Acid/base Definitions. Acid/Base Definitions. Acid / Base Chemistry. Acid/Base Definitions. Identifying Acids and Bases



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Acids Identifying Acids and Bases Acid (anhydrides) contains H+ ions as the cation, with and other element as the anion Non-metal oxide H2SO4 HI P2O5 Bases Base (anhydrides) Contains OH- as the anion Combined with a cation A metal with oxygen Ca(OH)2 Al2O3 Salts Ionic compound Formed from a cation and an anion Metal and non-metal Mg F 2 KCl LiBr Acid/base Definitions Arrhenius: Acids: Any substance which releases H + ion in water solution. Bases: Any substance which releases OH - ion in water solution. Acid/Base Definitions Bronsted - Lowry: Acids: Any substance which donates a proton. Bases: Any substance which accepts a proton. Acid/Base Definitions Lewis: Acids: Any substance which can accept an electron pair. Bases: Any substance which can donate an electron pair. Hydrogen ion concentration = [H+] Hydroxide ion concentration = [OH-] Pure H 2 0 neutral substance ph=7 and [H+]=[OH-] H 2 O will self-dissociate H 2 0 > H + + OH - Note: the number of H2O molecules that will self dissociate is very small; only 2 molecules per a billion, therefore [H2O] is unchanged. 1

- Kw - is the ion product constant for water - Kw = 1.0 x 10-14 M 2 Kw= [H+] [OH-] FOR H2O:Kw= [1.0 x 10-7 M][1.0 x 10-7 M] - there is an indirect relationship between [H+] and [OH-] Calculate [OH - ] if [H + ] is 1.0 x 10-2 M Kw= [H+] [OH-] [OH - ]= Kw = 1.0 x 10-14 M 2 [H + ] 1.0 x 10-2 M [OH - ]= 1.0 X 10 12 M Calculate [H + ] if [OH - ] is 6.5 x 10-11 M Kw= [H+] [OH-] [H + ]= Kw = 1.0 x 10-14 M 2 [H + ]= [OH - ] 6.5 x 10-11 M [H + ] will tell us if a soln is acidic or basic. Acidic solutions Neutral solutions Basic solutions ph= 1-6 -1-6 [H+]= 1.0 x 10 M 1.0 x 10 M [H+] > [OH-] ph= 7 [H+] = [OH-] -7 1.0 x 10 M for both ph= 8-14 -8-14 [H+]= 1.0 x 10 M 1.0 x 10 M [H+] < [OH-] If [H+]= 1.0 x 10-9 M, will the solution be basic or acidic? Basic Why? because the ph= 9 and this solution contains less [H+] than [OH-] 2

If [OH - ]= 2.63 x 10-4 M, will the solution be basic or acidic or neutral? Why? Acid / Base Formulas ph= -log [H+] poh= -log [OH-] ph + poh =14 [H + ] = antilog (-ph) [OH - ] = antilog (-poh) How to use the calculator!! Calculate the ph of a solution with [H + ] = 0.001 M. Is this acidic, basic, or neutral? Acid / Base Formulas Calculate the ph of a solution with [OH - ] = 7.9 x 10-3 M. Is this acidic, basic, or neutral? Calculate the ph of a solution with [H + ] = 4.2 x 10-6 M. Acidic, basic, or neutral? What is the [OH - ]? Acid / Base Formulas A solution with a ph of 9.1 has what [H + ]? 2 nd log 9.1 ENTER What is the [OH - ]? What is the poh? Acids, Bases, and Their Conjugates Water is amphoteric has basic + acidic properties. Conjugate Acid: Substances formed in a reaction when a base gains a H+ Conjugate Base: Substances formed in a reaction when an acid donates H+ Acids, Bases, and Their Conjugates NH3 + H2O -- NH4 + + OH - HCl + H2O ---> 3

Identifying monoprotic, diprotic, and triprotic: Acids, Bases, and Their Conjugates CO 3-2 + H 2 O ---> HCO 3 - + OH - monoprotic: HCl H + + Cl - diprotic: Ca(OH) 2 + 2HNO 3 --> Ca(NO 3 ) 2 + 2HOH H2SO4 H + + HSO4-1 HSO4-1 H + + SO4-2 triprotic: H3PO4 H + + H2PO4-1 H2PO4-1 H + + HPO4-2 HPO4-2 H + + PO4-3 Strengths of Acids and Bases Strong acids and bases will completely dissociate ex. HCl --> H + + Cl - The six strong acids - HCl, HBr, HI, H 2 SO 4, HNO 3, HClO 4 All others are weak. Strong Bases: Groups I and II with OH - except Be Strengths of Acids and Bases Weak acids and bases slightly dissociate ex. HClO <--> ClO - + H + Strengths of Acids and Bases Strong Acids: HCl, HI, HBr, H2SO4, HNO3, HClO4 Strong Bases: Groups I and II with OH- except Be Electrolytes and Non-Electrolytes Strong Electrolytes: - Compounds that conduct electricity in aqueous solution - Substances that completely dissociate - Strong acids and bases - Soluble salts 4

Electrolytes and Non- Electrolytes Weak Electrolytes: - Slight electron conductivity - Slight dissociation in aqueous - Weak acids and bases - Insoluble salts if in molten state Electrolytes and Non- Electrolytes NonElectrolytes: - Don t conduct electricity - No ions to carry current - Organic, molecular H2S03 Examples of Electrolytes and Nonelectrolytes: weak acid/weak elect. CsOH strong base/strong elect. HCl04 strong acid/strong elect. C6H1206 nonelectrolyte- covalent, molecular, organic b/c carbon Examples of Electrolytes and Nonelectrolytes: Al(OH)3 weak base/weak electrolyte HF weak acid/weak electrolyte CCl4 nonelectrolyte- molecular compound NaN03 soluble salt/strong electrolyte S03 acid anhydride/weak electrolyte Neutralization Acid + Base > Salt + H 2 O Neutralization is found using titration Titration: Analytical method used to determine the concentration of acids, bases in a neutralization reaction. Neutralization Reactions: General Equation: Acid + Base Salt + Water HCl + NaOH NaCl + HOH Acid Base Salt Water 5

Neutralization Reactions: Titration: Analytical method used to determine the concentration of acids and bases in a neutralization reaction. 1. Measured volume of an acid or base with unknown concentration. Put in Erlenmeyer. 2. Add a few drops of indicator to the Ernlenmeyer. 3. A measured volume of titrant is delivered into the flask through a buret (base or acid) 4. Concentration of the titrant is known Standard Solution. 5. Neutralization occurs when indicator is revealed End Point 6. Equivalence Point MOL ACID = MOL BASE Titration Curve Neutralization practice problems: 1. What volume of calcium hydroxide in a 3.0 M solution is needed to neutralize 45.0 ml of 2.6 M perchloric acid? Neutralization practice problems: 2. Determine the [Ba(OH) 2 ] solution if 300.0 ml was titrated to neutrality with 220.5 ml of 6.0 M solution of phosphoric acid? Neutralization practice problems: 1. What volume of calcium hydroxide in a 3.0 M solution is needed to neutralize 45.0 ml of 2.6 M perchloric acid? M CR M vol mol mol vol G: 45 ml Ca(OH)2 + HCl04 2 HOH + Ca(Cl04)2 2.6 M HCl04 0.45 L HCl04 x 2.6 mol HCl04 x 1 mol Ca(OH)2 x 1 L Ca(OH)2 3.0 M Ca(OH)2 1 L Ca(OH)2 2 mol HCl04 3 mol Ca(OH)2 U:? vol Ca(OH)2 2. Determine the [Ba(OH)2] solution if 300.0 ml was titrated to neutrality with 220.5 ml of 6.0 M solution of phosphoric acid M CR vol mol mol G: 220.5 ml 3Ba(OH)2 + 2H3P04 Ba3(P04)2 + 6HOH 6.0 M H3P04 0.2205 L H3P04 x 6.0 mol H3P04 x 3 mol Ba(OH)2 300.0 ml Ba(OH)2 1 L H3P04 2 mol H3P04 U:? M Ba(OH)2 = 1.985 mol Ba(OH)2 1.985 mol Ba(OH)2 0.300L Ba(OH)2 = 6.62 M Ba(OH)2 6