15 Acids, Bases, and Salts. Lemons and limes are examples of foods that contain acidic solutions.

Transcription

1 15 Acids, Bases, and Salts Lemons and limes are examples of foods that contain acidic solutions.

2 Chapter Outline 15.1 Acids and Bases 15.2 Reactions of Acids and Bases 15.3 Salts 15.4 Electrolytes and Nonelectrolytes 15.5 Introduction to ph 15.6 Neutralization 15.7 Writing Net Ionic Equations 15.8 Acid Rain

3 Arrhenius Acids Arrhenius Acid: An acid solution contains an excess of H + ions. Common Properties of Acids 1. Sour taste 2. Turns litmus paper pink 3. Reacts with: Metals to produce H 2 gas Bases to yield water and a salt Carbonates to give carbon dioxide

4 Arrhenius Bases Arrhenius Bases: A basic solution contains an excess of OH ions. Common Properties of Bases 1. Bitter/caustic taste 2. Turns litmus paper blue 3. Slippery, soapy texture 4. Neutralizes acids

5 Brønsted-Lowry Acids and Bases A Brønsted-Lowry acid is a proton (H + ) donor. A Brønsted-Lowry base is a proton (H + ) acceptor. HCl (g) + H 2 O (l) 2 H 3 O + (aq) + Cl - (aq) Acid Base Hydronium ion (hydrated H + ) Conjugate Acid/Base Pairs: Two species that differ from each other by the presence of one hydrogen ion. conjugate pair 2 HCO 3 - (aq) + H 2 O (l) Base Acid H 2 CO 3 (aq) + OH - (aq) Conjugate Acid Conjugate Base conjugate pair 1

6 Brønsted-Lowry Acids and Bases Identify the base in the following reaction and the compound s conjugate acid. H 2 PO 4 - (aq) + H 2 O (l) HPO 4 2- (aq) + H 3 O + (aq) Base a. H 2 PO - 4 (aq) b. H 2 O (l) c. HPO 2-4 (aq) d. H 3 O + (aq) Conjugate Acid a. H 2 PO - 4 (aq) b. H 2 O (l) c. HPO 2-4 (aq) d. H 3 O + (aq)

7 Lewis Acid-Bases Lewis Acid: electron pair acceptor. Lewis Base: electron pair donor.

8 Summary of the Acid/Base Theories The theory that best explains the reaction of interest is used.

9 Reactions of Acids 1. Reactivity with Metals Acids react with any metals above hydrogen in the activity series. 2 HCl (aq) + Mg (s) MgCl 2 (s) + H 2 (g) acid + metal In general: salt + hydrogen 2. Reactivity with Bases Also called a neutralization reaction. 2 HCl (aq) + Ca(OH) 2 (aq) CaCl 2 (aq) + 2 H 2 O (l) In general: acid + base salt + water

10 Reactions of Acids 3. Reactivity with Metal Oxides 2 HCl (aq) + Na 2 O (s) 2 NaCl (aq) + H 2 O (l) acid + metal oxide (base) In general: salt + water 4. Reactivity with Metal Carbonates 2 HCl (aq) + Na 2 CO 3 (aq) 2 NaCl (aq) + H 2 O (l) + CO 2 (g) acid + carbonate (base) In general: salt + water + carbon dioxide

11 Base Reactions Bases can be amphoteric (act as either Brönsted acids or bases) As a base: Zn(OH) 2 (aq) + 2 HBr (aq) As an acid: Zn(OH) 2 (aq) + 2 NaOH (aq) ZnBr 2 (aq) + 2 H 2 O (l) Na 2 Zn(OH) 4 (aq) NaOH and KOH can also react with metals. 2 NaOH (aq) + 2 Al (s) + 6 H 2 O (l) 2 NaAl(OH) 4 (aq) + 3 H 2 (g) In general: base + metal + water salt + hydrogen

12 Salts Salts: products from acid-base reactions. HCl (aq) + NaOH (aq) NaCl (aq) + H 2 O (l) Sodium chloride (table salt) Salts are ionic compounds. Salts contain a cation (a metal or ammonium ion) derived from the base and an anion (excluding oxide or hydroxide ions) derived from the acid. Salts are generally crystalline compounds with high melting and boiling points.

13 Salts Practice Which salt forms in the reaction of aluminum oxide and hydrobromic acid? 3 Al 2 O 3 (s) + 18 HBr (aq) 6 AlBr 3 (aq) + 9 H 2 O (l) a. AlBr b. AlBr 3 c. Al 2 Br d. Al 2 Br 3

14 Electrolytes and Nonelectrolytes Electrolytes: compounds that conduct electricity when dissolved in water. Nonelectrolytes: substances that do not conduct electricity when dissolved in water. Comparing Solution Conductivity (Distilled water) (Sugar solution) (Salt solution)

15 Electrolytes and Nonelectrolytes Ion movement causes conduction of electricity in water. 3 classes of compounds, acids, bases, and salts are electrolytes because they produce ions in water when they dissolve.

16 Electrolytes and Nonelectrolytes Practice Which compound will not dissociate in water? a. HCl b. KBr c. NaOH d. CH 3 OH a. HCl (acid, dissociates) b. KBr (salt, dissociates) c. NaOH (base, dissociates) d. CH 3 OH (organic, does not dissociate)

17 Dissociation of Electrolytes Salts dissociate into their respective cations and anions when dissolved in water. The negative end of the water dipole is attracted to the positive Na + ion. NaCl (s) Na+ (aq) + Cl - (aq) Hydrated sodium (purple) and chloride (green) ions When NaCl dissolves in water, each ion is surrounded by several water molecules. The permanent dipoles in the water molecules cause specific alignment around the ions.

18 Electrolyte Ionization Ionization: process of ion formation in solution. Ionization results from the chemical reaction between a compound and water. Acids ionize in water, producing the hydronium ion (H 3 O + ) and a counter anion. HCl (g) + H 2 O (l) H 3 PO 4 (aq) + H 2 O (l) H 3 O + (aq) + Cl - (aq) H 2 PO 4 - (aq) + H 3 O + (aq) Bases ionize in water, producing the hydroxide ion (OH - ) and a counter cation. NH 3 (aq) + H 2 O (l) OH - (aq) + NH 4 + (aq)

19 Strong and Weak Electrolytes Strong electrolytes: undergo complete ionization in water. Example: HCl (strong acid) Weak electrolytes: undergo incomplete ionization in water. Example: CH 3 COOH (weak acid) HCl (left) is 100% ionized. CH 3 COOH exists primarily in the unionized form.

20 Strong and Weak Electrolytes Double arrows indicate incomplete ionization (typically weak electrolytes). HF (aq) + H 2 O (l) F - (aq) + H 3 O + (aq)

21 Salts Salts can dissociate into more than 2 ions, depending upon the compound. A 1 M solution of NaCl produces a total of 2 M of ions. NaCl (s) Na + (aq) + Cl - (aq) 1M 1M 1M A 1 M solution of CaCl 2 produces a total of 3 M of ions. CaCl 2 (s) Ca 2+ (aq) + 2 Cl - (aq) 1M 1M 2M

22 Salts Practice a. 0.75M b. 1.5M c. 3.0M d. 4.5M What is the concentration of bromide ion in a 1.5 M solution of magnesium bromide? For every one mole of MgBr 2, 2 moles of Br - ionize. MgBr 2 (s) Mg 2+ (aq) + 2 Br - (aq) 1.5 M 1.5M 3.0M

23 Colligative Properties of Electrolyte Solutions Colligative properties: depend only on the number of moles of dissolved particles present. This must be taken into consideration when calculating freezing point depression or boiling point elevation due to the presence of solute particles. Example: What is the boiling point elevation of a 1.5 m aqueous solution of CaCl 2? (K b for water is ºC/m). Because CaCl 2 is a strong electrolyte, 3 mol of ions (1 mol Ca 2+ and 2 mol Cl - ions) will be present in the solution. 3 mol ions ºC ΔT b = 1.5 m CaCl 2 = 2.3 ºC 1 mol CaCl 2 1 m

24 What is the boiling point of a 2.0 m aqueous solution of NaCl? (K b for water = ºC/m) a ºC b ºC c ºC d ºC Colligative Properties of Electrolytes Practice Because NaCl is a strong electrolyte, 2 mol of ions (1 mol Na + and 1 mol Cl - ions) will be present in the solution. 2 mol ions ºC ΔT b = 2.0 m NaCl = 2.05 ºC 1 mol NaCl 1 m T b = ºC ºC = ºC

25 Autoionization of Water Pure water auto(self) ionizes according to the reaction: H 2 O (l) + H 2 O (l) H 3 O + (aq) + OH (aq) Based on the reaction stoichiometry: Concentration H 3 O + = Concentration OH = 1 x 10 7 M [H 3 O + ] x [OH ] = (1 x 10 7 ) 2 = 1 x When acid or base is present in water, [H 3 O + ] and [OH - ] change. In acidic solutions, [H 3 O + ] > [OH ]. In basic solutions, [H 3 O + ] < [OH ].

26 Introduction to ph ph = - log[h 3 O + ] In pure water, [H 3 O + ] = 1 x 10-7 M, so ph = - log(1 x 10-7 ) = 7 Neutral High H Increasing acidity Increasing basicity 3 O + Low H 3 O + Low OH - High OH - The ph scale

27 ph

28 ph Practice By what factor is a solution of ph = 3 more acidic than a solution with a ph = 5? a. 2 b. 20 c. 200 d. 100 ph = 5 means [H 3 O + ] = 1 x 10-5 M ph = 3 means [H 3 O + ] = 1 x 10-3 M Factor = 1 x 10-3 M 1 x 10-5 M = 100

29 ph Calculations ph = - log[h 3 O + ] [H 3 O + ] = 10 -ph Generalizations [H 3 O + ] = 1 x 10-5 M Exponent = ph ph = 5 If exactly 1 [H 3 O + ] = 2 x 10-5 M The ph is between the exponent and next lowest whole number ph = 4.7 If a number between 1 and 10

30 ph Calculations Calculate the ph of a M [H 3 O + ] solution. ph = - log(0.015) = 1.8 Note: the digits to the left of the decimal place in the ph reflect the power of ten from the [H 3 O + ]. The number of decimal places for the mantissa must equal the number of significant figures in the original number.

31 ph Calculations Practice What is the ph of a M HCl solution? a b c d. 1.7 ph = - log(0.020) = 1.7

32 ph Calculations Practice What is the ph of a NaOH solution with a [OH - ] = 2.5 x M? a. 3.4 b c d [H 3 O + ] x [OH - ] = 1 x [H 3 O + ] = 1 x /[OH - ] = 1 x /2.5 x = 4.0 x 10-4 M ph = - log(4.0 x 10-4 ) = 3.4

33 Neutralization HCl General Reaction acid + base salt + water (aq) + NaOH (aq) NaCl Example Overall Ionic Equation: (aq) + H 2 O (l) H + (aq) + Cl - (aq) + Na + (aq) + OH - (aq) Na + (aq) + Cl - (aq) + H 2 O (l) All species are included; soluble compounds shown as ions. Net Ionic Equation: H + (aq) + OH - (aq) H 2 O (l) Spectator ions (green) are removed from both sides.

34 Titration Titration: experiment where the volume of one reagent (titrant) required to react with a measured amount of another reagent is measured. Titrations allow the amount of an acid or base present in a sample to be determined. Indicators are used to signal the endpoint of a titration, the point when enough titrant is added to react with the acid/base present. Burets deliver measured amounts of the titrant into a solution of the unknown reagent. Endpoint

35 Titration Calculations If ml of M HCl is used to titrate ml of NaOH, what is the molarity of the base? Reaction HCl Knowns (aq) + NaOH (aq) NaCl ml of M HCl ml NaOH (aq) + H 2 O (l) Solving for Molarity of base Calculate mol HCl ml 1000 ml soln 1 mol NaOH = mol NaOH 1 mol HCl Molarity NaOH = mol NaOH L soln = M NaOH

36 Titration Calculations What is the molarity of a NaOH solution if ml of base is required to titrate g of oxalic acid (H 2 C 2 O 4 )? Reaction H 2 C 2 O 4 (aq) + 2 NaOH (aq) NaC 2 O 4 (aq) + 2 H 2 O (l) Knowns g H 2 C 2 O ml NaOH Calculate.243 g H 2 C 2 O 4 1 mol H 2 C 2 O g H 2 C 2 O 4 Solving for Molarity of base 2 mol NaOH = mol NaOH 1 mol H 2 C 2 O 4 Molarity NaOH = mol NaOH L soln = M NaOH Always check reaction stoichiometry!

37 What is the concentration of a nitric acid solution if 10.0 ml of the solution is neutralized by 3.6 ml of 0.20 M NaOH? a M Reaction HNO 3 (aq)+naoh(aq) NaNO 3 (aq)+2 H 2 O(l) b M c M d. 5.6 M Calculate Titration Calculations Practice.60 ml NaOH Knowns Solving for 0.20 mol NaOH 1000 ml NaOH 3.60 ml, 0.20 M NaOH ml HNO 3 Molarity of acid 1 mol HNO 3 = mol HNO 3 1 mol NaOH Molarity HNO 3 = mol HNO L soln = M HNO 3

38 Acid Rain Acid rain: atmospheric precipitation more acidic than typical. General Process for Acid Rain Formation: 1. Emission of nitrogen or sulfur oxides. 2. Transportation of these chemicals throughout the atmosphere. 3. Chemical reaction of the oxides with water. This forms sulfuric and nitric acids. 4. Precipitation carries the acids to the ground.

39 Learning Objectives 15.1 Acids and Bases Compare the definitions of acids and bases, including Arrhenius, Brønsted-Lowry, and Lewis acids/bases Reactions of Acids and Bases Describe the general reactions of acids and bases Salts Explain how a salt is formed and predict the formula of a salt given an acid and base precursor.

40 Learning Objectives 15.4 Electrolytes and Nonelectrolytes Describe properties, ionization, dissociation, and strength of electrolytes and compare them to nonelectrolytes Introduction to ph Calculate the ph of a solution from the hydrogen ion concentration Neutralization Describe a neutralization reaction and do calculations involving titrations.