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Water Ionization, pH Calculation, Neutralization, Acids, Titrations, Study Guides, Projects, Research of Chemistry

An in-depth exploration of acids and bases, including self-ionization of water, Arrhenius and Bronsted-Lowry models, characteristics, calculating pH and pOH, neutralization reactions, monoprotic and polyprotic acids, and titrations. Topics cover the definitions, dissociation, and reactions of acids and bases, as well as the importance of pH and pOH in determining acidity and basicity.

What you will learn

  • What is a neutralization reaction and how does it occur?
  • What is the self-ionization of water and how does it explain water's behavior as a weak electrolyte?
  • What are monoprotic and polyprotic acids, and how do they differ in their behavior?
  • What are the differences between the Arrhenius and Bronsted-Lowry models of acids and bases?
  • How do you calculate pH and pOH for simple and complex concentrations?

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2021/2022

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NAME: ___________________________________
UNIT #11: Acids and Bases
pH and pOH
Neutralization Reactions
Oxidation and Reduction
1. SELF-IONIZATION OF WATER
a) Water molecules collide, causing a very small number to ionize in a reversible reaction:
H
2
O(l) + H
2
O(l) ↔ H
3
O
+
(aq) + OH
-
(aq)
water molecules hydronium ion hydroxide ion
b) The hydronium ion (H
3
O
+
) consists of a water molecule attached to a hydrogen ion
(H
+
) by a covalent bond; thus, H
3
O
+
and H
+
can be used interchangeably in a chemical
equation to represent a hydrogen ion in aqueous solution.
Simplified equation for self-ionization of water:
H
2
O(l) ↔ H
+
(aq) + OH
-
(aq)
c) Water is considered neutral since it produces equal numbers of H
+
and OH
-
ions.
Hydrogen ion concentration, shown as [H
+
], for water = 1.0 x 10
-7
M
Hydroxide ion concentration, shown as [OH
-
], for water = 1.0 x 10
-7
M
Self-ionization produces a tiny number of ions but explains how pure water can behave
as a very weak electrolyte.
d) Water is the usual solvent for acids and bases; the dissociation of acids or bases in an
aqueous solution increases either the [H
+
] or [OH
-
], resulting in an aqueous solution
that is no longer neutral.
2. DEFINITION OF ACIDS AND BASES
a) Arrhenius Model of Acids and Bases
1. Acid: A substance that contains hydrogen and ionizes to produce hydrogen ions in
aqueous solutions.
Ex. HCl(g) → H
+
(aq) + Cl
-
(aq)
2. Base: A substance that contains a hydroxide group and dissociates to produce
hydroxide ions in aqueous solution.
Ex. NaOH(s) → Na
+
(aq) + OH
-
(aq)
3. The Arrhenius model is limited because it does not describe all bases such as NH
3
(ammonia) and NaHCO
3
(baking soda/sodium bicarbonate). These substances
dissociate to form OH
-
in an aqueous solution but do not have the hydroxide
group in their formula and therefore, would not meet the definition of an
Arrhenius base.
NH
3
(aq) + H
2
O(l) ↔ NH
4+
+ OH
-
NaHCO
3
(s) + H
2
O(l) ↔ H
2
CO
3
(aq) + Na
+
(aq) + OH
-
(aq)
sodium bicarbonate carbonic acid
pf3
pf4
pf5
pf8
pf9
pfa
pfd
pfe
pff
pf12
pf13
pf14
pf15
pf16
pf17
pf18
pf19

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NAME: ___________________________________

UNIT #11: Acids and Bases

pH and pOH

Neutralization Reactions

Oxidation and Reduction

1. SELF-IONIZATION OF WATER

a) Water molecules collide, causing a very small number to ionize in a reversible reaction: H 2 O(l) + H 2 O(l) ↔ H 3 O+(aq) + OH-(aq) water molecules hydronium ion hydroxide ion b) The hydronium ion (H 3 O+) consists of a water molecule attached to a hydrogen ion (H+) by a covalent bond; thus, H 3 O+^ and H+^ can be used interchangeably in a chemical equation to represent a hydrogen ion in aqueous solution. Simplified equation for self-ionization of water: H 2 O(l) ↔ H+(aq) + OH-(aq) c) Water is considered neutral since it produces equal numbers of H+^ and OH-^ ions. Hydrogen ion concentration , shown as [H+] , for water = 1.0 x 10-7M Hydroxide ion concentration , shown as [OH-] , for water = 1.0 x 10-7M Self-ionization produces a tiny number of ions but explains how pure water can behave as a very weak electrolyte. d) Water is the usual solvent for acids and bases; the dissociation of acids or bases in an aqueous solution increases either the [H+] or [OH-], resulting in an aqueous solution that is no longer neutral.

  1. DEFINITION OF ACIDS AND BASES a) Arrhenius Model of Acids and Bases 1. Acid : A substance that contains hydrogen and ionizes to produce hydrogen ions in aqueous solutions. Ex. HCl(g) → H+(aq) + Cl-(aq) 2. Base: A substance that contains a hydroxide group and dissociates to produce hydroxide ions in aqueous solution. Ex. NaOH(s) → Na+(aq) + OH-(aq) 3. The Arrhenius model is limited because it does not describe all bases such as NH 3 (ammonia) and NaHCO 3 (baking soda/sodium bicarbonate). These substances dissociate to form OH-^ in an aqueous solution but do not have the hydroxide group in their formula and therefore, would not meet the definition of an Arrhenius base. NH 3 (aq) + H 2 O(l) ↔ NH 4 +^ + OH- NaHCO 3 (s) + H 2 O(l) ↔ H 2 CO 3 (aq) + Na+(aq) + OH-(aq) sodium bicarbonate carbonic acid

b) Bronsted-Lowry Model of Acids and Bases

  1. Acid: A substance that is a hydrogen ion (H+) donor; also referred to as a proton donor since H+^ consists of 1 proton only.
  2. Base: A substance that is a hydrogen ion (H+) or proton acceptor.
  3. Using the example cited above: NH 3 (aq) + H 2 O(l) ↔ NH 4 +^ + OH- Bronsted-Lowry acid : H 2 O since it donates a H+^ to NH 3 Bronsted-Lowry base : NH 3 since it accepts a H+^ from H 2 O NaHCO 3 (s) + H 2 O(l) ↔ H 2 CO 3 (aq) + Na+(aq) + OH-(aq) Bronsted-Lowry acid : H 2 O since it donates a H+^ to NaHCO 3 , with Na+^ released Bronsted-Lowry base : NaHCO 3 since it accepts a H+^ from H 2 O, with Na+^ released

3. CHARACTERISTICS OF ACIDS AND BASES

a) Acids

  1. pH between 0 and 6.9; the lower the pH value, the stronger the acid.
  2. React with certain metals (aluminum, magnesium and zinc) to produce H 2 gas.
  3. Acids are corrosive.
  4. Acids taste sour. Ex. Carbonic and phosphoric acids give carbonated beverages their sharp taste. Citric and ascorbic acids give lemons and grapefruit their tart taste. Acetic acid makes vinegar taste sour.
  5. Causes blue litmus paper to turn pink.
  6. Conducts electricity. b) Bases
  7. pH between 7.1 and 14; the higher the pH value, the stronger the base.
  8. Taste bitter. Ex. Soap has a bitter taste.
  9. Feel slippery Ex. Soap has a slippery texture.
  10. Causes red litmus paper to turn blue.
  11. Conducts electricity. c) Water: has a neutral pH of 7.0 but can act as a very weak acid or base due to the tiny degree of self-ionization.
  12. CALCULATING pH and pOH FROM SIMPLE CONCENTRATIONS a) pH and pOH : Since [H+] and [OH-] concentrations are usually very small numbers expressed in scientific notation, scientists have adopted a shorthand method to express these concentrations in an aqueous solution. This is known as the pH scale and ranges from 0 to 14. b) pH = -log[H+] and pOH = -log[OH-] Additionally, pH + pOH = 14 c) For concentrations of H+^ and OH-^ that are 1.0 x 10n, calculating the pH or pOH of the solution is relatively easy. The log (n) is the number to which 10 is raised.

b) HONORS ONLY

  1. For pH and pOH values that are not whole numbers, the calculation is more difficult. This requires a calculator with a LOG and ANTILOG function.
  2. If pH = -log[H+], then multiplying both sides of the equation by -1 gives: -pH = log [H+] To calculate [H+] using this equation, you must take the antilog of both sides of the equation: antilog (-pH) = [H+]; likewise, antilog (-pOH) = [OH-]
  3. pH = 3.6 and pOH = 10. antilog -3.6 = [H+] and antilog -10.4 = [OH-] In your calculator, enter the antilog of -3.6 to calculate the [H+]. In a TI-83 or equivalent, punch the 2nd^ button, followed by the LOG button and the left parenthesis “(“ will appear; punch in (-), followed by 3.6, followed by the right parenthesis “)” and press ENTER. ANSWER: [H+] = 2.5 x 10- In your calculator, enter the antilog of -10.4 to calculate the [OH-] using the same method. ANSWER: [OH-] = 3.98 x 10-
  4. NEUTRALIZATION REACTIONS a) A neutralization reaction is a reaction in which an acid and a base react in aqueous solution to produce a salt and water. Acid + Base → Salt + Water b) It is a double replacement reaction where the H+^ ion in the acid replaces the metal cation in the base (forming H 2 O) and the metal cation of the base replaces the H+^ ion in the acid (forming an ionic salt). Examples: HNO 3 + KOH → KNO 3 + H 2 O nitric acid potassium hydroxide potassium nitrate water 2HCl + Mg(OH) 2 → MgCl 2 + 2H 2 O hydrochloric acid magnesium hydroxide magnesium chloride water H 2 SO 4 + 2NaOH → Na 2 SO 4 + 2H 2 O sulfuric acid sodium hydroxide sodium sulfate water
  5. MONOPROTIC AND POLYPROTIC ACIDS a) Acids that deliver 1 H+^ to solutions are termed monoprotic. Ex. HNO 3 nitric acid → H+(aq) + NO 3 - (aq) HCl hydrochloric acid → H+(aq) + Cl-(aq) b) Acids that deliver more than 1 H+^ to solutions are termed polyprotic. Ex. H 2 SO 4 sulfuric acid → 2H+(aq) + SO 4 2-(aq) this is diprotic since it delivers 2H+ H 3 PO 4 phosphoric acid → 3H+(aq) + PO 4 3-(aq) this is triprotic since it delivers 3H+

9. TITRATIONS

a) A titration is a method for determining the unknown concentration of an acid or base solution by reacting it with a known volume and concentration of an opposing base or acid. b) Neutralization occurs when all H+^ ions in solution have bonded with all OH-^ ions in solution to produce water, a compound with a neutral pH of 7.0. c) Titrations are performed by carefully adding an acid in a burette to a base in a receiving flask until the pH of the base is neutralized to 7.0. The receiving flask contains a chemical “indicator” which turns color when a neutral pH in the base is achieved. The titration is complete at this point, referred to as the endpoint. The same process can be performed by adding a base to an acid in the receiving flask. d) Equation for calculating an unknown concentration of acid/base from a titration: (Macid)(Vacid)( # of H+) = (Mbase)(Vbase)( # of OH-) Where Macid = Molarity of acid Vacid = Volume of acid Mbase = Molarity of base Vbase = Volume of base

of H+^ = H+^ ions delivered to solution by acid

of OH-^ = OH-^ ions delivered to solution by base

e) Examples:

  1. If it takes 45 mL of 1.0M NaOH solution to neutralize 57 mL of HCl, what is the concentration of the HCl? NOTE: HCl delivers 1 H+^ ion and NaOH delivers 1 OH-^ ion based on their formulas. (Macid)(Vacid)( # of H+) = (Mbase)(Vbase)( # of OH-) (Macid)(57 mL)(1) = (1.0M)(45 mL)(1) Macid = 0.79M
  2. If if takes 67 mL of 0.5M H 2 SO 4 to neutralize 15 mL of Al(OH) 3 , what is the concentration of the Al(OH) 3? NOTE: H 2 SO 4 delivers 2 H+^ ions and Al(OH) 3 delivers 3 OH-^ ions based on their formulas. (Macid)(Vacid)( # of H+) = (Mbase)(Vbase)( # of OH-) (0.5M)(67 mL)(2) = (Mbase)(15 mL)(3) Mbase = 1.49M
  3. STRENGTHS OF ACIDS AND BASES a) Strong acids and bases are strong electrolytes; they completely dissociate in solution, delivering the maximum number of ions to solution. Strong Acids: HCl, HBr, HNO 3 , H 2 SO 4 Strong Bases: Group 1 & 2 metals bound to OH-; NaOH, KOH, RbOH, Ca(OH) 2 b) Weak acids and bases are weaker electrolytes; they do not dissociate completely in solution and deliver smaller numbers of ions to solution. Weak Base: Fe(OH) 2 Weak Acids: H 2 CO 3 (carbonic acid in soda), H 2 S, HF

e) The oxidizing agent is the substance that facilitates oxidation by accepting lost electrons; it is the substance that is reduced. f) The reducing agent is the substance that facilitates reduction by losing electrons; it is the substance that is oxidized. g) Analyzing redox reactions

  1. A complete balanced equation is given. To determine which substances are oxidized and reduced, the reaction should be divided into half reactions.
  2. Example: Complete chemical equation: 2KBr + Cl 2 → 2KCl + Br 2 Looking at this in terms of ions gives: 2K+Br-^ + Cl 2 (neutral molecule) → 2K+Cl-^ + Br 2 (neutral molecule) Since the 2 potassium ions in this reaction maintain a charge of 1+, it has not been oxidized or reduced. 2Br-^ becomes Br 2 , which means each of the bromine ions have lost their extra electron to become Br 2. Cl 2 becomes 2Cl-, which means each of the chlorine atoms gained an electron to become 2Cl-. Net ionic equation : 2Br -^ + Cl 2 → Br 2 + 2Cl – Half reaction : 2Br -^ → Br 2 + 2e–^ Bromine is oxidized and acts as the reducing agent. Half reaction : Cl 2 + 2e -^ → 2Cl –^ Chlorine is reduced and acts as the oxidizing agent. NOTE: The 2 electrons were transferred from bromine to chlorine and thus, oxidation and reduction must occur simultaneously.

APPENDIX: pH SCALE WITH COMMON SUBSTANCES