Acids and Bases Chapter 14/15 Brendan Thomas Justin Simmons Sam Choi Alina Wieprecht

Properties Acids Bases Sour Taste Conduct electricity React with many metals to produce H2 Turn litmus paper red Start with H (HCl, etc) Lemons, oranges, grapefruits Bitter taste Conduct electricity (dissociate in water to create ions) Do NOT react with metals Turn litmus paper blue Often end in OH (hydroxide) Ivory soap, dish detergent Sometimes called alkalis Smooth and slippery

Scientists Arrhenius Arrhenius Acid: Arrhenius Base: A substance that dissociates to produce hydrogen ions (H+) in water Arrhenius Base: A substance that dissociates to produce hydroxide ions (OH-) in water Keyword: Dissociation

Scientists (cont.) Bronsted- Lowry Bronsted-Lowry Acid: This theory assumes that H+ is a proton with no electrons. Bronsted-Lowry Acid: Any substance that can donate H+ ions (a proton donor) Bronsted-Lowry Base: Any substance that can accept H+ ions (a proton acceptor) Keyword: Donate/accept H+ ions

Scientists (cont.) Lewis Lewis Acid: Lewis Base: Any substance such as the H+ ion that can accept a pair of nonbonding electrons (electron pair acceptor) Lewis Base: Any substance such as the OH- ion that can donate a pair of nonbonding electron (electron pair donor) Keyword: donate/accept electron pairs

Naming Binary and Ternary Acids Binary Acids Acids that consist of two elements 1) H+ 2) nonmetal Use the stem of the name of the nonmetal hydro(stem)ic acid Examples HI _________________ HBr _________________ Tertiary Acids that consist of three elements 1) H+ 2) O2- 3) nonmetal If the negative ion ends in “ite” (stem)ous acid If the negative ion ends in “ate” (stem)ic acid Examples H2CO3 ________________ H2SO3 ____________________ Hydroiodic acid Carbonic acid Hydrobromic acid Sulfurous acid

Writing Acid-Base Reactions in Aqueous Solutions Acid-base aqueous reactions deal with the dissociation of strong and weak acids and bases. +1 - HClO3 H + ClO3 NaOH Na+1 + OH-1 HCl _____________ KOH _____________ H+1 + Cl-1 K+1 + OH-1

Writing Neutralization Reactions The acid and base will neutralize each other and form a product of salt and water. Na3PO4 H2O H3PO4 + 3NaOH  ________ + ___________ N2CO3 2 NaOH 2 NaOH + H2CO3 ________ + ___________

Calculating Hydronium and Hydroxide Ion Concentrations Hydronium [H3O+] Hydroxide [OH-] Given pH is 4.35, calculate the hydronium ion concentration. 4.35 = - log[H3O+] - 4.35 = log[H3O+] 10^(-4.35) = [H3O+] [H3O+] = 4.47 x 10^(-5) M Given pOH is 8.7, calculate the hydroxide ion concentration. 8.4 = -log[OH-] - 8.4 = log[OH-] 10^(-8.4) = [OH-] [OH-] = 3.9 x 10^(-9) M Units: M

More Hydronium and Hydroxide Ion Calculations Given pH is 7.3, calculate the hydronium ion concentration. 7.3 = - log[H3O+] - 7.3 = log[H3O+] 10^(-7.3) = [H3O+] [H3O+] = 5.01 x 10^(-8) M Given pOH is 6.7, calculate the hydroxide ion concentration. 6.7 = -log[OH-] - 6.7 = log[OH-] 10^(-6.7) = [OH-] [OH-] = 2 x 10^(-7) M Units: M

pH and pOH Calculations Given [H3O+] is 2.4 x 10^(-3) M, calculate pH. pH = - log [2.4 x 10^(-3) M] pH = 2.6 Given [OH-] is 7.8 x 10^(-5) M, calculate pOH. pOH = - log [7.8 x 10^(-5) M] pOH = 4.1

More pH and pOH Calculations Given [OH-] = 5.9 x 10^(-6) M, calculate pH. pOH = -log [5.9 x 10^(-6) M] pOH = 5.23 pH = 14 – 5.23 = 8.77 Given [H3O+] = 9 x 10^(-2) M, calculate pOH. pH = -log [9 x 10^(-2)] pH = 1.05 pOH = 14 – 1.05 = 12.95

Titration Is the controlled addition and measurement of the amount of the solution of known concentration required to react completely with a measured amount of a solution of unknown concentration So, in English: when you’re using an indicator, the solution will turn a certain color once you’ve reached the equivalence point (where the two liquids are present in chemically equivalent amounts) *using an indicator only works with strong acids and bases

Calculate the Molarity of an unknown acid or base using titration data -2 Need 20 mL of 1.0 x 10 mol NaOH to reach the end point in the titration of 10 mL of HCl HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l) Calculating Moles of NaOH -2 1.0 x 10 mol NaOH x 1L x 20.0mL NaOH = 2.0 x 10-4 mol NaOH used 1L NaOH 1000mL Mole Ratio -4 -4 2.0 x 10 mol NaOH x 1 mol HCl = 2.0 x 10 mol HCl 1 mol NaOH Calculating Molarity of HCl Solution -4 -2 2.0 x 10 mol HCl x 1000mL = 2.0 x 10 mol HCl = 2.0 x 10 M HCl -2 10.0mL HCl 1L 1L HCl

Calculate the Molarity of an unknown acid or base using titration data -3 Need 35.0 mL of 6.0 x 10 mol KOH to reach an end point in titration of 15.0 mL of HBr HBr(aq) + KOH(aq) KBr(aq) + H2O(l) Calculating Moles of KOH -3 6.0 x 10 mol KOH x 1L x 35.0mL = 2.1 x 10 mol KOH used -4 1L KOH 1000mL Mole Ratio -4 -4 2.1 x 10 mol KOH x 1 mol HCl = 2.1 x 10 mol HBr 1 mol KOH Calculating Molarity of HBr Solution -4 -2 2.1 x 10 mol HBr x 1000mL = 1.4 x 10 mol HBr = 1.4 x 10 M HBr -2 15.0mL HBr 1L 1L HBr

Picture Websites https://science7acidbase.wikispaces.com/Litmus+Paper http://commons.wikimedia.org/wiki/File:PH_scale_2.png http://www.sciencebuddies.org/science-fair-projects/project_ideas/Chem_p045.shtml