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Acids and Bases 5. Boon Chemistry January 24 & 25, 2013. Catalyst. Take out your homework please. Think back to the investigations you completed this week. When a person takes Tums (or another antacid) what happens to the pH in their stomach? Why?. Objectives
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Acids and Bases 5 Boon Chemistry January 24 & 25, 2013
Catalyst • Take out your homework please. Think back to the investigations you completed this week. When a person takes Tums (or another antacid) what happens to the pH in their stomach? Why? • Objectives • I can demonstrate my knowledge of acid/base chemistry on the Unit 7 exam. • Agenda • Catalyst andHomework Review • Unit 7 Exam • Flashcards: Thermochemistry
Exit Slip: Answers Standard IE 1e/5d • What is the [H3O+] in a 0.0040 M solution of NaOH (a strong base)? • (a) 2.5 x 10-2 M (c) 4.0 x 10-12 M • (b) 2.5 x 10-12 M (d) 1.0 x 10-7 M • Why? 1.0 x 10-14 = [H3O+] (4.0 x 10-3) Solve by dividing. • As the [H3O+] of a solution increases, the value of • (a) log [H3O+] increases .(c) the solution’s pH decreases. • (b) –log[H3O+] decreases. (d) all of the above • Why? Test each answer. They are all true. • What is the [OH-] in a sample of lime juice with a pH of 2.0? • (a) 1.0 x 10-2 M (c) 1.0 x 10-10 M • (b) 1.0 x 10-7 M (d) 1.0 x 10-12 M • Why? Read carefully. If pH is 2, then pOH is 12. This corresponds to 1.0 x 10-12 M
Exit Slip answers continued (4) If the pH of a solution is increased from 2.0 to 4.0, the [H3O+] concentration (a) decreases by a factor of 2. (b) decreases by a factor of 100. (c) increases by a factor of 3. (d) increases by a factor of 1000. Why? As pH goes up, [H3O+] goes down. Each time you move one space on the pH scale you are changing the [H3O+] by a factor of 10. Moving 2 spaces is 10x10=100. (5) What is the pH of household ammonia in which the [H3O+] is 1.0 x 10-12 M? (a) 2 (c) 10 (b) 7 (d) 12 Why? 10-12 corresponds to pH=12. pH = -log[H3O+]
Homework Review page 544 1-4, 545 1-4 • [H3O+] = 5.0 x 10-3 M pH = -log[H3O+] pH = -log[5.0 x 10-3 ] pH = 2.3 2. [H3O+] = 0.2 M pH = -log[H3O+] pH = -log[2.0 x 10-1] pH = 0.70 • [OH-] = 2.0 x 10-3 pOH = -log[OH-] pOH = -log[2.0 x 10-3 ] pOH = 2.7 pOH + pH = 14; pH = 11.3 4. [OH-] = 0.35 M pOH = -log[0.35] pOH = 0.45 pOH + pH = 14 pH = 13.55
Homework Review page 544 1-4, 545 1-4 • pH = 3.3 [H3O+] = 10-pH [H3O+] = 10-3.3 [H3O+] = 5.0 x 10-4 • pH = 11.7; pOH = 14 - 11.7 = 2.3 [OH-] = 10-pOH = 10-2.3 = 5.0 x 10-3 • pH = 8.1; pOH = 14 - 8.1 = 5.9 [H3O+] = 10-8.1 = 7.9 x 10-9 [OH-] = 10-5.9 = 1.3 x 10-6 4. pH = 7.67; pOH = 14-7.67 = 6.33 [OH-] = 10-6.33 = 4.7 x 10-7 This is outside the normal range, the patient should be concerned.
Acid/Base Exam • Expectations/Instructions: • Work independently and silently. • Write on your test. • Be sure to answer every question. There are multiple choice and short answer. • The Tools: • Use your pH foldable worksheet. • Use your own calculator. • If you finish early: • Turn in your test and all your pH worksheets, handouts, etc. • Begin the flashcard assignment for Unit 8.
Thermochemistry Flashcards • Words • Endothermic • Exothermic • Melting point • Boiling point • Freezing point • Condensation • Fusion • vaporization • Sublimation • Evaporation • Heat • Temperature • Energy • Enthalpy • Entropy • Joule • Specific heat • calorie • Front of card: • Vocabulary word • Picture/diagram • Back of card: • definition • Sentence using the word • Resources: • Textbook Glossary • Words are also posted on the wall.
Homework • Due Next Class: Complete your flashcards