1. 1 Chapter 10 Acids & Bases

2. 2 CHAPTER OUTLINE  General Properties General Properties  Arrhenius Acids & Bases Arrhenius Acids & Bases  Brønsted-Lowery Acids & Bases Brønsted-Lowery Acids & Bases  Strength of Acids & Bases Strength of Acids & Bases  Ionization of Water Ionization of Water  pH Scale pH Scale

3. 3 GENRAL PROPERTIES OF ACIDS & BASES  Many common substances in our daily lives are acids and bases.  Oranges, lemons and vinegar are examples of acids. In addition, our stomachs contain acids that help digest foods.  Antacid tablets taken for heartburn and ammonia cleaning solutions are examples of bases.

4. 4 GENRAL PROPERTIES OF ACIDS & BASES  General properties associated with acids include the following:  sour taste  change color of litmus from blue to red  react with metals to produce H 2 gas  react with bases to produce salt & water

5. 5 GENRAL PROPERTIES OF ACIDS & BASES  General properties associated with bases include the following:  bitter taste  change color of litmus from red to blue  slippery, soapy feeling  react with acids to produce salt & water

6. 6 ARRHENIUS ACIDS & BASES  The most common definition of acids and bases was formulated by the Swedish chemist Svante Arrhenius in 1884. According to the Arrhenius definition,  Acids are substances that produce hydronium ions (H 3 O + ) in aqueous solution. Commonly written as HCl (g) + H 2 O (l) H 3 O + (aq) + Cl – (aq) HCl (g) H + (aq) + Cl – (aq) H2OH2O

7. 7 ARRHENIUS ACIDS & BASES According to the Arrhenius definition,  Bases are substances that produce hydroxide ion (OH - ) in aqueous solution. NaOH (s) Na + (aq) + OH – (aq) H2OH2O NH 3 (aq) + H 2 O (l) NH 4 + (aq) + OH – (aq)

8. 8 BRØNSTED-LOWRY ACIDS & BASES  The Arrhenius definition of acids and bases is limited to aqueous solutions.  A broader definition of acids and bases was developed by Br ø nsted and Lowry in the early 20 th century.  According to Br ø nsted-Lowry definition, an acid is a proton donor, and a base is a proton acceptor. HCl (g) + H 2 O (l) → H 3 O + (aq) + Cl – (aq) AcidBase NH 3 (aq) + H 2 O (l) → NH 4 + (aq) + OH – (aq) AcidBase A substance that can act as a Brønsted-Lowry acid and base (such as water) is called amphiprotic.

9. 9 BRØNSTED-LOWRY ACIDS & BASES  In Brønsted-Lowry definition, any pair of molecules or ions that can be inter-converted by transfer of a proton is called conjugate acid-base pair. HCl (g) + H 2 O (l) → H 3 O + (aq) + Cl – (aq) AcidBase Conjugate acid Conjugate base

10. 10 BRØNSTED-LOWRY ACIDS & BASES AcidBase NH 3 (aq) + H 2 O (l) → NH 4 + (aq) + OH – (aq) Conjugate acid Conjugate base

11. 11 AcidBase H 2 O + Cl   HCl + OH  Conjugate acid Conjugate base Example 1: Identify the conjugate acid-base pairs for each reaction shown below:

12. 12 AcidBase C 6 H 5 OH + C 2 H 5 O   C 6 H 5 O  + C 2 H 5 OH Conjugate acid Conjugate base Example 1: Identify the conjugate acid-base pairs for each reaction shown below:

13. 13 HS  Example 2: Write the formula for the conjugate acid for each base shown: + H +  H 2 S NH 3 + H +  NH 4 + CO 3 2  + H +  HCO 3 

14. 14 HI Example 3: Write the formula for the conjugate base for each acid shown: - H +  I  CH 3 OH - H +  CH 3 O  HNO 3 - H +  NO 3 

15. 15 ACID & BASE STRENGTH  Strong acids and bases are those that ionize completely in water.  Strong acids and bases are strong electrolytes.  According to the Arrhenius definition, the strength of acids and bases is based on the amount of their ionization in water. 100 1M

16. 16 ACID & BASE STRENGTH  Weak acids and bases are those that ionize partially in water.  Weak acids and bases are weak electrolytes. 100 ~1~1~1~1 1M~0.01M

17. 17 IONIZATION OF STRONG vs. WEAK ACIDS Ionizes completely Ionizes partially

18. 18 COMMON ACIDS Strong AcidsWeak Acids HCl Hydrochloric acid HC 2 H 3 O 2 Acetic acid HBr Hydrobromic acid H 2 CO 3 Carbonic acid HI Hydroiodic acid H 3 PO 4 Phosphoric acid HNO 3 Nitric acid HF Hydrofluoric acid H 2 SO 4 Sulfuric acid HCN Hydrocyanic acid H2SH2S Hydrosulfuric acid

19. 19 COMMON BASES Strong BasesWeak Bases LiOH Lithium hydroxide NH 3 Ammonia NaOH Sodium hydroxide CO(NH 2 ) 2 Urea Ca(OH) 2 Calcium hydroxide KOH Potassium hydroxide Ba(OH) 2 Barium hydroxide

20. 20 COMPARISON OF ACIDS & BASES

21. 21 IONIZATION OF WATER  Water can act both as an acid and a base.  In pure water, one water molecule donates a proton to another water molecule to produce ions. H 2 O + H 2 O  H 3 O + + OH – AcidBase Conjugate acid Conjugate base

22. 22 IONIZATION OF WATER  In pure water, the transfer of protons between water molecules produces equal numbers of H 3 O + and OH – ions.  The number of ions produced in pure water is very small, as indicated below: [H 3 O + ] = [OH – ] = 1.0 x 10 -7 M  When the concentrations of H 3 O + and OH – are multiplied together, the ion-product constant (K w ) is formed. K w =[H 3 O + ][OH – ]=[1.0x10 -7 ][1.0x10 -7 ]=1.0x10 -14  All aqueous solutions have H 3 O + and OH – ions.  An increase in the concentration of one of the ions will cause an equilibrium shift that causes a decrease in the other one.

23. 23 ACIDIC & BASIC SOLUTIONS  When [H 3 O + ] and [OH – ] are equal in a solution, it is neutral.  When [H 3 O + ] is greater than [OH – ] in a solution, it is acidic.  For example, if [H 3 O + ] is 1.0 x 10 –4 M, then [OH – ] would be 1.0 x 10 –10 M.

24. 24 ACIDIC & BASIC SOLUTIONS  When [OH - ] is greater than [H 3 O + ] in a solution, it is basic.  For example, if [OH - ] is 1.0 x 10 –6 M, then [H 3 O + ] would be 1.0 x 10 –8 M.

25. 25 ACIDIC & BASIC SOLUTIONS [H 3 O + ]=[OH - ] Neutral [H 3 O + ]>[OH - ] Acidic [H 3 O + ]<[OH - ] Basic

26. 26 Example 1: Calculate the [OH – ] in a solution with [H 3 O + ]=2.3x10 –4 M. Classify the solution as acid or basic. [OH  ]= [H 3 O + ] KwKw 2.3x10  4 1.0x10  14 = = 4.3x10  11 Solution is acidic [H 3 O + ] > 1.0x10  7 [OH  ] < 1.0x10  7

27. 27 Example 2: Calculate the [H 3 O + ] in a solution with [OH  ]=3.8x10 –6 M. Classify the solution as acid or basic. [H 3 O + ]= [OH  ] KwKw 3.8x10  6 1.0x10  14 = = 2.6x10  9 Solution is basic [OH  ] > 1.0x10  7 [H 3 O + ] < 1.0x10  7

28. 28 Example 3: Calculate the [OH – ] in a solution with [H 3 O + ]=5.8x10 –8 M. Classify the solution as acid or basic. [OH  ]= [H 3 O + ] KwKw 5.8x10  8 1.0x10  14 = = 1.7x10  7 Solution is basic [H 3 O + ] < 1.0x10  7 [OH  ] > 1.0x10  7

29. 29 Example 4: Calculate the [H 3 O + ] in a solution with [OH  ]=1.3x10 –2 M. Classify the solution as acid or basic. [H 3 O + ]= [OH  ] KwKw 1.3x10  2 1.0x10  14 = = 7.7x10  13 Solution is basic [OH  ] > 1.0x10  7 [H 3 O + ] < 1.0x10  7

30. 30 pH SCALE  The acidity of a solution is commonly measured on a pH scale.  The pH scale ranges from 0-14, where acidic solutions are less than 7 and basic solutions are greater than 7. pH = -log [H 3 O + ]

31. 31 pH SCALE Acidic solutionspH < 7H 3 O + > 1x10 -7 Neutral solutionspH = 7H 3 O + = 1x10 -7 Basic solutionspH > 7H 3 O + < 1x10 -7

32. 32 Example 1: The [H 3 O + ] of a liquid detergent is 1.4x10 –9 M. Calculate its pH. pH = -log [H 3 O + ]= -log [1.4x10 -9 ]= -(-8.85) pH = 8.85 2 significant figures The number of decimal places in a logarithm is equal to the number of significant figures in the measurement. Solution is basic

33. 33 Example 2: The pH of black coffee is 5.3. Calculate its [H 3 O + ]. [H 3 O + ] = antilog (-pH)= 10 –pH = 10 -5.3 [H 3 O + ] = 5 x 10 -6 1 significant figure Solution is acidic

34. 34 Example 3: The [H 3 O + ] of a solution is 3.5 x 10 –3 M. Calculate its pH. pH = -log [H 3 O + ]= -log [3.5x10 -3 ]= -(-2.46) pH = 2.46 2 significant figures Solution is acidic

35. 35 Example 4: The pH of tomato juice is 4.1. Calculate its [H 3 O + ]. [H 3 O + ] = antilog (-pH)= 10 –pH = 10 -4.1 [H 3 O + ] = 8 x 10 -5 1 significant figure Solution is acidic

36. 36 Example 5: The [OH  ] of a cleaning solution is 1.0 x 10  5 M. What is the pH of this solution? [H 3 O + ] = = 1.0 x 10  9 M K w =[H 3 O + ][OH – ] pH =  log[H 3 O + ]=  log(1.0x10  9 ) = 9.00 2 sig figs Solution is basic

37. 37 Example 6: The pH of a solution is 11.50. Calculate the [H 3 O + ] for this solution. [H 3 O + ] = antilog (-pH)= 10 –pH = 10 -11.50 [H 3 O + ] = 3.2 x 10 -12 Solution is basic

38. 38 THE END