Acids and Bases L

Properties of acids Sour Conduct electricity Strong electrolytes (some) Weak electrolytes (some) React with metals to form hydrogen gas React with hydroxides to form water and salts

Properties of bases Bitter Feel slippery Can be strong or weak electrolytes React with acids to form water and salts Change indicators (litmus turns blue)

Naming Compound that forms “–OH” in water. Naming ionic compounds Acids (H) Bases If the  ends in “-ide” the name is “Hydro-,” the root of , the suffix “-ic” then “acid” Ex: HCl = Hydro-chlor-ic acid HF = Hydro-flor-ic acid If  ends in “-ate” the name is the root of , and the suffix “-ic” Ex: H2SO4 = sulferic acid If  ends in “-ite” the name is the root of , and the suffix “-ous” Ex: HNO2 = Nitrous acid (SEE NEXT) Compound that forms “–OH” in water. Naming ionic compounds

Lets Practice!! Name an acid or base when given the formula and vice versa. A. H2SO4 E. Magnesium Hydroxide B. HCl F. Phosphoric Acid C. NaOH G. Sulfurous Acid D. HNO3 H. Hydrobromic Acid

Acid Base theories Arrhenius Acid – creates H+/H3O+ as the only positive ion HCl(aq)  H+(aq) + Cl-(aq) Base – creates OH- as the only negative ion NaOH(aq)  Na+(aq) + OH-(aq) Neutralization – mixing of an acid with a base that produces salt and water HCl(aq) + NaOH(aq)  H+(aq) + Cl-(aq) + Na+(aq) + OH-(aq)  NaCl(aq) + H2O(l) H+(aq) + OH-(aq)  H2O(l)

NH3(g) + HCl(g)  NH4+ + Cl- NH4Cl(s) Acid Base Theories Arrhenius: Limitations NH3(g) + HCl(g)  NH4+ + Cl- NH4Cl(s) Since this reaction doesn’t produce hydronium or hydroxide what is it??? Bronsted- Lowery: Acid – proton donator Base – proton acceptor Coordinate covalent bond is formed One atom provides both electrons for a covalent bond (the base)

The great proton grab Bronsted-Lowery Acids and bases form conjugate acid-base pairs Whichever base is stronger will exist in higher concentrations at equilibrium.

Water Amphiprotic – can be either an acid or base Self ionization Water falls apart into ions H2O H+ + OH- Very small amount [H+ ] = [OH-] = 1 x 10-7M A neutral solution. In water Kw = [H+ ] [OH-] = 1 x 10-14 Kw is called the water dissociation constant. In water, If you know either [H] or [OH] you can calculate the other

Ion Product constant H2O H+ + OH- Kw is constant in every aqueous solution [H+] x [OH-] = 1 x 10-14M2 If [H+] > 10-7 then [OH-] < 10-7 If [H+] < 10-7 then [OH-] > 10-7 If we know one, we can determine the other. If [H+] > 10-7 acidic [OH-] < 10-7 If [H+] < 10-7 basic [OH-] > 10-7

Examples: 2. [H+] = 1.0 X 10-10 3. [OH-] = 1.0 X 10-2 Is The solution Acidic, Basic or Neutral? What is the [OH-]/ [H+]? 1. [H+] = 1.0 X 10-5 2. [H+] = 1.0 X 10-10 3. [OH-] = 1.0 X 10-2 4. [H+] = 1.0 X 10-7 5. [OH-] = 1.0 X 10-7

Logarithms Powers of ten. A shorthand for big or small numbers. pH = -log[H+] in neutral pH = - log(1 x 10-7) = 7 in acidic solution [H+] > 10-7, 100→10-6.99 pH < -log(10-7) pH < 7 in base pH > 7 pH > -log(10-7) [H+] < 10-7, 10-7.01→10-14

pH and pOH pOH = - log [OH-] [H+] x [OH-] = 1 x 10-14M2 pH + pOH = 14

1 3 5 7 9 11 13 14 Basic 100 10-1 10-3 10-5 10-7 10-9 10-11 10-13 10-14 Acidic Neutral [OH-] pH [H+] pOH

Examples: Calculating pH from [H+] [H+] = 1.0 X 10-6 mol/L [H+] = .0001M [OH-] = 1.0 X 10-2 mol/L [OH-] = 1.0 X 10-11 mol/L Using pH to find [H+] A. 4.0 B. 11.0 C. 8.0

Strength vs Concentration Strong: dissociates fully in water Example: HCl, HBr, HClO4, HNO3, H2SO4 For every mole of acid placed in solution you get a mole of H+ Monoprotic – can donates only 1 proton polyprotic – can donate more then 1 proton Weak: dissociated partially in water Established equilibrium (Ka or Kb) Donates one proton but can usually give more

Using Ka/b Ka – acid ionization constant If you know the initial concentration of an acid and the formula you can calculate [H+] given Ka Same equation as Keq Ex: CH3COOH + H2O  CH3OOH- + H3O+ What is [H3O+], if 2.0M of acetic acid was dissolved in water and has a Ka of 1.8 x 10-5?

Ka = [CH3OOH-][H3O+] [CH3COOH] Ka = 1.8 x 10-5 = [CH3OOH-][H3O+] CH3COOH + H2O  CH3OOH- + H3O+ What is [H3O+], if 2.0M of acetic acid was dissolved in water and has a Ka of 1.8 x 10-5? Ka = [CH3OOH-][H3O+] [CH3COOH] Ka = 1.8 x 10-5 = [CH3OOH-][H3O+] 1.8 x 10-5 = x(x) 2M 1.8 x 10-5 = X2

CH3COOH + H2O  CH3OOH- + H3O+ What is [H3O+], if 2 CH3COOH + H2O  CH3OOH- + H3O+ What is [H3O+], if 2.0M of acetic acid was dissolved in water and has a Ka of 1.8 x 10-5? 1.8 x 10-5 = X2 2M 3.6 x 10-5 = X2 √(3.6 x 10-5) = √(x2) = [H3O+] 6.0 x 10-3 = [H3O+] NOTE: if an acid has a Ka it is WEAK! Strong Acids: HCl, HBr, HI, HNO3, HClO3, HClO4, H2SO4

Indicators – dyes that change color in the presence of an acid or base Hydrangea Red Cabbage If grown in acidic soil turns pink If grown in basic soil turns blue Chopped and boiled, remaining liquid Turns pink in acids Turns green in bases

Indicators (common) Active ingredient in laxatives Litmus Paper Phenolphthalein Red Litmus paper Tests for bases Acid – no change Base – turns blue Blue Litmus paper Tests for acids Acid – turns red Base – no change Neutral Litmus paper Tests for both Active ingredient in laxatives Acid – Clear and colorless Bases – Pink Used in titrations

Titrations Procedure for calculating the pH of an acid/base by reacting it with a known concentration of base/acid (Macid)(Vacid)(# of H+) = (Mbase)(Vbase)(# of OH-)

Buffers or buffer solutions Resist change in pH caused by the addition of acids or bases Contains something that reacts with both acids and bases Mixtures of weak acids and bases Conjugate pairs Most common in organic systems H2CO3/HCO3- Non-conjugate acid-base pairs Most common in pharmaceutical (Alka-seltzer) NH4+/CH3COO- Amphoteric species Water HCO3-, HPO4-2 in blood pH

Oxidation and Reduction reactions LEO goes GER the loss of electrons from atoms of a substance LEO- Loss of Electrons is Oxidation Na → Na+ + e- Na loses an electron and thus, is oxidized. Oxidizing agent- substance that facilitates oxidation by accepting lost electrons; it is the substance that is reduced. the gain of electrons by atoms of a substance GER- Gain of Electrons is Reduction Cl2 + 2e- → 2Cl- Cl2 gains electrons and thus, is reduced. Reducing agent- the substance that facilitates reduction by losing electrons; it is the substance that is oxidized.

2KBr + Cl2 → 2KCl + Br2 Net ionic equation: 2Br - + Cl2 → Br2 + 2Cl – Half reaction: 2Br - → Br2 + 2e – 2Br– is oxidized to become Br2. 2Br– is the reducing agent Half reaction: Cl2 + 2e - → 2Cl – Cl2 is reduced to become 2Cl – Cl2 is the oxidizing agent