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

© 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.

© 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

© 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.

© 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.

© 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.

© 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.

© 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

© 2013 Pearson Education, Inc. Chapter 10, Section 5 9 Guide to Balancing an Equation for Neutralization

© 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

© 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

© 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.

© 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

© 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

© 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

© 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, Answer is B. 3, 2, 1, 6. Solution

© 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

© 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.

© 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.

© 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.

© 2013 Pearson Education, Inc. Chapter 10, Section 5 21 Guide to Calculations for an Acid–Base Titration

© 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 M NaOH are required to neutralize 10.0 mL HCl? Analyze the Problem.

© 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 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

© 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 M NaOH are required to neutralize 10.0 mL HCl? Step 3 State equalities and conversion factors. 1 L NaOH and mole NaOH 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

© 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 M NaOH are required to neutralize 10.0 mL HCl? Step 4 Set up the problem to calculate the needed quantity L NaOH x mole NaOH x 1 mole HCl 1 L NaOH 1 mole NaOH = mole HCl = M HCl L HCl

© 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 ( L) of 1.00 M KOH. A L B L C L Learning Check

© 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 ( 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

© 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 ( 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

© 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 ( 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 moles H 2 SO 4 1 L H 2 SO 4

© 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 ( L) of 1.00 M KOH. Step 4 Set up the problem to calculate the needed quantity 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 = L H 2 SO moles H 2 SO 4 Answer is A.