Chapter 19 Acids Bases and Salts Anything in

Chapter 19 Acids, Bases, and Salts Anything in black letters = write it in your notes (‘knowts’)

19. 1 – Acid-Base Theories Acids Taste sour Dissolve active metals to produce hydrogen gas Turns litmus paper RED Bases Taste bitter Feels slippery on skin (dissolves oils on skin) Turns litmus paper BLUE Have you seen the litmus paper yet? ? These are experimental definitions, they do not explain (theory) how an acid is different from a base.

Arrhenius defined an acid and base theoretically. Svante Arrhenius (1857 – 1927)

First, a vocab word… Dissociate to split or separate from another

Arrhenius Definition (~1887) ACID – substance that dissociates in water to form hydrogen ions (H+). HCl (aq) H+ (aq) + Cl- (aq) BASE – substance that dissociates in water to form hydroxide ions (OH-). Na. OH (aq) Na+ (aq) + OH- (aq)

When an acid is placed in water, H+ ions are produced. Hydrogen ions can also be thought of as H 3 O+ ions. H 3 O+ = hydronium ion HCl (aq) H+ (aq) + Cl- (aq) or equivalently, HCl + H 2 O H 3 O + + Cl-

Brønsted-Lowry Definition (~1923) ACID – donates H+ BASE – accepts H+ Johannes Bronstad (1879 – 1947) Thomas Lowry (1874 – 1936) B-L definition covers more examples than the Arrhenius definition.

ammonia ammonium ion water donates a H+ and so is a B-L acid ammonia accepts a H+ and so is a B-L base

Conjugate Acid – formed when a base accepts a H+ Conjugate Base – formed when an acid donates a H+

EXAMPLES… ACIDS donate H+, BASES accept H+ label each as acid, base, c. acid, c. base HCO 3 - + H 2 O acid + base HF + HCO 3 - base + c. base + + HCO 3 - acid H 3 O + + c. acid F- + H 2 CO 3 base c. base + c. acid + OH- CO 3 -2 H 2 O + c. acid + c. base CO 3 -2

Amphoteric – substance that can be an acid or a base – depending on what it reacts with. Water is amphoteric

ACIDS donate H+, BASES accept H+ label each reactant as an acid and base, label the products as conjugate acids or conjugate bases. HNO 3 + H 2 O H 3 O+ + NO 3 CH 3 COOH + H 2 O H 3 O+ + CH 3 COONH 3 + H 2 O NH 4+ + OHH 2 O + CH 3 COO- CH 3 COOH + OH-

Lewis Acids and Bases (not covered)

Acid/Base Indicators Litmus Acid – red, Base – blue, Neutral - colorless Phenolphthalein Acid – colorless, Base – pink, Neutral - colorless Cabbage Acid – red/pink, Base – yellow/green, Neutral – blue/purple

Assignment: Chapter 19 Worksheet #1 (FRONT)

19. 2 – Hydrogen Ions and Acidity Molarity (M) – unit used to express the concentration of a solution mol solute (mol) Molarity = liters of soln (L) anything in [brackets] means the concentration in molarity [H+] = ‘the hydrogen ion concentration’ [OH-] = ‘the hydroxide ion concentration’

Self-Ionization of Water ionizes to produce a small amount of H+ and OH - ions. H 2 O H+ + OH- In pure water at 25 C [H+] = [OH-] = 1 x 10 -7 M

Ion-product constant for water (Kw) Kw = [H+][OH-] = 1. 0 x 10 -14 remember…anything in [brackets] represents the concentration in molarity A solution is acidic if [H+] > 1. 0 x 10 -7 M …or if the p. H of the solution is below 7

Just as the mole was used to simplify large numbers of atoms, p. H is used to simplify small concentration numbers p. H = ‘power of the hydrogen ion’ p. H = -log[H+] Instead of writing out numbers like these… [H+] = 1 x 10 -7 M [H+] = 2. 4 x 10 -4 M p. H = 7. 00 [H+] = 7. 3 x 10 -10 M p. H = 9. 14 p. H = 3. 62 we can write number like these

Instead of saying “This solution has a hydronium ion concentration of 2. 4 x 10 -4 M”. We can just say “This solution has a p. H of 3. 62”. Not only is p. H an easier number to talk about, p. H is understood by most people, whereas molarity is not. The p. H scale is used to describe how acidic or basic (alkaline) a substance is.

Examples Pure water has [H+] = 1. 00 x 10 -7 M The p. H of water would be p. H = -log[H+] p. H = - log [1. 00 x 10 -7] p. H = 7

Examples [H+] = 2. 3 x 10 -5 M. Calculate the p. H = - log [H+] p. H = - log [2. 3 x 10 -5] p. H = 4. 64

Examples [H+] = 1. 0 x 10 -5 M. Calculate the p. H = - log [H+] p. H = - log [1. 0 x 10 -5] p. H = 5. 0

Examples p. H = 4. 2. Calculate [H+] p. H = - log [H+] -4. 2 = log [H+] 10 -4. 2 = 10 log [H+] 10 -4. 2 = 6. 31 x 10 -5 M = [H+]

Summary of p. H Acidic Neutral Basic [H+] 100 M 10 -7 M p. H 0 7 14 10 -7 M 100 M [OH-] 10 -14 M p. H = - log [H+] 10 -14 M [H+] = 10 -p. H Kw = [H+][OH-] = 1. 0 x 10 -14

19. 3 – Strengths of Acids and Bases Strong acids & bases completely dissociate (split into ions) in water; Weak acids & bases partly dissociate and only form a small amount of ions.

HCl is a strong acid, HCl (aq) H+ (aq) + Cl- (aq) it completely ionizes Concentration Initial at Equilibrium [HCl] [H 3 O+] [Cl-] 0. 10 M 0. 10 M

CH 3 COOH is a weak acid, CH 3 COOH (aq) CH 3 COO- (aq) + H+ (aq) it forms a small amount of ions Concentration Initial at Equilibrium [CH 3 COOH] 0. 10 M 0. 0987 M [CH 3 COO-] [H 3 O+] 0 0 1. 34 x 10 -3 M 0. 00134 M

Chemical Equilibrium – occurs when forward rxn rate equals reverse rxn rate; dynamic CH 3 COOH + H 2 O CH 3 COO- + H 3 O + Le Chatelier’s Principle – at equilibrium, a rxn will shift forward or backward in response to any change in conditions (temp, pressure, concentration) Increase [CH 3 COOH], rxn shifts the rxn to the right. Increase [CH 3 COO-], rxn hifts rxn to the left.

Le Chatelier’s Principle Example 1 CH 3 COOH + H 2 O CH 3 COO- + H 3 O + 1. Add methyl orange to acetic acid; divide into 3’s red p. H 4. 0, orange p. H 5. 0, yellow p. H 6. 0 2. Add Na+CH 3 COO- to acetic acid Doing this, increases [CH 3 COO-] and causes a shift in the rxn to the left; increasing the p. H. 3. Add Na. OH to the acetic acid Doing this, decreases [H 3 O+] causing a shift in the rxn to the left; increasing the p. H.

Le Chatelier’s Principle Example 2 Na. Cl (aq) Na+ + Cl- What will happen when drops of HCl (aq) are added to a saturated solution of Na. Cl? HCl (aq) H+ + Cl- the rxn will shift to the left because the [Cl -] concentration increased.

Le Chatelier’s Principle Example 3 Co(H 2 O)6+2 (aq) + 4 Cl- (aq) + heat Co. Cl 4 -2 (aq) + 6 H 2 O 1. Add ~ 3 m. L of conc. HCl to about 2 m. L of 0. 1 M Co. Cl 2 2. Add water to reverse the rxn 3. Add ~ 2 m. L of 0. 1 M Ag. NO 3(aq). Ag+ (aq) + Cl- (aq) Ag. Cl (s)

Buffer – solution that maintains a fairly stable p. H when small amounts of acid or base are added. Buffer solutions consist of a weak acid and its conjugate base or a weak base and its conjugate acid.

Acetic Acid/Acetate Ion Buffer CH 3 COOH CH 3 COO- The acetate ion reacts with any added acid. CH 3 COO- + H+ CH 3 COOH The acetic acid reacts with any added base. CH 3 COOH + OH- CH 3 COO- + H 2 O

Universal p. H Indicator Color Chart

Blood is buffered at p. H b/w 7. 35 - 7. 45. This is done mainly by the carbonic acid/bicarbonate buffer. Carbonic acid (H 2 CO 3) neutralizes added bases. Bicarbonate ion (HCO 3 -) neutralizes added acids. H 2 CO 3 (aq) H 2 O + CO 2 Hold your breath, CO 2 level builds up in blood, causing [H 2 CO 3] ↑, p. H ↓ Hyperventilate, CO 2 level drops in blood, causing [H 2 CO 3] ↓, p. H ↑

Questions What determines if an acid or base is strong or weak? What occurs during chemical equilibrium? Describe Le Chateliers Principle. CH 3 COO- + H+ CH 3 COOH Which direction will the rxn shift if [CH 3 COO-] increases? What would happen to the p. H of the solution? What is a buffer? What two things must a buffer contain? How does a buffer work?

Le Chatelier’s Principle and Reversible Reactions ANALYSIS Part 1 1. (a) H 2 SO 4 caused the solution to turn from yellow to orange. (b) Adding H 2 SO 4 caused the reaction to shift to the right. (c) Adding Na. OH caused the reaction to shift to the left. Part 2 1. (a) Fe+3 (ferric or iron(III)) & NO 3 -1 (nitrate) (b) K+1 (potassium) & SCN-1 (thiocyanate) (c) The Fe(SCN)+2 caused the solution to turn dark red. (d) Adding Fe 3(SO 4)2 increased the concentration of Fe(SCN)+2. (e) Adding Fe(NO 3)3 increased the concentration of Fe(SCN)+2. (f) Adding Na. OH decreased [Fe(SCN)+2] (g) Reducing [SCN-] caused the rxn to shift to the left (yellow)

19. 4 – Neutralization Reactions Neutralization Rxn – complete rxn of a strong base with a strong acid A neutralization rxn will produce a salt and water. Acid HCl + Base + Na. OH Salt + H 2 O Na. Cl + H 2 O

Titration – determining the concentration of an unknown solution using a solution whose concentration is known. Standard – solution that has a known concentration. Equivalence Point – point where the amount of acid equals the amount of base End Point – point where the indicator changes color

ASSIGNMENT: Read Section 19. 4 (p. 672 – 675) Answer Questions #35 -43

EXAMPLE 10. 0 m. L of 0. 5 M HCl solution is added to 20. 0 m. L of Na. OH of unknown concentration. What is the concentration of the Na. OH? HCl + Na. OH 0. 5 M 10. 0 m. L x. M 20. 0 m. L Na. Cl + H 2 O Since the reaction of HCl and Na. OH is 1: 1 and twice the volume of Na. OH was used, the Na. OH must half as strong as HCl; [0. 25 M].

EXAMPLE What volume of 0. 10 M KOH is required to neutralize 20. 0 m. L of 0. 20 M H 2 SO 4 solution? H 2 SO 4 + 2 KOH 0. 20 M 0. 10 M 20. 0 m. L x m. L K 2 SO 4 + 2 H 2 O Since KOH requires twice as many moles as H 2 SO 4, you should double your answer.

19. 5 – Salts in Solution not covered…

Write the balanced chemical equation for each neutralization reaction Sulfuric acid + magnesium hydroxide Phosphoric acid + calcium hydroxide Nitric acid + ammonium hydroxide

Chapter 19 Quiz #2 1. What color will litmus paper be in an acidic solution? 2. What color will phenolphthalein indicator be in an basic solution? 3. What does [H+] mean? 4. What two products are always formed in an acid-base neutralization reaction?

5. Explain the difference between a strong acid and a weak acid. 6. Explain why the p. H of pure water is 7. 00 7. What is a buffer? 8. How is the molarity of a solution calculated?

9. A student titrated 10. 0 m. L of an HCl solution. The titration required 23. 3 m. L of 0. 24 M Na. OH solution. a. Which solution was the standard? b. Which solution was more concentrated? c. Convert both volumes to liters d. Calculate the number of moles of Na. OH that reacted. e. Calculate the number of moles of HCl that reacted. f. Calculate the molarity of the HCl solution.

10. Calculate the p. H of solutions with the following hydrogen ion concentrations. a. [H+] = 1. 23 x 10 -4 M b. [H+] = 3. 42 x 10 -7 M 11. Calculate the hydrogen ion concentrations of solutions with the given p. H. a. p. H = 3. 14 b. p. H = 9. 2