2. Acids, Bases, and Salts Chapter 23

3. Properties of Acids Acid Property #1. The word acid comes from the Latin word acere, which means "sour." All acids taste sour. Acid Property #2. In 1663, Robert Boyle wrote that acids would make a blue vegetable dye called "litmus" turn red. Acid Property #3. Acids destroy the chemical properties of bases.

4. Acid Property #4. Acids conduct an electric current. (electrolytes) Acid Property #5. Upon chemically reacting with an active metal, acids will evolve hydrogen gas (H 2 ).

5. Properties of Bases Base Property #1. The word "base" has a more complex history and its name is not related to taste. All bases taste bitter. Base Property #2. Bases are substances which will restore the original blue color of litmus after having been reddened by an acid. Base Property #3. Bases destroy the chemical properties of acids.

6. Base Property #4. Bases conduct an electric current. (electrolytes) Base Property #5. Bases feel slippery, sometimes people say soapy. This is because they dissolve the fatty acids and oils from your skin and this cuts down on the friction between your fingers as you rub them together.

7. Properties of A Salt A salt is the combination of a cation(+ ion) and an anion (- ion). Salts are products of the reaction between acids and bases. Solid salts usually make crystals. If a salt dissolves in water solution, it usually dissociates into the anions and cations that make up the salt.

8. The Acid Base Theory The three main theories regarding acids and bases are: 1. Arrhenius 3. Lewis 2. Brønsted-Lowry

9. Arrhenius Theory – late 1890s Acid - any substance which delivers hydrogen ion (H + ) to the solution. HA ---> H + + A¯ Base - any substance which delivers hydroxide ion (OH ¯ ) to the solution. XOH ---> X + + OH¯ When acids and bases react, they neutralize each other, forming water and a salt: HA + XOH ---> H 2 O + XA

10. Problems with Arrhenius Theory The theory did not explain why ammonia (NH 3 ) was a base. The solvent has no role to play in this theory. We know however, that an acid added to benzene will not dissociate. Solvents are vital. The end result of mixing certain acids and bases can be a slightly acidic or basic solution. Arrhenius had no explanation for this phenomenon.

11. Brønsted – Lowry Theory – Early 1920s Two chemists, independent of one another, proposed a new definition of an acid and a base. An acid is a substance from which can donate a proton. A base is a substance that can accept a proton from an acid. *Can occur in any solvent.

12. Reactions Based on Bronsted - Lowry Reactions that proceed to a large extent: HCl + H 2 O H 3 O + + Cl¯ HCl - this is an acid, because it has a proton available to be transferred. H 2 O - this is a base, since it gets the proton that the acid lost. Now, here comes an interesting idea: H 3 O + - this is an acid, because it can give a proton. (hydronium) Cl¯ - this is a base, since it has the capacity to receive a proton.

13. HCl + H 2 O H 3 O + + Cl¯ Notice that each pair (HCl and Cl¯ as well as H 2 O and H 3 O + differ by one proton (symbol = H + ). These pairs are called conjugate pairs. HNO 3 + H 2 O H 3 O + + NO 3 ¯ The acids are HNO 3 and H 3 O + and the bases are H 2 O and NO 3 ¯.

14. Bases and Conjugate Acid BaseName Conjugate acid Name CH 3 OO - Acetate ionCH 3 COOHAcetic acid NH 3 AmmoniaNH 4 + Ammonium H 2 PO 4 - Dihydrogen phosphate ion H 3 PO 4 Phosphoric acid HSO 4 - Hydrogen sulfate ion H 2 SO 4 Sulfuric acid OH - Hydroxide ionH20H20water NO 3 - Nitrate ionHNO 3 Nitric acid H2OH2OwaterH30+H30+ Hydronium ion

15. Weak/Strong Conjugate Bases The stronger the acid the weaker the conjugate base: HCl (SA) Cl - (weak conjugate base) The weaker the acid the stronger the conjugate base: H 3 PO 4 (WA)H 2 PO 4 - (strong conjugate base)

16. Water is considered an amphoteric substance because it can act as either an acid or a base. EX: HSO 4 -1, HPO 4 -2

17. Lewis Theory –Early 1920s Remember drawing Lewis Dot Structures for ionic and covalent compounds? Lewis Theory focuses on the nature of electrons rather than proton transfer. An acid as an electron pair acceptor and a base as an electron pair donor. Lewis Theory is much more general and apply to reactions that do not involve hydrogen or hydrogen ions.

18. Draw the Lewis Dot Structure for each molecule. The molecule with a lone e- pair on the central atom is the Lewis base. The other is the Lewis acid. It will accept the e- pair.

19. Identify each as a Lewis Acid or Lewis Base: 1. NH 3 7. Fe +2 2. PCl 3 3. H 2 O 4. AlCl 3 5. SO 3 6. O -2

20. Strong Acids and Bases Strong acids are those that ionized completely in water. The dissociation of a strong base looks like the diagram at the right in that it dissociates into positive and negative ions.

21. Acids 1. Monoprotic- one hydrogen to donate EX: HCl 2. Polyprotic- two to three hydrogens to donate. EX: diprotic: H 2 SO 4 triprotic: H 3 PO 4 *H + exists as H 3 O + in aqueous solutions.

22. Weak Acids and Bases Some acids and bases ionize only slightly in water. These are considered weak. The most important weak base is ammonia.

23. Acidic Neutral Solution

24. Strong Acids *HNO 3 - nitric acid *HCl - hydrochloric acid *H 2 SO 4 - sulfuric acid *HClO 3 - chloric acid *HClO 4 - perchloric acid *HBr - hydrobromic acid *HI - hydroiodic acid

25. Strong Bases *LiOH - lithium hydroxide *NaOH - sodium hydroxide *KOH - potassium hydroxide *RbOH - rubidium hydroxide *CsOH - cesium hydroxide *Ca(OH) 2 - calcium hydroxide *Sr(OH) 2 - strontium hydroxide *Ba(OH) 2 - barium hydroxide

26. Neutralization Reactions The word "neutralization" is used to describe the reaction of an acid plus a base because the acid and base properties of H + and OH - are destroyed or neutralized. In the reaction, H + and OH - combine to form water and a salt.

27. When acids and bases are equal in strength and concentration, a neutral (pH = 7) solution is formed. A neutralization reaction is a type of double replacement reaction. EX: a. HCl + NaOH --> b. H 2 SO 4 + Fe(OH) 3 -->

28. Calculations of Neutralization Reactions Utilyze equation: M A V A = M B V B M A and V A = Molarity and Volume of Acid M B and V B = Molarity and Volume of Base *Volumes may remain in milliliters *Must adjust the Molarity of all acids or bases based on the number of moles of H + or OH - they contribute. (See next slide)

29. *Add Adjust M A of H 2 SO 4 by 2 (times 2 moles H+). Only acid that needs adjusted. Adjust M B by 2 for the soluble group 2 hydroxides: Ca, Sr, Ba. If finding Molarity of these substances, divide molarity answer by 2 to get actual molarity. (opposite process)

30. 1. How many mL of a 1.5 M HCl acid is needed to neutralize a 500. mL 1.5 M NaOH solution? 2. How many mL of a 2.0 M H 2 SO 4 acid is needed to neutralize a 500. mL 1.5 M KOH solution? 3. How many mL of a 0.750 M HNO 3 acid is needed to neutralize a 275 mL 1.5 M Ca(OH) 2 solution? 4. Calculate concentration of acid if 300. mL of H 2 SO 4 is used to neutralize a 500. mL 2.50 M NaOH solution.

31. Net Ionic Equations For aqueous reactions, it is common to write equations in the ionic form. Standard form: NaOH + HCl  NaCl + H 2 O Ionic form: Na + + OH - + H + + Cl -  Na + + Cl - + H 2 O Notice* substances occurring in molecular form (H 2 O) are written as molecules. * ionic substances are written as ions if soluble.

32. Now take the ionic equation and cancel out any spectator ions on the product and reactant side of the equation. The result will be a net ionic equation. Na + + OH - + H + + Cl -  Na + + Cl - + H 2 O OH - + H +  H 2 O

33. Rules for Writing Net Ionic Equations Rule 1Binary Acids: HCl, HBr, and HI are strong: all other binary acids and HCN are weak. Strong acids are written in ionic form; weak acids are written in molecular form. Rule 2Ternary Acids: If the number of oxygen atoms in an inorganic acid molecule exceeds the number of hydrogen atoms by two or more, the acid is strong. We will consider all organic carboxylic acids as weak. Strong: HClO 3, HClO 4, H 2 SO 4, HNO 3, H 2 SeO 4 Weak: HClO, H 3 AsO 4, H 2 CO 3, H 4 SiO 4, HNO 2

34. Rule 3 Polyprotic Acids: (acids that contain more than one ionizable hydrogen atom. EX: H 2 SO 4, H 3 PO 4, H 2 CO 3 ). In the second and subsequent ionizations H 2 SO 4 is always weak, even though it is a strong acid. Rule 4 Bases: Hydroxides of Group 1 and 2 (Ca, Sr, and Ba) are soluble and strong. All others including ammonia, hydroxlamine, and organic bases are weak.

35. Rule 5Salts: Salts are written in ionic form if soluble, and in undissociated form if insoluble. *Know the solubility rules. Rule 6Oxides: Oxides (except Group I metals) are always written in molecular or undissociated form. MgO, ZnO (H 2 O) Rule 7Gases: Gases are always written in molecular form. CO 2, NH 3, O 2,

36. Practice Net Ionic Equations 1. AgNO 3 (aq) + H 2 SO 4 (aq)  2. H 4 SiO 4 (aq) + NaOH (aq)  3. HBr (aq) + KOH (aq)  4. HCl(aq) + Cr(NO 3 ) 2 (aq) + HgCl 2 (aq)  CrCl 3 (aq) + Hg 2 Cl 2 (s) + HNO 3 (aq) 5. H 2 CO 3 (aq) + NaOH(aq) 

37. pH and pOH The scale is measured on a log scale of 0 to 14, with each unit representing a ten-fold change.

38. pH Scale The pH scale is a measure of hydronium ion [H 3 O + ] concentration (acid molarity) as well as the hydroxide [OH-] (base molarity) concentration. Hydronium ion concentration indicates acidity and will have a pH of 0-6.9. Hydroxide ion concentration indicates basicity and will have a pH of 7.1-14. The higher the [H 3 O + ], the higher the acidity, the higher the [OH-], the higher the basicity.

40. Calculating pH The concentration (M or mol/L) of H 3 O + is expressed in powers of 10, from 10 -14 to 10 0. Scientists use pH which is the negative log of acid concentration, [H 3 O + ], and can be a negative value if molarity is greater than 1. pH = -log[H 3 O + ]  acid molarity [H+]

41. EX: 0.50M HCl is added to water to make a final volume of 1 liter. What is the pH of this solution? Step 1: Identify acid concentration, [H 3 O + ], in mol/L 0.50 M Step 2: Place value in equation and solve. pH = -log[0.50] = 0.30 acidic

42. Practice pH Calculations Find pH of the following solutions if [H 3 O + ] is: 1. 1.00 x 10 -3 2. 6.59 x 10 -6 3. 9.47 x 10 -10

43. Calculate pH of Strong Acids If you have the strong polyprotic acid, H 2 SO 4, you must adjust the molarity by the number of moles of H + contributed (2 moles of H + ). EX: 0.250M H 2 SO 4 is added to water to make a final volume of 1 liter. What is the pH of this solution? Step 1: Identify [H 3 O + ] in mol/L 0.250 M Step 2: Recognize that H 2 SO 4 contributes 2 moles H +. Step 3: Place value in equation and solve. pH = -log[0.250 x 2] = 0.301 acidic

44. Calculate pH of Weak Acids Weak acids will not dissociate 100%. A dissociation factor will be included in these problems. EX: Calculate pH of a 0.150 M HNO 2 solution (dissociation is 5.00%). Step 1: 0.150 M Step 2: Convert % to decimal: 5.00%  0.0500 Step 3: pH = -log[0.150 x 0.0500] = 2.13 acidic

45. Calculate H 3 0 + and H + Concentration from pH If given pH you can calculate the hydronium or hydrogen ion (acid) concentration by performing the anti-log function. EX: pH is 2.00. 10^(-2.00) = 1.00 x 10 -2 M Find [H 3 O + ] if the pH is: 1. 6.678 3. 10.0 2. 2.533 4. 2.56 Remember the unit for concentration is M.

46. pOH You can calculate the pH of a solution if you know the concentration of hydroxide ion [OH-] (base molarity). If we use the ion product constant of water (1.00 x 10 -14 )we can derive this equation: [pH]x[pOH] = 1.00 x 10 -14 Working with this equation leads to: pH + pOH = 14

47. EX: Find the pH of a solution with an [OH - ] of 1.0 x 10 -8 M. Step 1: Calculate pOH by using equation: pOH = -log[OH - ] pOH = -log[1.0x10 -8 ] = 8.0 Step 2: Subtract the pOH from 14 to find pH: 14 – pOH = pH 14 – 8.0 = 6.0 acidic

48. EX: Find pH of a 0.750 M KOH solution. Step 1: Identify that your substance is a base and you need to calculate pOH. pOH = -log[0.750] = 0.125 Step 2: 14 – 0.125 = 13.875 pH basic

49. Practice pH Calculations Using pOH Find the pH of the following solutions with [OH] of: 1. 1.00 x 10 -4 M 2. 2.64 x 10 -13 M 3. 5.67 x 10 -2 M 4. 3.45 x 10 -11 M

50. Calculate [OH-] from pH or pOH If given pH: 14 – pH = pOH 10^(-pOH) = [OH-] If given pOH: 10^(-pOH) = [OH-] Remember unit for concentration is M.

51. Summary of pH and pOH pH = -log[H + ] or [H 3 O + ] (acid molarity) pOH = -log[OH - ] (base molarity) pH + pOH = 14 [H + or H 3 O + ] (acid molarity) = 10^(-pH) [OH - ] (base molarity) = 10^(-pOH) *Hints: 1. Identify initial substance as acid or base. 2. Label all concentrations with M unit. 3. Adjust molarity of strong and weak acids/bases.

52. To check an answer for concentration, use equation [OH - ][H 3 O + ] = 1.00 x 10 -14 EX: A problem gives [OH - ] as 2.30 x 10 -4 M. You find the pOH to be 3.64 and pH is 10.36. You find [H 3 O + ] by 10^(-10.36) to be 4.365 x 10 -11 M. Check answer by dividing 1.00 x 10 -14 by the given [OH-] 2.30 x 10 -4. You get 4.35 x 10 -11. This answer is comparable to the original and is acceptable.

53. pH Indicators acid-base indicator: A substance that indicates the degree of acidity or basicity of a solution through characteristic color changes. The narrower the pH range, the better the indicator. Refer to the following chart to identify common indicators and their pH ranges.

55. Titrations Titration is a standard laboratory method of quantitative/chemical analysis which can be used to determine the concentration of an unknown reactant (acid or base). An acid or base of known concentration (a standard solution) and volume is used to react with a measured volume of an unknown concentration of an acid or base.

57. Using a buret to add the [unknown], it is possible to determine the exact amount (V) that has been consumed when the endpoint is reached. The endpoint is the point at which the titration is stopped. This is classically a point at which the number of moles of [unknown] is equal to the number of moles of [known].

59. Many methods can be used to indicate the endpoint of a reaction; titrations often use visual indicators. In simple acid-base titrations a pH indicator may be used, such as phenolphthalein, which turns (and stays) pink when a certain pH (pH > 7) is reached or exceeded. Methyl orange can also be used, which is red in acids and yellow in alkalis (bases).

60. Any indicator that changes color along the steep portion of the titration curve is suitable for the titration. Methyl violet changes color too soon, and alizarin yellow R too late.

61. Due to the logarithmic nature of the pH curve, the transitions are generally extremely sharp, and thus a single drop of unknown just before the endpoint can change the pH by several points - leading to an immediate color change in a chosen indicator.

62. Titrations of Strong Acids and Strong Bases Titration of a strong acid and a strong base will result in an equivalence point of 7 because a neutral salt water solution is formed.

63. Titration of a Strong Acid and a Weak Base Titration of a strong acid and a weak base will result in an equivalence point of less than 7 because an acidic salt water solution is formed.

64. Titration of a Weak Acid and a Strong Base Titration of a weak acid and a strong base will result in an equivalence point of greater than 7 because a basic salt water solution is formed.

65. Titrations of polyprotic acids result in titration curves with more than one equivalence point. The titration to the right is that of H 3 PO 4

66. Titrations of weak acids and weak bases require calculations and result in titration curves without a sharp transition.