1. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 14.2 Writing Equations for Neutralization Reactions Write a molecular equation for the reaction between aqueous HCl and aqueous Ca(OH) 2. Write a molecular equation for the reaction that occurs between aqueous H 3 PO 4 and aqueous NaOH. Hint: H 3 PO 4 is a triprotic acid, meaning that 1 mol of H 3 PO 4 requires 3 mol of OH - to completely react with it. SKILLBUILDER 14.2 Writing Equations for Neutralization Reactions FOR MORE PRACTICE Example 14.12; Problems 45, 46. Solution : We must identify the acid and the base and know that they react to form water and a salt. Notice that Ca(OH) 2 contains 2 mol of OH – for every 1 mol of Ca(OH) 2 and will therefore require 2 mol of H + to neutralize it. We first write the skeletal reaction. We then balance the equation.

2. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro You are given the volume of an unknown HCl solution and the volume of a known NaOH solution required to titrate the unknown solution. You are asked to find the concentration of the unknown solution. You will need two equations. The first is the equation for the neutralization reaction of HCl and NaOH (which you should write using your knowledge of acid–base reactions). The second is simply the definition of molarity EXAMPLE 14.4 Acid–Base Titration The titration of 10.00 mL of an HCl solution of unknown concentration requires 12.54 mL of a 0.100 M NaOH solution to reach the endpoint. What is the concentration of the unknown HCl solution? Given: 010.00 mL HCl solution 12.54 mL of a 0.100 M NaOH solution Find : concentration of HCl solution (mol/L) Equation :

3. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro The solution map has two parts. In the first part, use the volume of NaOH required to reach the endpoint to calculate the number of moles of HCl in the solution. The final conversion factor comes from the balanced neutralization equation. In the second part, use the number of moles of HCl and the volume of HCl solution to determine the molarity of the HCl solution. EXAMPLE 14.4 Acid–Base Titration Continued Calculate the moles of HCl in the unknown solution by following the first part of the solution map. To get the concentration of the solution, divide the number of moles of HCl by the volume of the HCl solution in L. (Note that 10.00 mL is equivalent to 0.01000 L.) The unknown HCl solution therefore has a concentration of 0.125 M Solution Map : Solution :

4. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 14.4 Acid–Base Titration Continued The titration of a 20.0-mL sample of an H 2 SO 4 solution of unknown concentration requires 22.87 mL of a 0.158 M KOH solution to reach the endpoint. What is the concentration of the unknown H 2 SO 4 solution? SKILLBUILDER 14.4 Acid–Base Titration FOR MORE PRACTICE Example 14.14; Problems 53, 54, 55, 56, 57, 58.

5. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 14.5 Determining [H 3 O + ] in Acid Solutions What is the H 3 O + concentration in each of the following solutions? (a) 1.5 M HCl (b) 3.0 M HC 2 H 3 O 2 (c) 2.5 M HNO 3 Solution : (a) Since HCl is a strong acid, it completely ionizes. The concentration of H 3 O + will be 1.5 M. [H 3 O + ] = 1.5 M (b) Since HC 2 H 3 O 2 is a weak acid, it partially ionizes. The calculation of the exact concentration of H 3 O + is beyond the scope of this text, but we know that it will be less than 3.0 M. [H 3 O + ] < 3.0 M (c) Since HNO 3 is a strong acid, it completely ionizes. The concentration of H 3 O + will be 2.5 M. [H 3 O + ] = 2.5 M What is the H 3 O + concentration in each of the following solutions? (a) 0.50 M HCHO 2 (b) 1.25 M HI (c) 0.75 M HF SKILLBUILDER 14.5 Determining [H 3 O + ] in Acid Solution s FOR MORE PRACTICE Example 14.15; Problems 61, 62.

6. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 14.6 Determining [OH – ] in Base Solutions What is the OH – concentration in each of the following solutions? (a) 2.25 M KOH (b) 0.35 M CH 3 NH 2 (c) 0.025 M Sr(OH) 2 Solution : (a) Since KOH is a strong base, it completely dissociates into K + and OH – in solution. The concentration of OH - will be 2.25 M. [OH – ] = 2.25 M (b) Since CH 3 NH 2 is a weak base, it only partially ionizes water. We cannot calculate the exact concentration of OH - but we know it will be less than 0.35 M. [OH – ] < 0.35 M (c) Since Sr(OH) 2 is a strong base, it completely dissociates into Sr 2+ (aq) and 2 OH - (aq). Sr(OH) 2 forms 2 mol of OH - for every 1 mol of Sr(OH) 2. Consequently, the concentration of OH - will be twice the concentration of Sr(OH) 2. [OH – ] = 2(0.025 M) = 0.050 M What is the OH – concentration in each of the following solutions? (a) 0.055 M Ba(OH) 2 (b) 1.05 M C 5 H 5 N (c) 0.45 M NaOH SKILLBUILDER 14.6 Determining [OH – ] in Base Solutions FOR MORE PRACTICE Example 14.16; Problems 65, 66.

7. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 14.8 Calculating pH from [H 3 O + ] Calculate the pH of each of the following solutions and indicate whether the solution is acidic or basic. (a) [H 3 O + ] = 1.8 × 10 –4 M (b) [H 3 O + ] = 7.2 × 10 –9 M Solution : To calculate pH, simply substitute the given [H 3 O + ] into the pH equation. (a) pH = – log [H 3 O + ] = – log 1.8 × 10 –4 = – (– 3.74) = 3.74 Since the pH < 7, this solution is acidic. (b) pH = – log [H 3 O + ] = – log(7.2 × 10 –9 ) = – (– 8.14) = 8.14 Since the pH > 7, this solution is basic. Calculate the pH of each of the following solutions and indicate whether the solution is acidic or basic. (a) [H 3 O + ] = 9.5 × 10 –9 M (b) [H 3 O + ] = 6.1 × 10 –3 M SKILLBUILDER 14.8 Calculating pH from [H 3 O + ]

8. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 14.9 Calculating [H 3 O + ] from pH Calculate the H 3 O + concentration for a solution with a pH of 4.80. Solution : To find the [H 3 O + ] from pH, we must undo the log function. Use either Method 1 or Method 2.

9. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 14.12 Writing Equations for Neutralization Reactions Write a molecular equation for the reaction between aqueous HBr and aqueous Ca(OH) 2 Solution : Skeletal equation: Balanced equation:

10. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 14.14 Acid–Base Titrations A 15.00-mL sample of a NaOH solution of unknown concentration requires 17.88 mL of a 0.1053 M H 2 SO 4 solution to reach the endpoint in a titration. What is the concentration of the NaOH solution? Given: 15.00-mL NaOH 17.88 mL of a 0.1053 M H 2 SO 4 solution Find : concentration of NaOH solution (mol/L) Equation : Solution Map :

11. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 14.14 Acid–Base Titrations Continued Solution : The unknown NaOH solution has a concentration of 0.2510 M.

12. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 14.15 Determining [H 3 O + ] in Acid Solutions What is the H 3 O + concentration in a 0.25 M HCl solution and in a 0.25 M HF solution? Solution : In the 0.25 M HCl solution (strong acid), [H 3 O + ] = 0.25 M. In the 0.25 M HF solution (weak acid), [H 3 O + ] < 0.25 M.

13. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 14.16 Determining [OH – ] in Base Solutions What is the OH – concentration in a 0.25 M NaOH solution, in a 0.25 M Sr(OH) 2 solution, and in a 0.25 M NH 3 solution? Solution : In the 0.25 M NaOH solution (strong base), [OH – ] = 0.25 M. In the 0.25 M Sr(OH) 2 solution (strong base), [OH – ] = 0.50 M. In the 0.25 M NH 3 solution (weak base), [OH – ] < 0.25 M.

14. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 14.17 Finding the Concentration of [H 3 O + ] or [OH – ] form K w Calculate [OH – ] in a solution with [H 3 O + ] = 1.5 × 10 –4 M. Solution :

15. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 14.18 Calculating pH from [H 3 O + ] Calculate the pH of a solution with [H 3 O + ] = 2.4 × 10 –5 M. Solution : pH = – log [H 3 O + ] = – log(2.4 × 10 –5 ) = – (–4.62) = 4.62

16. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 14.19 Calculating [H 3 O + ] from pH Calculate the [H 3 O + ] concentration for a solution with a pH of 6.22. Solution : Method 1: Inverse Log Function [H 3 O + ] = invlog(– pH) = invlog(– 6.22) = 6.0 × 10 –7 Method 2: 10 x Function [H 3 O + ] = 10 –pH = 10 –6.22 = 6.0 × 10 –7