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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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Presentation on theme: "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."— Presentation transcript:

1 Lecture 23 10/28/05

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

3 Titration of a weak acid with strong base 50 mL of 0.02 M MES [2-(n-morpholino)ethanesulfonic acid] pK a = 6.15 with 0.100 M NaOH.

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6 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 = V eq, Y = ? Before Equivalence point X = ½ V eq, Y = pK a After Equivalence point X = ?, Y = ?

7 How does pK a or concentration change the titration curve?

8 Titration of 10.0 mL of 0.100 M B (base) pK b1 = 4.00 pK b2 = 9.00 with 0.100 M HCl

9 Titration of 10.0 mL of 0.100 M B (base) with 0.100 M HCl pK b1 = 4.00 pK b2 = 9.00

10 Finding endpoint with pH electrode

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12 Titration of H 6 A with NaOH

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