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Lecture 223/17/06 Research Club Meeting  Today at 1:30  TSB 155  Free pizza Research Colloquium  April 4  Deadline March 27.

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Presentation on theme: "Lecture 223/17/06 Research Club Meeting  Today at 1:30  TSB 155  Free pizza Research Colloquium  April 4  Deadline March 27."— Presentation transcript:

1 Lecture 223/17/06 Research Club Meeting  Today at 1:30  TSB 155  Free pizza Research Colloquium  April 4  Deadline March 27

2 Quiz 1) For the titration of 75 mL of 0.15 M HCl with 0.33 M NaOH, fill in the table. x-axis (mL of NaOH) y-axis (pH) Initial point Equivalence point 2) Why does adding water to a buffer solution not change the pH?

3 After Equivalence point (NaOH titrating HF) X-axis Y-axis

4 Building a titration curve via calculations Comparison of a titration of a strong acid vs. a weak acid Region x-axis (mL of acid/base) 0.1 M NaOH y-axis (pH) 100 mL of 0.1 M HCl y-axis (pH) 100 mL of 0.1 M HF Initial pH 012.10 Before the equivalence point (1/2 equiv. point) 501.473.16 equivalence point 10077.93 After the equivalence point 15012.3

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7 Titration of a polyprotic acid http://www.creative-chemistry.org.uk/alevel/module4/documents/N-ch4-05.pdf

8 Sample (100 mL – 0.1 M) Titrant (0.2 M) Initial pH pH at equivalence point pH at 2X equivalence point X-axis at equivalence point Strong acid Strong base Weak acid Strong base Strong acid Weak base Strong acid Give examples of: Strong acid Strong base Weak acid Weak base

9 Give the identity and concentration of the titrated sample (50 mL) represented by the blue line. KbKb Hydroperoxide4.2 x 10 -3 Ammonia1.8 x 10 -5 Hypochlorite2.86 x 10 -7 Pyridine1.5 x 10 -9 Aniline4 x 10 -10

10 Which solution is a buffer? Explain. a) A mixture of sodium acetate to acetic acid b) A mixture of sodium nitrate and nitric acid

11 In lab, you titrate a 25-mL sample of HCN with 0.075 M NaOH. If it takes 37.5 mL to reach the equivalence point, what is the concentration of the HCN?

12 Which of these combinations would be the best to buffer the pH at approximately 9? a) CH 3 COOH (K a = 1.8 x 10 -5 ) / NaCH 3 COO b) HCl (K a = very large) / HCl c) NH 3 (K b = 1.8 x 10 -5 ) / NH 4 +


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