1. Chapter 16 Acids and Bases

2. The Arrhenius Model acidhydrogen ions, H +,  An acid is any substance that produces hydrogen ions, H +, in an aqueous solution.  Example: when hydrogen chloride gas is dissolved in water, the following ions are produced. HCl (g)  H + (aq) + Cl - (aq) basehydroxide ions, OH -,  A base is an substance that produces hydroxide ions, OH -, in an solution.  Example: When solid sodium hydroxide is dissolved in water the following ions are produced. NaOH (s)  Na + (aq) + OH - (aq) Copyright © Houghton Mifflin Company 16-2

3. Another Theory: Brønsted-Lowry Model  The Arrhenius model is limiting in its classification of acids and bases, suggesting there is only one kind of acid or base. A more general definition was suggested by a Danish chemist named Johannes Brønsted and an English chemist named Thomas Lowry.  The Brønsted-Lowry Model states:  An acid is a proton (H + ) donor  A base is a proton (H + ) acceptor. Copyright © Houghton Mifflin Company 16-3

4. Brønsted-Lowry Model  Let’s look at a general reaction for an acid (HA) in water. HA (aq) + H 2 O (l)  H 3 O + (aq) + A - (aq)  In the reaction which reactant was the proton donor (acid)?  Which reactant was the proton acceptor (base)?  If we look at the products, we see that now we have role reversal.  Which product is a proton donor?  Which product is a proton acceptor?  In this case, the proton donor in the products side is known as the conjugate acid, and the proton acceptor is known as the conjugate base. Copyright © Houghton Mifflin Company 16-4

5. Practice In the following reactions identify the acid, base, conjugate acid, and conjugate base. 1. HF (aq) + H 2 O (l)  H 3 O + (aq) + F - (aq) 2. H 2 S (aq) + H 2 O (l)  H 3 O + (aq) + HS - (aq) 3.NH 3(aq) + H 2 O (aq)  NH 4 + (aq) + OH - (aq) 4.HSO 4 - + C 2 O 4 2-  SO 4 2- + HC 2 O 4 - 5. Write the conjugate bases for the following acids. a. HClO 4 b. HNO 3 c. HC 2 H 3 O 2 Copyright © Houghton Mifflin Company 16-5

6. 16.2 Acid Strength dissociation.  When we put an acid or base in water, the compound breaks apart into it’s respective ions is called dissociation. degree which a compound dissociates determines the strength of the acid or base.  The degree which a compound dissociates in water determines the strength of the acid or base.  Strong acids and bases completely ionizedcompletely dissociated  Strong acids and bases are substances that are completely ionized, or completely dissociated in solution.  Weak acids and bases ionize or dissociate partially  Weak acids and bases are substances that only ionize or dissociate partially in solution. Copyright © Houghton Mifflin Company 16-6

7. Figure 16.1: Representation of the behavior of acids of different strengths in aqueous solution. Copyright © Houghton Mifflin Company 16-7

8. Acids vs Conjugate Bases Copyright © Houghton Mifflin Company 16-8 Acid Conjugate Base HClO 4 ClO 4 - HII-I- HBrBr - HClCl - H 2 SO 4 HSO 4 - HNO 3 NO 3 - H3O+H3O+ H2OH2O HSO 4 - SO 4 2- HFF-F- HNO 2 NO 2 - HCOOHHCOO - CH 3 COOHCH 3 COO - NH 4 + NH 3 HCNCN - H2OH2OOH - NH 3 NH 2 - Acid Strength Increases Base Strength Increases Strong Acids Strong Bases Weak Acids Weak Bases

9. 16.3 Water as an Acid and a Base  A substance is said to be amphoteric if it can behave either as an acid or as a base.  Water is the most common amphoteric substance. Let’s look at the ionization of 2 water molecules. H 2 O (l) + H 2 O (l) ↔ H 3 O + (aq) + OH - (aq)  In this reaction one water molecule acts as an acid, and one acts as a base.  At 25°C the concentrations of these ions have been calculated repeatedly to be [H 3 O + ]=[OH - ]= 1.0 x 10 -7 M Copyright © Houghton Mifflin Company 16-9

10. Ion-Product Constant, K w [H 3 O + ][OH - ] = 1.0 x 10 -14 = K w We call this constant, 1.0 x 10 -14, K w or the ion-product constant for water. K w = [H + ][OH - ] = 1.0 x 10 -14 In any aqueous solution at 25°C, no matter what it contains, the product of [H + ] and [OH - ] must always equal 1.0 x 10 -14. This means if the [H + ] goes up, the [OH - ] must go down so the product does not change. [H + ] = the concentration of H ions Copyright © Houghton Mifflin Company 16-10

11. 3 Types of Situations  In an aqueous solution, there are 3 possible situations. Each one always maintains a K w =1.0 x 10 -14. 1.A neutral solution, where [H + ]=[OH - ] 2.An acidic solution, where [H + ]>[OH - ] 3.A basic solution, where [H + ]<[OH - ] Example: Calculate the [H + ] and determine the type of solution if you have 1.0 x 10 -5 M OH - What information do you know from this equation? K w = [H + ][OH - ] Rearrange to find [H + ] = K w = 1.0 x 10 -14 = ____ [OH - ] 1.0 x 10 -5 Is this solution acidic, basic, or neutral? Copyright © Houghton Mifflin Company 16-11

12. Practice Problems Determine the [H + ] and [OH - ] and whether the solution is acidic, basic or neutral. 1.10.0 M H + 2.1.0 x 10 -7 M OH - 3.3.4 x 10 -4 M H + 4.2.6 x 10 -8 M H + Copyright © Houghton Mifflin Company 16-12

13. 16.4 The pH Scale  Calculating the [H + ] and [OH - ] brings a lot of very small numbers which is rather inconvenient. In 1909 a French chemist, named Soren Sorensen, came up with the pH scale.  pH, literally means “power of hydrogen”  When calculating the pH we use the following: pH = - log [H + ]  Calculate the pH of a solution with [H + ] = 1.0 x 10 -7 M  Acidic solution pH < 7.00  Basic solution pH > 7.00  Neutral solution pH= 7.00 Copyright © Houghton Mifflin Company 16-13

14. The pH Scale On the pH scale, each increase of 1 unit equals a power of ten change in the [H + ]. A solution with a pH of 3, has a [H + ] = 1 x 10 -3 M, which is 10 greater than a solution with pH of 4, or [H + ] of 1 x 10 -4 M and 100 times greater than a pH of 5 or 1 x 10 -5 =[H + ]. As the [H + ] increases, the pH decreases. The pH scale runs values from 0 to 14, zero being very acidic, 14 being very basic, and 7 being neutral. Copyright © Houghton Mifflin Company 16-14

15. Figure 16.3: The pH scale. Use the scale to determine if the following are acidic, neutral or basic: a. grapefruit pH=3.2 b. orange juice pH=3.5 c. urine pH=4.8-7.5 (depends on H 2 O) d. saliva pH= 6.4-6.9 e. milk pH=6.5 f. gastric juice in stomach pH=2.0 g. blood pH=7.35-7.45 h. tears pH=7.4 i. milk of magnesia pH=10.6 j. ammonia pH=11.4 k. Draino pH=12.0 Copyright © Houghton Mifflin Company 16-15

16. Calculating pOH Sometimes we find it easier to calculate the pOH, since we are given the [OH - ]. pOH = -log [OH - ] Calculate the pOH of [OH - ] = 1.0 x 10 -2 M The pOH is not commonly used to describe a solution, so we must convert the pOH to the pH pH + pOH = 14 What is the pH of our pOH solved at the top of this slide? Copyright © Houghton Mifflin Company 16-16

17. Calculating pH and pOH Practice  Calculate the pH and pOH for the following: 1.[H + ] = 1.0 x 10 -4 M 2.[OH - ] = 1.0 x 10 -3 M 3.[H + ] = 5.60 x 10 -12 M 4.Determine whether each of the solutions above is acidic, basic or neutral. the number of decimal places for a log must equal the number of significant figures in the original number.  NOTE: When determining sig figs for logs, the number of decimal places for a log must equal the number of significant figures in the original number. Copyright © Houghton Mifflin Company 16-17

18. Calculating the [H + ] from the pH Sometimes the pH is available, but we need to know the [H + ] of the solution. In this case we need to find the inverse log of the –pH. [H + ] = inverse log (-pH) On your calculator you may enter the –pH, then push the inverse key and then the log. Your calculator may also have a 10 x key, typically above the log key. With this key you would enter your –pH, then push the 10 x key. Calculate the [H + ] if the pH is 5.00. The same sequence is used for [OH - ] and pOH. Copyright © Houghton Mifflin Company 16-18

19. Practice Problems The pH of a human blood sample was measured to be 7.41. What is the [H + ] in this blood? The pOH of a liquid drain cleaner was found to be 10.50. What is the [OH - ] for this cleaner? Copyright © Houghton Mifflin Company 16-19

20. The pH curve for the titration of 25.0 mL of 0.200 M HNO 3 with 0.100 M NaOH. If you graphed the results from the computer simulation, it would look like this graph. Copyright © Houghton Mifflin Company 16-20

21. Neutralization Reaction Net Ionic Equation  Since the acid and base in the reaction are strong, they will both dissociate completely in water. Write the following as an ionic equation: HCl (aq) + NaOH (aq)  NaCl (aq) + H 2 O (l) H + (aq) +Cl - (aq) +Na + (aq) +OH - (aq)  Na + +Cl - + H 2 O (l) We can then rewrite this as a NIE: H + (aq) + OH - (aq)  H 2 O (l) neutralization reaction.  The above is a neutralization reaction. It is called so because if equal amounts of H + and OH - are available for reaction, a neutral solution will result. Copyright © Houghton Mifflin Company 16-21

22. Calculating Volume in Neutralization Reactions What volume of 0.100 M HCl is needed to neutralize 25.0 mL of a 0.350 M NaOH solution? M acid V acid (#H + ) = M base V base (#OH - ) (0.100M) ?V (1) = 0.35M(.025L)(1) V= (0.35M)(.025L) 0.100 M V= 0.0875L or 87.5mL OR YOU CAN CALCULATE USING STOICHIOMETRY! Copyright © Houghton Mifflin Company 16-22

23.  Step 1: Write a balanced net ionic equation for the reaction. H + (aq) + OH - (aq)  H 2 O (l)  Step 2:Calculate the moles of reactants.  Step 2: Calculate the moles of reactants. 25.0 mL NaOH x 1L x 0.350 mol NaOH 1 1000 mL 1 L NaOH 8.75 x 10 -3 mol OH - = 8.75 x 10 -3 mol OH -  Step 3:Determine which reactant is limiting.  Step 3: Determine which reactant is limiting.  This problem requires the addition of just enough H + ions to react exactly with the OH - ions present.  The number of moles of OH - ions present determines the number of moles of H + ions. OH - are limiting.  Therefore the OH - are limiting. Copyright © Houghton Mifflin Company 16-23

24.  Step 4:Calculate the moles of H + required.  Step 4: Calculate the moles of H + required. 8.75 x 10 -3 mol H +  8.75 x 10 -3 mol OH - x 1 mol H + = 8.75 x 10 -3 mol H + 1 1 mol OH -  Step 5: Calculate the volume of 0.100 M HCl required.  We know that HCl:H + is a 1:1 ratio, so the [H + ] is 0.100 M.  Volume x 0.100 mol H + = 8.75 x 10 -3 mol H + 1L  Now we must solve for volume.  Volume = 8.75 x 10 -2 L or 87.5 mL Copyright © Houghton Mifflin Company 16-24

25. Practice Problem 1.Calculate the volume of 0.10 M HNO 3 needed to neutralize 125 mL of 0.050 M KOH. Copyright © Houghton Mifflin Company 16-25