curve from 4 points Initial pH X = 0 mL, Y = ? Equivalence point X = V eq, Y = 7 Before

50 mL of 0.02 M MES with 0.100 M NaOH.Construct curve from 4 points

• Initial pH• X = 0 mL, Y = ?

• Equivalence point• X = Veq, Y = ?

• Before Equivalence point• X = ½ Veq, Y = pKa

• After Equivalence point• X = ?, Y = ?

Page 7: Lecture 23 10/28/05. 50 mL of 0.02 M KOH with 0.1 M HBr Construct curve from 4 points Initial pH X = 0 mL, Y = ? Equivalence point X = V eq, Y = 7 Before

How does pKa or concentration change the titration curve?

Page 8: Lecture 23 10/28/05. 50 mL of 0.02 M KOH with 0.1 M HBr Construct curve from 4 points Initial pH X = 0 mL, Y = ? Equivalence point X = V eq, Y = 7 Before

• Titration of 10.0 mL of 0.100 M B (base) • pKb1 = 4.00

• pKb2 = 9.00

• with 0.100 M HCl

Page 9: Lecture 23 10/28/05. 50 mL of 0.02 M KOH with 0.1 M HBr Construct curve from 4 points Initial pH X = 0 mL, Y = ? Equivalence point X = V eq, Y = 7 Before

Titration of 10.0 mL of 0.100 M B (base) with 0.100 M HClpKb1 = 4.00 pKb2 = 9.00

Page 10: Lecture 23 10/28/05. 50 mL of 0.02 M KOH with 0.1 M HBr Construct curve from 4 points Initial pH X = 0 mL, Y = ? Equivalence point X = V eq, Y = 7 Before

Finding endpoint with pH electrode

Page 11: Lecture 23 10/28/05. 50 mL of 0.02 M KOH with 0.1 M HBr Construct curve from 4 points Initial pH X = 0 mL, Y = ? Equivalence point X = V eq, Y = 7 Before
Page 12: Lecture 23 10/28/05. 50 mL of 0.02 M KOH with 0.1 M HBr Construct curve from 4 points Initial pH X = 0 mL, Y = ? Equivalence point X = V eq, Y = 7 Before

Titration of H6A with NaOH

Page 13: Lecture 23 10/28/05. 50 mL of 0.02 M KOH with 0.1 M HBr Construct curve from 4 points Initial pH X = 0 mL, Y = ? Equivalence point X = V eq, Y = 7 Before

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