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General, Organic, and Biological Chemistry Fourth Edition Karen Timberlake 10.5 Reactions of Acids and Bases Chapter 10 Acids and Bases © 2013 Pearson Education, Inc. Lectures
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© 2013 Pearson Education, Inc. Chapter 10, Section 5 2 Acids and Metals Acids react with certain metals to produce hydrogen gas and the metal salt. metal acid metal salt hydrogen gas Magnesium reacts rapidly with acid and forms a salt of magnesium and H 2 gas.
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© 2013 Pearson Education, Inc. Chapter 10, Section 5 3 Acids and Carbonates Acids react with carbonates or bicarbonates (hydrogen carbonate), to produce carbon dioxide gas, water, and an ionic compound (salt). The acid reacts with CO 3 2− to produce carbonic acid, H 2 CO 3, which breaks down rapidly to CO 2 and H 2 O. metal acid carbon dioxide metal salt water
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© 2013 Pearson Education, Inc. Chapter 10, Section 5 4 Acid Rain Acid rain is a term given to precipitation, such as rain, snow, hail, or fog, that has a pH of 5.6 or less. is formed when sulfur impurities from coal and oil react with water and oxygen gas to form H 2 SO 4. degrades marble statues and limestone structures. interferes with photosynthesis, killing plants and trees.
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© 2013 Pearson Education, Inc. Chapter 10, Section 5 5 Acid Rain A marble statue in Washington Square Park has been eroded by acid rain. Acid rain has severely damaged forests in Eastern Europe.
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© 2013 Pearson Education, Inc. Chapter 10, Section 5 6 Learning Check Write the products and the balanced chemical equation for each of the following reactions of acids.
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© 2013 Pearson Education, Inc. Chapter 10, Section 5 7 Solution Write the products and the balanced chemical equation for each of the following reactions of acids.
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© 2013 Pearson Education, Inc. Chapter 10, Section 5 8 Neutralization is the reaction of an acid, such as HCl and a base, such as NaOH. acid base salt water The net ionic equation shows that H + combines with OH − to form H 2 O, leaving the ions Na+ and Cl- in solution Crossing out spectator ions we get Neutralization Reactions
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© 2013 Pearson Education, Inc. Chapter 10, Section 5 9 Guide to Balancing an Equation for Neutralization
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© 2013 Pearson Education, Inc. Chapter 10, Section 5 10 Write the balanced equation for the neutralization of magnesium hydroxide and nitric acid. Step 1 Write the reactants and products. Step 2 Balance the H in the acid with the OH in the base. Balancing Neutralization Reactions
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© 2013 Pearson Education, Inc. Chapter 10, Section 5 11 Write the balanced equation for the neutralization of magnesium hydroxide and nitric acid. Step 3 Balance the H 2 O with the H and OH. Step 4 Write the salt from the remaining ions. Balancing Neutralization Reactions
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© 2013 Pearson Education, Inc. Chapter 10, Section 5 12 Learning Check Write the balanced equation for the reaction of the base KOH with the strong acid, H 2 SO 4.
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© 2013 Pearson Education, Inc. Chapter 10, Section 5 13 Write the balanced equation for the reaction of the base KOH with the strong acid, H 2 SO 4. Step 1 Write the reactants and products. Step 2 Balance the H in the acid with the OH in the base. Solution
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© 2013 Pearson Education, Inc. Chapter 10, Section 5 14 Write the balanced equation for the reaction of the base KOH with the strong acid, H 2 SO 4. Step 3 Balance the H 2 O with the H and OH. Step 4 Write the salt from the remaining ions. Solution
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© 2013 Pearson Education, Inc. Chapter 10, Section 5 15 Select the correct group of coefficients for the following neutralization equations. 1. A. 1, 3, 3, 1 B. 3, 1, 1, 1 C. 3, 1, 1, 3 2. A. 3, 2, 2, 2 B. 3, 2, 1, 6 C. 2, 3, 1, 6 Learning Check
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© 2013 Pearson Education, Inc. Chapter 10, Section 5 16 Select the correct group of coefficients for the following neutralization equations. 1. Answer is C. 3, 1, 1, 3. 2. Answer is B. 3, 2, 1, 6. Solution
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© 2013 Pearson Education, Inc. Chapter 10, Section 5 17 Acid – Base Titration Titration is a laboratory procedure often used to determine the molarity of an acid. uses a base, such as NaOH, to neutralize a measured volume of an acid. Base (NaOH) Acid solution
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© 2013 Pearson Education, Inc. Chapter 10, Section 5 18 Indicator An indicator is added to the acid in the flask. causes the solution to change color when the acid is neutralized.
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© 2013 Pearson Education, Inc. Chapter 10, Section 5 19 End Point of Titration At the end point, moles of OH − equal moles of H 3 O + in the acid, and the indicator has a faint, permanent pink color.
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© 2013 Pearson Education, Inc. Chapter 10, Section 5 20 Concentration of the Acid From the measured volume of the NaOH solution at the end point and its molarity, we calculate the number of moles of NaOH used, the moles of acid in the flask, and the concentration of the acid.
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© 2013 Pearson Education, Inc. Chapter 10, Section 5 21 Guide to Calculations for an Acid–Base Titration
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© 2013 Pearson Education, Inc. Chapter 10, Section 5 22 Calculating Molarity from a Titration with a Base What is the molarity of an HCl solution if 18.5 mL of a 0.225 M NaOH are required to neutralize 10.0 mL HCl? Analyze the Problem.
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© 2013 Pearson Education, Inc. Chapter 10, Section 5 23 Calculating Molarity from a Titration with a Base What is the molarity of an HCl solution if 18.5 mL of a 0.225 M NaOH are required to neutralize 10.0 mL HCl? Step 1 Write the balanced equation for the neutralization. Step 2 Write a plan to calculate molarity or volume. liters molarity moles mole–mole moles divide molarity of NaOH NaOH NaOH factor HCl by liters
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© 2013 Pearson Education, Inc. Chapter 10, Section 5 24 Calculating Molarity from a Titration with a Base What is the molarity of an HCl solution if 18.5 mL of a 0.225 M NaOH are required to neutralize 10.0 mL HCl? Step 3 State equalities and conversion factors. 1 L NaOH and 0.225 mole NaOH 0.225 mole NaOH 1 L NaOH 1 mole of NaOH = 1 mole of HCl 1 mole NaOH and 1 mole HCl 1 mole HCl 1 mole NaOH
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© 2013 Pearson Education, Inc. Chapter 10, Section 5 25 Calculating Molarity from a Titration with a Base What is the molarity of an HCl solution if 18.5 mL of a 0.225 M NaOH are required to neutralize 10.0 mL HCl? Step 4 Set up the problem to calculate the needed quantity. 0.0185 L NaOH x 0.225 mole NaOH x 1 mole HCl 1 L NaOH 1 mole NaOH = 0.00416 mole HCl = 0.416 M HCl 0.0100 L HCl
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© 2013 Pearson Education, Inc. Chapter 10, Section 5 26 Calculate the volume in liters of 2.00 M H 2 SO 4 required to neutralize 50.0 mL (0.0500 L) of 1.00 M KOH. A. 0.0125 L B. 0.0500 L C. 0.0200 L Learning Check
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© 2013 Pearson Education, Inc. Chapter 10, Section 5 27 Solution Calculate the volume in liters of 2.00 M H 2 SO 4 required to neutralize 50.0 mL (0.0500 L) of 1.00 M KOH. H 2 SO 4 (aq) + 2KOH(aq) K 2 SO 4 (aq) + 2H 2 O(l) Step 1 Write the balanced equation for the neutralization. H 2 SO 4 (aq) + 2KOH(aq) K 2 SO 4 (aq) + 2H 2 O(l) Step 2 Write a plan to calculate molarity or volume. liters molarity moles mole–mole moles molarity liters of KOH KOH KOH factor H 2 SO 4 H 2 SO 4 H 2 SO 4
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© 2013 Pearson Education, Inc. Chapter 10, Section 5 28 Solution Calculate the volume in liters of 2.00 M H 2 SO 4 required to neutralize 50.0 mL (0.0500 L) of 1.00 M KOH. Step 3 State equalities and conversion factors. 1 L KOH and 1.00 mole KOH 1.00 mole KOH 1 L KOH 2 moles of KOH = 1 mole of H 2 SO 4 2 moles KOH and 1 mole H 2 SO 4 1 mole H 2 SO 4 2 moles KOH
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© 2013 Pearson Education, Inc. Chapter 10, Section 5 29 Solution Calculate the volume in liters of 2.00 M H 2 SO 4 required to neutralize 50.0 mL (0.0500 L) of 1.00 M KOH. Step 3 State equalities and conversion factors. 1 L of H 2 SO 4 = 2.00 moles of H 2 SO 4 1 L H 2 SO 4 and 2.00 moles H 2 SO 4 2.00 moles H 2 SO 4 1 L H 2 SO 4
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© 2013 Pearson Education, Inc. Chapter 10, Section 5 30 Solution Calculate the volume in liters of 2.00 M H 2 SO 4 required to neutralize 50.0 mL (0.0500 L) of 1.00 M KOH. Step 4 Set up the problem to calculate the needed quantity. 0.0500 L KOH x 1.00 mole KOH x 1 mole H 2 SO 4 1 L KOH 2 moles KOH x 1 L H 2 SO 4 = 0.0125 L H 2 SO 4 2.00 moles H 2 SO 4 Answer is A.
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