Chemical Equilibrium. Chapter 17: Keeping the chemical themes straight. 2NO 2 (g) N 2 O 4 (g) K c = [NO 2 ]2 [N 2 O 4 ]

Transcription

1 Chapter 17: Chemical Equilibrium 1. Describe the steps necessary to make ml of a 2.80 M KI solution? 1 L 2.80 mol KI 166 g KI grams KI = 500. ml x x x = 232 g KI 1000 ml 1 L soln 1 mol KI Weigh out 232 g KI, place in volumetric flask, dilute to 1 L mark. 2. Rubbing alcohol contains 142 ml of isopropyl alcohol (C 3 H 7 OH Molar Mass = 39.1 g/mol and density 0.78 g/ml) and 58.0 g of water. What is the mole fractions of alcohol? (3pt) mol IPA = 142 ml X 0.78 g/ml x 39.1 g/mol = 2.83 mol IPA mol IPA = 58 g water x 1 mol/18.0 g = 3.22 mol H2O 2.83 mol IPA χipa = 3.22 mol H2O mol IPA 3. In order to provide antifreezing protection to 20.0 C using ethylene glycol in water what is the molality of the aqueous solution? (Molar mass EG (62.07 g/mol) Kf water = 1.86 C/m) = (3pt) ΔT b = Kb m = (1.86 C/m) m 20.0 C/ 1.86 C/m = m = m Keeping the chemical themes straight Kinetic theory tells us how fast or slow a chemical reaction will occurs (reaction rates). Thermodynamics tells us about the energetics of a process(enthalpy and internal energy). Equilibrium theory tells us the extent that a chemical reaction will occur (must be experimentally determined..i.e. predetermined by the big guy). Chemical equilibrium is a property of all reactions, and occurs when the when the rate of the forward and reverse chemical reactions are equal! N 2 O 4 (g) k r Start with pure NO 2 Start with pure N 2 O 4 Start with a mixture equilibrium equilibrium Concentration k f 2NO 2 (g) Concentration equilibrium Concentration The equilibrium constant, Kc is a unitless number that summarizes the extent of a chemical reaction. It is related to the equilibrium concentrations of reactants and products. N 2 O 4 (g) Unitless Equilibrium Constant 2NO 2 (g) K c = [NO 2 ]2 [N 2 O 4 ] Balanced chemical equation [Products] raised to stoichiometric coefficient [Reactants] raised to stoichiometric coefficient Time Time Time At equilibrium there is no change in the net concentration after time (the reaction is still happening however!). Kc is temperature dependent and does not depend on the initial concentrations of either products or reactants.

2 Kc gives us information on whether a reaction is productfavored or reactantfavored or both (extent) Equilibrium Constant aa + bb <==> cc + dd Kc = [C] c [D] d [A] a [B] b [Products] raised to stoichiometric coefficient [Reactants] raised to stoichiometric coefficient We can write the equilibrium constant, K, in two different ways depending on whether liquid phase (molarity => Kc) or gas phase (partial pressure => Kp). The two are related for any chemical reaction. N2(g) + 3H2(g) <==> 2NH3(g) K c = [NH 3 ]2 [N 2 ][[H 2 ] 3 K p = (P NH 3 )2 P N2 (P H2 ) 3 VERY SMALL Kc VERY LARGE Uses Molarity! Uses partial pressure in atmospheres, not torr or mmhg! Reactant Favored Both reactants and products Productfavored Kp = Kc (RT) n or Kc = Kp (RT) n We distinguish two types of equilibrium expressions: homogeneous and heterogeneous. 1. Homogenous equilibriumreactants and products are have the same phase. (i.e. all liquids, or all gases). All reactants and products are included in the equilibrium expression. 3ClO2 (aq) 2ClO3(aq) + Cl (aq) K c = [ClO 3] 2 [Cl ] [ClO 3 ] 3 CH 4 (g) + 2H2S (g) CS2 (g) + 4H 2 (g) K p = P CS 2 (P H2 ) 4 P CH4 (P H2 S) 2 Solids and pure liquids solvents do not change concentration (Just as density of a material does not change whether you have a little or a lot of the substance) T = 650 C Same [CO2] Same Kc 2. Heterogenous equilibrium reactants and products have different phases...solids reacting with gases or liquids. We exclude solids and pure liquids from the equilibrium expression. CaCO 3 (s) CaO(s) + CO 2 (g) K p = P CO2 Different CaO and CaCO3 Pure solids and pure liquids are excluded from the equilibrium constant expression. C(s) + H 2 O(g) <==> CO(g) + H 2 (g) [CO][H K c = 2 ] P CO P H = 2 (RT) (1) [H 2 O] CH 3 COOH(aq) + H 2 O(l) P H 2O K c = [CH 3 COO ][H 3 O + ] [CH 3 COOH] CH 3 COO (aq) + H 3 O + (aq) Determine whether the following reactions are homogeneous or heterogeneous, and write equilibrium equation for these reactions: K c = [CO]2 (a) CO 2 (g) + C(s) <==> 2 CO(g) [CO 2 ] (b) Hg(l) + Hg 2+ (aq) <==> Hg 2+ 2 (aq) K c = [Hg2+ 2 ] [Hg 2+ ] (c) 2 Fe(s) + 3H 2 O(g) <==> Fe 2 O 3 (s) + 3 H 2 (g) (d) 2 H 2 O(l) <==> 2 H 2 O(g) (e) N 2 (g) + 3 H 2 (g) <==> 2 NH 3 (g) (f) 2 NH 3 (g) <==> N 2 (g) + 3 H 2 (g) K c = [H 2 ]3 [H 2 O] 3 K c = [NH 3 ]2 [H 2 ] 3 [N 2 ] Suppose for reaction a Kp > what does this imply? Suppose for reaction e Kc < what does this imply

3 The reaction quotient Q is used to predict the direction of a chemical reaction. N 2 O 4 (g) 2NO 2 (g) Q c = [NO 2 ]2 [N 2 O 4 ] Substitute the initial concentrations of the reactants and products K c and compare it to Kc. If Q c > K c reaction proceeds from right to left If Q c = K c the system is at equilibrium If Q c < K c reaction proceeds from left to right At equilibrium Write the reaction quotient, Q c, for each of the following reactions: (a) The decomposition of dinitrogen pentoxide, N 2 O 5 (g) NO 2 (g) + O 2 (g) (b) The combustion of propane gas, C 3 H 8 (g) + O 2 (g) CO 2 (g) + H 2 O(g) Write the reaction quotient, Q c, for each of the following reactions: (a) The decomposition of dinitrogen pentoxide, N 2 O 5 (g) NO 2 (g) + O 2 (g) (b) The combustion of propane gas, C 3 H 8 (g) + O 2 (g) SOLUTION: CO 2 (g) + H 2 O(g) (a) 2N 2 O 5 (g) 4NO 2 (g) + O 2 (g) Q c = [NO 2 ] 4 [O 2 ] [N 2 O 5 ] 2 (b) C 3 H 8 (g) + 5O 2 (g) 3CO 2 (g) + 4H 2 O(g) Q c = Q c is the same as K c except Q is not at equilibrium [CO 2 ] 3 [H 2 O] 4 [C 3 H 8 ][O 2 ] 5 Using Qc to see if a reaction is at equilibrium You are given a mixture containing 1.57 mol N 2, 1.92 mol H 2, and 8.13 mol NH 3 in a 20.0 L vessel at 500 K. The equilibrium constant K c for the production of NH3 from N2 and H2 = 170 for the reaction. Is reaction mixture at equilibrium? If not, what direction will the reaction occur? N H 2 2NH3 Given: Kc = 170 Initial [N 2 ] = 1.57 mol/20.0 L = M; Initial [H 2 ] t = 1.92 mol/20.0 L = M Initial [NH 3 ] t = 8.13mol/20.0 L = M Using Qc to see if a reaction is at equilibrium 1. Balance the chemical equation N 2(g) + 3 H 2(g) 2NH3(g) 2. Write the equilibrium expression for the reaction K c = [NH 3 ] 2 /[N 2 ][H 2 ] 3 3. Calculate Q c by substituting initial concentrations Q c = [NH 3 ] t2 /[N 2 ] t [H 2 ] 3 t = (0.46) 2 /(0.0785)(0.096) 3 = Compare Q c/p to K c. Is Q > K or is Q < K? Comparing Q and K To Determine Rxt Direction For the reaction N 2 O 4 (g) 2NO 2 (g), K c = 0.21 at 100 o C. At a point during the reaction, [N 2 O 4 ] = 0.12 M and [NO 2 ] = 0.55 M. Has the reaction reached equilibrium? If not, in which direction is it progressing? PLAN: Write an expression for Q c, substitute with the values given, and compare the Q c with the given K c. Q c > K c (2370 >> 170) 5. Determine which way the reaction must go to attain equilibrium. therefore the reaction mixture is not at equilibrium and the reaction will proceed from right to left to reduce Q to 170.

4 Comparing Q and K To Determine Reaction Direction For the reaction N 2 O 4 (g) 2NO 2 (g), K c = 0.21 at 100 o C. At a point during the reaction, [N 2 O 4 ] = 0.12 M and [NO 2 ] = 0.55 M. Has the reaction reached equilibrium? If not, in which direction is it progressing? Q c = [NO 2] 2 [N 2 O 4 ] = (0.55)2 = 2.5 > 0.21 (0.12) Computing K From Equilibrium Data Sample Integrative Problem Methane (CH 4 ) reacts with hydrogen sulfide to yield H 2 and carbon disulfide, a solvent used in manufacturing. What is the value of K p at 1000 K if the partial pressures in an equilibrium mixture at 1000 K are 0.20 atm of CH 4, 0.25 atm of H 2 S, 0.52 atm of CS 2, and 0.10 atm of H 2? Q c > K c, therefore the reaction is not at equilibrium and will proceed from right to left, from products to reactants, until Q c = K c. Computing K From Equilibrium Data Sample Integrative Problem Methane (CH 4 ) reacts with hydrogen sulfide to yield H 2 and carbon disulfide, a solvent used in manufacturing. What is the value of K p at 1000 K if the partial pressures in an equilibrium mixture at 1000 K are 0.20 atm of CH 4, 0.25 atm of H 2 S, 0.52 atm of CS 2, and 0.10 atm of H 2? Balance equation Equilibrium laws can be reversed, scaled and added. 1. Kc of a reaction in the reverse direction is the reciprocal of the Kc in the forward reaction. N 2 O 4 (g) 2 NO 2 (g) 2 NO 2 (g) N 2 O 4 (g) K eq = [N 2O 4 ] [NO 2 ] 2 = 1 K eq = Products raised to stoichiometric coeff divided by reactants raised to stoich coeff! 2. Kc for a reaction multiplied by a constant is the equilibrium constant raised to the power of the constant. K p = (0.52)(0.10) 4 /(0.20)(0.25) 2 = Plug in the equilibrium numbers in ATM! N 2 O 4 (g) 2 NO 2 (g) Now multiply the equation by 2 K c is squared! Note that partial pressure must be in units of atmosphere! 2 N 2 O 4 (g) 4 NO 2 (g) 3. Equilibrium If a reaction can in Multiple be expressed Reactions as the sum of two or more reactions, Kc for the overall reaction is given by the product of the equilibrium constants of the individual reactions. A + B C + D A + B C + D E + F E + F K c K c K c K c `= [C][D] [A][B] K c `` = [E][F] [C][D] Summary What happens to Kc or Kp when we: Change the Direction of the Reaction 2. Multiply Stoichiometric Coefficients by a Constant 3. Sum Chemical Reactions (Hess s Law) aa + bb cc + dd K c = [C]c [D] d [A] a [B] b cc + dd aa + bb K = 1/K c K c = K c K c = [E][F ] [A][B] = [E][F ] [C][D] [C][D] [A][B] 2. n aa + bb cc + dd K c = (K c ) n 3. For a sequence of equilibria, K overall = K 1 x K 2 x K 3 x

5 Sample At 1565 K Integrative suppose we Problem have the following elementary reactions giving a net reation (3). What is the equilibrium constant for the overall reaction? Sample At 1565 K Integrative suppose we Problem have the following elementary reactions giving a net reation (3). What is the equilibrium constant for the overall reaction? 2CO2(g) <==> 2CO(g) + O2(g) 2H2O(g) <==> 2H2(g) + O2(g) CO2(g) + H2 <==> CO(g) + H2O (g) K2 = 1.3 X CO2(g) <==> 2CO(g) + O2(g) 2H2O(g) <==> 2H2(g) + O2(g) 2CO2(g) <==> 2CO(g) + O2(g) 2H2(g) + O2(g) <==> 2H2O(g) K = Reverse ==> Take Inverse 2CO2(g) + 2H2 <==> 2CO(g) + 2H2O (g) K 3 = K1 X K = Divide equation by 2 ============> Square root of K 3 CO2(g) + H2 <==> CO(g) + H2O (g) K 3 = = 11.1 Review and Perspective in Equilibrium How Do We Use The Equilibrium Constant? Kc = [C]c [D] d [A] a [B] b 1. Write the Equilibrium Expression, Kc or Kp from equation. 2. Hess s Law with K 3. Predict The Direction of the Reaction: Qc vs Kc OR Kp = (Pc)c (Pd) d (Pa) a (Pb) b Le Châtelier s Principle If a chemical reaction at equilibrium experiences a change in concentration, temperature, volume, or partial pressure, then the equilibrium will shift to counteract the imposed change according to the law of mass action. 4 Possible Disturbances to consider 1. Concentration Changes 2. Pressure or Volume changes (gas) 3. Temperature Changes 4. Addition of a catalyst 3. Using Equilibrium Values, calculate Kc or one unknown 4. Given K, determining the equilibrium concentrations 1. Changes in Concentration Change aa + bb Increase concentration of product(s) Decrease concentration of product(s) Increase concentration of reactant(s) Decrease concentration of reactant(s) Qc = [C]c [D] d [A] a [B] b cc + dd Shifts the Equilibrium left right right left We can examine Q c vs Kc to predict what will happen Le Chatlier Principles: Effect of a Change in Concentration To improve air quality and obtain a useful product, chemists often remove sulfur from coal and natural gas by treating the fuel contaminant hydrogen sulfide with O 2 ; 2H 2 S(g) + O 2 (g) 2S(s) + 2H 2 O(g) What happens to (a) [H 2 O] if O 2 is added? (b) [H 2 S] if O 2 is added? (c) [O 2 ] if H 2 S is removed? (d) [H 2 S] if sulfur is added? PLAN: We can write and compare Q with K when the system is disturbed to see in which direction the reaction will progress. Q = [H 2 O] 2 [H 2 S] 2 [O 2 ]

6 Le Chatlier Principles: Effect of a Change in Concentration To improve air quality and obtain a useful product, chemists often remove sulfur from coal and natural gas by treating the fuel contaminant hydrogen sulfide with O 2 ; 2H 2 S(g) + O 2 (g) 2S(s) + 2H 2 O(g) (a) When O 2 is added, Q decreases and the reaction progresses to the right to come back to K. So [H 2 O] increases. (b) When O 2 is added, Q decreases and the reaction progresses to the right to come back to K. So [H 2 S] decreases. (c) When H 2 S is removed, Q increases and the reaction progresses to the left to come back to K. So [O 2 ] decreases. (d) Sulfur is not part of the Q (K) expression because it is a solid. Therefore, as long as some sulfur is present the reaction is unaffected. [H 2 S] is unchanged. 2. Pressure and Volume Effects In Gas Reactions For anything change in equilibrium to occur there must be a difference in the number of moles on each side of the equation! 1N2O4 (g) Cl 2 (g) + I 2 (g) Stress or Change 2NO2 (g) 2ICl(g) CHANGE WILL OCCUR NO CHANGE Shifts the Equilibrium Increase P (decrease V) to side with fewest moles of gas Decrease pressure (increase V) to side with largest moles of gas Increase volume (decrease P) to side with most moles of gas Decrease volume (increase P) to side with fewest moles of gas Le Chatlier s Principle: Affect of volume (pressure) on the equilibrium position How would you change the volume of each of the following reactions to increase the yield of products? (a) CaCO 3 (s) (b) S(s) + 3F 2 (g) (c) Cl 2 (g) + I 2 (g) PLAN: CaO(s) + CO 2 (g) SF 6 (g) 2ICl(g) When gases are present, a change in volume will affect the concentration of the gas. If the volume decreases (pressure increases), the reaction will shift to fewer moles of gas and vice versa. Le Chatlier s Principle: Affect of volume (pressure) on the equilibrium position How would you change the volume of each of the following reactions to increase the yield of products? (a) CaCO 3 (s) (b) S(s) + 3F 2 (g) (c) Cl 2 (g) + I 2 (g) CaO(s) + CO 2 (g) (a) CO 2 is the only gas present. To increase its yield, decrease the pressure which = increasing the volume. SF 6 (g) (b) There are more moles of gaseous reactants than products, so decrease the volume (increase the pressure) to shift the reaction to the right. 2ICl(g) (c) There are an equal number of moles of gases on both sides of the reaction. Therefore, a change in volume will have no effect. 3. Changes in Temperature There are 2 cases: Think of heat as reactant or as product! 1. When H < 0 => Exothermic => think heat is a product A (g) + B (g) C (g) + D (g) + heat Is the amount of NO(g) formed from the given amounts of N2 and O2 greater at higher of lower temperatures? N2 (g) + O2 (g) 2NO (g) ΔH = kj 2. When H > 0 = Endothermic => think heat is reactant heat + A (g) + B (g) C (g) + D (g) What happens to K can be analyzed using massaction! Kc = [C]c [D] d [A] a [B] b Summary Changes in Temperature Change Exothermic Rx Increase temperature Decrease temperature K decreases K increases Endothermic Rx K increases K decreases

7 Is the amount of NO(g) formed from the given amounts of N2 and O2 greater at higher of lower temperatures? N2 (g) + O2 (g) 2NO (g) ΔH = kj TRANSLATION: ΔH = kj =>ENDO = HEAT IS REACTANT! Heat + N2 (g) + O2 (g) PRODUCTFAVORED SHIFT 2NO (g) Increasing the temperature by adding more heat to this reaction will increase the amount of NO formed and decrease the reactant side. Kc will increase in value because NO2 increases! Kc = [NO]2 [N2][O2] How does an increase in temperature affect the concentration of the underlined substance and K c for the following reactions? (a) CaO(s) + H 2 O(l) (b) CaCO 3 (s) (c) SO 2 (g) PLAN: Ca(OH) 2 (aq) ΔH o = 82 kj CaO(s) + CO 2 (g) ΔH o = 178 kj S(s) + O 2 (g) ΔH o = 297 kj Express the heat of reaction as a reactant or a product. Then consider the increase in temperature and its effect on K c. How does an increase in temperature affect the concentration of the underlined substance and K c for the following reactions? (a) CaO(s) + H 2 O(l) (a) CaO(s) + H 2 O(l) (b) CaCO 3 (s) (c) SO 2 (g) Ca(OH) 2 (aq) ΔH o = 82 kj Ca(OH) 2 (aq) + heat Increase in temp will shift reaction left, decrease [Ca(OH) 2 ], and decrease K c. (b) CaCO 3 (s) + heat CaO(s) + CO 2 (g) ΔH o = 178 kj CaO(s) + CO 2 (g) The reaction will shift right, resulting in an increase in [CO 2 ] and increase in K c. (c) SO 2 (g) + heat S(s) + O 2 (g) ΔH o = 297 kj S(s) + O 2 (g) The reaction will shift right, resulting in an decrease in [SO 2 ] and increase in K c. For the endothermic reaction: X(g) + Y2 <==> XY(g) + Y(g) the following pictures depict molecular scenes reactions mixtures at various stages in the reaction. Green = X ) If Kc = 2 at the temperature of the reaction which scene represents the mixture at equilibrium? 2) What will have to happen to the other two scenes to reach equilibrium (shift towards products or towards reactants?) 3) How will a rise in temperature alter the equilibrium [Y2]? 4. Adding a Catalyst to a system at equilibrium does not shift the position of an equilibrium system does not alter the value of Kc or Kp A catalyst lowers E a for both forward and reverse reactions and only achieves equilibrium at a faster rate. Catalyzed Pathway Uncatalyzed Pathway Catalyst lowers E a for both forward and reverse reactions.

8 Acids and Bases Chapter 18 Acids and bases have distinct properties. Acids: Acrid sour taste React with metals (Group I,II) to yield H 2 gas Changes plant dye litmus from blue to red React with carbonates and bicarbonates to produce CO2 gas 200 Million MT H 2SO 4 Acid Base The effects of acid rain on a statue of George Washington taken in 1935 (left) and 2001 (right) marble. CaCO3(s) + 2HCl(aq) ==> H2O(l) + CO2(g) + CaCl2(aq) Bases: Bitter taste Slippery feel 50 Million MT NaOH/yr 3 million containers Changes plant dye litmus from red to blue React and neutralizes the effects of acids Acids and bases are everywhere. Common Acids, soft drinks Most antiperspirants, water treatment plants, paper Common Bases Chemistry uses three definitions of acids and bases: 1) Arrehenius 2) BronstedLowry 3) Lewis Name Acid Definition Base Definition Arrhenius Acid/Base definitions arise from the reaction with water. Acid: a substance that has a covalent H atom in its formula, and releases a proton H + when dissolved in water. HCl(aq), HBr(aq), HI (aq), HNO3(aq), H2SO4(aq), HClO4(aq) Arrhenius Substance that donates H + Substances that donates OH H2O HA(g) <==> H + (aq) + A (aq) Brønsted Lowry Substances that donates H + Substances that accepts H + Lewis Electronpair acceptor Electronpair donor Base: a substance that contains OH in its formula, and releases hydroxide ions (OH ) when dissolved in water. NaOH(aq), KOH (aq), LiOH (aq), Mg(OH)2(aq), Ca(OH)2(aq) H2O MOH(s) <==> OH + M +

9 Acid and base equilibrium is represented in two ways semiconfusing ways. Don t let it distract you. H + has never been isolated in the lab, but chemists use it symbolically anyway. H + reacts with water to form the hydronium ion, H3O + 1. H2O HA(g) <==> H + + A H2O MOH(s) <==> OH + M + Lazy yet convenient way. H 2O written over arrow with H + as the acid. H + and H3O + are equivalent! 2. HA(g) + H2O(l) <==> H3O + (aq) + A (aq) MOH(s) + H2O(l) <==> OH (aq) + M + (aq) + H2O Reacting with H 2O in the equation with H 3O as acid. hydronium ion, H3O + The Arrhenius acidbase definition fails to cover all cases..so chemists invented the Bronsted Lowry definition. How Arrhenius Defintion Fails 1. NH3 or amines (like RNH2) don t have OH group but are bases. Chemistry uses three definitions of acids and bases: 1) Arrehenius 2) BronstedLowry 3) Lewis Name Acid Definition Base Definition Arrhenius Substance that donates H + Substances that donates OH 2. Many transition metals Cr, Al, Fe, Cu are acidic but have no H +. Brønsted Lowry Substances that donate H + Substances that accept H + 3. Arrhenius definition requires water...yet some acid base reactions occur without water as solvent. HCl(g) + NH3(g) => NH4Cl(s) gaseous without water Lewis Electronpair acceptor Electronpair donor A Bronsted acid is a substance that donates H + to a Bronsted base. A Bronsted base is a substance that accepts H + from a Bronsted acid. Generalized Bronsted Acid and Base HA + B <> A + BH + Acid a H + donor Base, a H + acceptor Conjugate Base Conjugate Acid Conjugate Acid Conjugate Base Conjugate pair Conjugate pair Conjugate Acid Conjugate Base A is called the conjugate base of the acid HA. Has a free pair of electrons to accept a proton. BH + is called the conjugate acid of the base B Both are called conjugate acidbase pairs!

10 Identifying the following conjugate acidbase pairs (H + donars and H+ acceptors). Amphoteric substances are compounds that can act an Bronsted acid (proton donar) or a base (a proton acceptor). Water is amphoteric! conjugate take away H + + acid 1 base 2 base 1 + acid 2 conjugate add H + reaction 1 HF + H 2 O F + H 3 O + reaction 2 HCOOH + CN HCOO + HCN Acid BaseH + acceptor Conjugate Acid Conjugate Base Solvent reaction 3 NH CO 3 reaction 4 H 2 PO + 4 OH reaction 5 H 2 SO 4 + N 2 H + 5 reaction 6 HPO + 4 SO 3 NH 3 + HCO 3 HPO 4 + H 2 O HSO + N 4 2 H 2+ 6 PO HSO 3 Base AcidH + donor Base Acid The stronger the acid (base) is, the weaker is its conjugate base (acid). strong acid conjugate base (very weak base) HCl(g) + H2O(aq) <===> Cl (aq) + H3O + (aq) The weaker the acidthe stronger its conjugate base. The stronger the acidthe weaker its conjugate base. Relative Strengths of Conjugate AcidBase Pairs Acid (HA) Conjugate Base, A strong base conjugate acid (very weak acid) NaOH(s) + H2O(l) <===> Na + + H2O + OH Weak acids (bases) give strong conjugate bases (acids). HCOOH + H2O <===> HCOO + H3O + weak acid basic in water NH3(aq) + H2O(aq) <===> NH4 + (aq) + OH (aq) weak base acidic in water An acidbase reaction will go in that direction that forms the weaker acidbase pair. In which direction will these reactions proceed? H 2 SO 4 (aq) + NH 3 (aq) NH 4+ (aq) + HSO 4 (aq) Given that Ka for HF = 6.8 X 10 4 and Ka for H2S = 9.0 X 10 8 In which direction will this reaction proceed? HCO 3 (aq) + SO 4 (aq) HSO 4 (aq) + CO 3 (aq) HS (aq) + HF(aq) F (aq) + H2S(aq)

11 In which direction will these reactions proceed? In which direction will these reactions proceed? H 2 SO 4 (aq) + NH 3 (aq) HCO 3 (aq) + SO 4 (aq) NH 4+ (aq) + HSO 4 (aq) HSO 4 (aq) + CO 3 (aq) H 2 SO 4 and NH 4 + are acids, NH 3 and HSO 4 are bases. H 2 SO 4 is stronger acid than NH 4+, and NH 3 stronger base than HSO 4, therefore NH 3 gets the proton H 2 SO 4 (aq) + NH 3 (aq) NH 4+ (aq) + HSO 4 (aq) stronger acid stronger base weaker acid weaker base HCO 3 and HSO 4 are acids, SO 4 and CO 3 are bases. HSO 4 is the stronger acid, and CO 3 is the stronger base. Therefore, CO 3 gets the proton HCO 3 (aq) + SO 4 (aq) HSO 4 (aq) + CO 3 (aq) weaker acid weaker base stronger acid stronger bas Water is usually a reactivesolvent with all acids and bases. acid + base <=> conjugate acid + base HClO4(l) + H2O(l) <==> H3O + (aq) + ClO4 (aq) NaOH(s) + H2O(l) <==> OH (aq) + Na + (aq) + H2O(l) An equilibrium constant called AcidDissociation Constant, Ka quantifies the extent of acid dissociation (i.e. acid strength). H2O HA(g) <==> H + + A K a = [H+ ][A ] [HA] Water s reaction can be viewed as acidbase reaction. HClO4(l) + H2O(l) <==> H3O + (aq) + ClO4 (aq) acid + base <==> conjugate acid + base Net Ionic Equation: H + (aq) + OH (aq) => H2O (aq) HA(g) + H2O(l) <==> H3O + (aq) + A (aq) K a = [H 3O + ][A ] [HA] KEY DOTS TO CONNECT large K a => higher [H 3 O + ] => stronger acid small K a => lower [H 3 O + ] => weaker acid Acid Strength Is Summarized in Tables of Ka and pka Acid Ionization Equilibrium Ka pka Hydroiodic Acid HI + H2O <==> H3O + + I 3 X Hydrochloric Acid HCl + H2O <==> H3O + + Cl 1.3 X Sulfuric Acid H2SO4 + H2O <==> H3O + + HSO Hydronium Ion H3O + H2O <==> H3O + + IO Nitric Acid HNO3 + H2O <==> H3O + + NO Iodic acid HIO3 + H2O <==> H3O + + IO X Chlorous Acid HClO2 + H2O <==> H3O + + ClO2 1.1 X Phosphoric Acid H3PO4 + H2O <==> H3O + + H2PO4 7.6 X Nitrous Acid HNO2 + H2O <==> H3O + + NO2 7.2 X Hydrofluoric Acid HF + H2O <==> H3O + + F 6.61 X Formic Acid HCOOH + H2O <==> H3O + + HCO2 1.8 X Acetic Acid CH3CO2H + H2O <==> H3O + + CH3CO2 1.8 X Hydrosulfuric Acid H2S + H2O <==> H3O + + HS 1.32 X Hypochlorous Acid HClO + H2O <==> H3O + + ClO 2.88 X Hydrocyanic Acid HCN + H2O <==> H3O + + CN 6.2 X Ka is the equilibrium constant that describes the strength of an acid and how much it dissociates. 1. Strongs acids dissociate completely into ions in water...i.e...k a >> 1. HA(g) + H 2 O(l) H 3 O + (aq) + A (aq) K a = [H 3O + ][A ] [HA] stronger acid => higher [H 3 O + ] => productfavored => larger K a 2. Weak acids dissociate very slightly. Ka << 1 HA(aq) + H 2 O(l) H 3 O + (aq) + A (aq) K a = [H 3O + ][A ] [HA] weaker acid => lower [H 3 O + ] => reactantfavored = lower K a

12 Strong acids and strong bases completely dissociate or completely ionize in water: Ka >> 1. HA(g) + H 2 O(l) H 3 O + (aq) + A (aq) Ka = [A ][H3O + ] [HA] productfavored Before dissociation After dissociation H3O + Weak acids and weak bases dissociate only to a slight extent in water Ka << 1. HA(aq) + H 2 O(l) H 3 O + (aq) + A (aq) Ka = [A ][H3O + ] [HA] reactantfavored Before dissociation After dissociation HA HA HA HA HA HA HA HA H3O + A Ka >>1 HA HA H3O + A Acids can have one, two or three acidic protons depending on their structure (and Ka1 Ka2 Ka3). Monoprotic acidsonly one H + available HCl H + + Cl Strong electrolyte, strong acid HNO 3 H + + NO 3 Strong electrolyte, strong acid CH 3 COOH H + + CH 3 COO Weak electrolyte, weak acid Diprotic acidstwo acidic H + available for reaction H 2 SO 4 H + + HSO 4 HSO 4 H + + SO 4 Strong electrolyte, strong acid Weak electrolyte, weak acid Kw is the equilibrium constant that describes the auto or selfionization of water. H 2 O(l) H 2 O(l) + H 2 O(l) Base H + (aq) + OH (aq) H 3 O + (aq) + OH (aq) Acid Acid Base K w = [H3O + ][OH ] = 1 X K w = [H + ][OH ] K w = [H 3 O + ][OH ] Triprotic acidsthree acidic H + H 3 PO 4 H + + H 2 PO 4 H 2 PO 4 H + + HPO 4 HPO 4 H + + PO 3 4 Weak electrolyte, weak acid This equilibrium constant for water is called the IonProduct of Water. Memorize the value. The product of [H3O + ][OH ] = 1 X a constant. Once we know one...we know the other! [H3O + ] [OH ] Kw 1 X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X ph is a defined logarithmic scale for measuring acidity. It is directly related to [H + ] concentration in solution. ph = log10 [H + ] = log10 [H3O + ] logarithmic form p = log10 [H + ] = 10 ph exponential form 1 ph unit represents a power of 10X in concentration (or an order of magnitude as it is called).

13 Acidic, basic and neutral solutions are identified by their ph values [H3O + ] ph [OH ] ph = log [H3O + ] connect these dots K w = [H3O + ][OH ] = The ionproduct of water is a constant! A solution with a ph of 4 is 10X lower in [H + ] than one with ph=3, 100X lower than ph=2 and 1000X lower than ph = 1. [H + ] log[h + ] = ph 0.5M M M M When concentration changes by factor of 10, ph changes by +/ 1 The amounts of H3O + and OH in aqueous solution determines whether a solution is acidic, neutral or basic. acidic solution ph < 7 WHEN [H 3 O + ] > [OH ] [H 3 O + ] = [OH ] [H 3 O + ] < [OH ] neutral solution ph = 7 Litmus Paper Color basic solution ph > 7 Log Knowing and the Exponential math can help you Forms go fast Logarithmic Form ph = log [H 3 O + ] 10 ph = [H 3 O + ] poh = log [OH ] 10 poh = [OH ] pk a = log K a Exponential Form 10 pka = pk a When H 3O +, OH, K a are large valuesthe pk s are small values. Strong acids have large H 3O and K a therefore small ph and small pk a Kw is related to ph via logarithm. K W = [H 3 O + ][OH ]= 1.0 X take log of both sides What is the concentration of OH ions in a HCl solution whose hydrogen ion concentration is 1.3 M? log Kw = log ([H 3 O + ][OH ]) = log(1.0 X ) log K W = log[h 3 O + ] + log[oh ] = log (10 14 ) pk W = ph + poh = (14) pk W = ph + poh = 14 K W = [H 3 O + ][OH ] = 10 14

14 What is the concentration of OH ions in a HCl solution whose hydrogen ion concentration is 1.3 M? Key: Strong acids dissociate completely stoichiometry dictates [H + ]. Memorize all strong acids (HCl, HBr, HI, HNO3, H2SO4, HClO4) A research chemist adds a measured amount of HCl gas to pure water at 25 o C and obtains a solution with [H 3 O + ] = 3.0 x 10 4 M. Calculate [OH ]. Is the solution neutral, acidic or basic? [H + ] = 1.3 M K w = [H + ][OH ] = 1.0 x [OH ] = K w 1 x 10 [H + = 14 = 7.7 x 10 ] M A research chemist adds a measured amount of HCl gas to pure water at 25 o C and obtains a solution with [H 3 O + ] = 3.0 x 10 4 M. Calculate [OH ]. Is the solution neutral, acidic or basic? Again, strong acids dissociate completely stoichiometry dictates [H + ] The ph of rainwater collected in a certain region of the northeastern United States on a particular day was What is the H + ion concentration of the rainwater? ph = log [H + ] antilog(ph) = antilog (log [H + ]) [H + ] = 10 ph = = 1.5 x 10 5 M SOLUTION: K w = [H + ][OH ] = 1.0 x The OH ion concentration of a blood sample is 2.5 x 10 7 M. What is the ph of the blood? K w [OH ] = = 1 x [H3O + = 3.3 x 10 ] 11 M 3.0 x 10 4 [H 3 O + ] > [OH ]; the solution is acidic. ph + poh = poh = log [OH ] = log (2.5 x 10 7 ) = 6.60 ph = poh = = 7.40 Problem: Calculating [H 3 O + ], ph, [OH ], and poh In a restoration project, a conservator prepares copperplate etching solutions by diluting concentrated nitric acid, HNO 3, to 2.0 M, 0.30 M, and M HNO 3. Calculate [H 3 O + ], ph, [OH ], and poh of the three solutions at 25 o C. PLAN: HNO 3 is a strong acid so [H 3 O + ] = [HNO 3 ]. Use K w to find the [OH ] and then convert to ph and poh. Problem: Calculating [H 3 O + ], ph, [OH ], and poh In a restoration project, a conservator prepares copperplate etching solutions by diluting concentrated nitric acid, HNO 3, to 2.0 M, 0.30 M, and M HNO 3. Calculate [H 3 O + ], ph, [OH ], and poh of the three solutions at 25 o C. For 2.0 M HNO 3, [H 3 O + ] = 2.0 M, and ph = log [H 3 O + ] = 0.30 = ph [OH ] = K w / [H 3 O + ] = 1.0 x /2.0 = 5.0 x M; poh = For 0.3 M HNO 3, [H 3 O + ] = 0.30 M and log [H 3 O + ] = 0.52 = ph [OH ] = K w /[H 3 O + ] = 1.0 x /0.30 = 3.3 x M; poh = For M HNO 3, [H 3 O + ] = M and log [H 3 O + ] = 2.20 = ph [OH ] = K w / [H 3 O + ] = 1.0 x /6.3 x 10 3 = 1.6 x M; poh = 11.80

15 AcidBase Equilibrium Problems 1 2 Strong Acids/Bases Weak Acid Weak Bases Know all strong acids and bases and you will recognize be able to recognize all weak acids/bases. 100% dissociation No need for ICE table. [H 3O + ] = [HA]. 1. Find K a /K b given equilibrium values 2. Find equilibrium values given K a Identify and write the correct equation 1. Simplify: If 100 x Ka/b < [A]0 then [HA] x = [HA] 2. Look for perfect square 3. Quadratic equation hydrohalic acids perchloric nitric sulfuric Group I hydroxides Group II hydroxides If an acid is a strong acid (base), it is 100% dissociatedeasy for calculations of ph, poh. HCl + H2O > Cl + H3O Determining concentrations and ph from K a and initial [HA] The ph of a solution containing M HF solution is What is the value of K a for HF at equilibrium? NaOH + H2O <> Na + + OH Weak acids (bases) has reversible equilibrium we must use an ICE Table and algebra! HCOOH + H2O <> HCOO + H3O + NH3 + H2O <> NH4 + + OH Determining concentrations and ph from K a and initial [HA] The ph of a solution originally containing M HF solution is What is the value of K a for HF at equilibrium? Initial Change Equilibrium [HF] = = M HF + H 2 O H 3 O + + F.250 M x M 0.250x K a = (9.2 x 10 3 ) 2 /0.241 = 3.52 x 10 4? M? M +x M +x M ph = = log [H 3 O + ] K a = [H 3O + ][F ] [HF] 10 ph = [H 3 O + ] = = 9.20 x 10 3 M x x There are 2 key simplifications that can help us solve equilibrium problems rapidly. 1. [H 3 O + ] from the autoionization of water is negligible in comparison to H + from an acid, HA in solution. 2. If 100 x Ka < [HA]initial then [HA]initial x = [HA] Why Does it Work? Because Ka values are known to about ± 5% accuracy. Check using the 5% Rule [H + ]equilbrium [HA] inital x 100 < 5% 3. If #2 fails then you must use the quadratic equation

16 Determining concentrations from K a and initial [HA] Propanoic acid (CH 3 CH 2 COOH, simplified as HPr) is an organic acid whose salts are used to retard mold growth in foods. What is the [H 3 O + ] of a 0.10 M aqueous solution of HPr (K a of HPr = 1.3 x 10 5 )? PLAN: 1. Write a balanced equation for acid in water. 2. Write the equation for Ka. 3. Setup and fill in the ICE Table 4. Use ph, initial solution concentration and ICE table to find K a. HPr(aq) + H 2 O(l) H 3 O + (aq) + Pr (aq) initial change x +x +x equilibrium 0.10 x x x K a = [H 3O + ][Pr ] [HPr] x = b ± b 2 4ac 2a = 1.3 x 10 5 = K a = [H 3O + ][Pr ] [HPr] If No x x = 1.3 x 10 5 = Can We Simplify? 100 x Ka < [HA]initial then [HA]initial x = [HA] 100 x 1.3 x 10 5 < [.10]? Yes thus: 0.10 x = 0.10 x K a = [H 3O + ][Pr ] [HPr] = 1.3 x 10 5 = x A Lewis Acid is any substance that is an electronpair acceptor (have an empty orbital for electrons). = 1.1 x 10 3 M = [H 3 O + ] solving for x Neutral Proton Donors H + BF 3 Lewis acids lack electrons and have an orbital. Checking assumption with 5% Rule: If [H + ]equilibrium divided by [HA]initial (x 100) is < 5% then assumption is valid. Some ewis Acids Cations %[HPr] diss = M 0.10M = 1.1% our assumption is valid << 5% Transition Metal Cations Al 3+, Ti 4+, Fe 3+, Zn 2+, Transition metal ions (Al 3+, Fe 3+, Cr 3+ ) react with water and act as Lewis acids in aqueous solution. Al(H 2 O) 6 (aq) electron density drawn toward Al solvent H 2 O acts as base Al(OH)(H 2 O) 5 (aq) + H + (aq) Acid Hydrolysis of Al 3+ H 2 O H 3 O + A Lewis Base is any substance that is an electronpair donor (nucleophile). Atoms having lone pairs of electrons is the distinguish ing feature H 2 O NH 3 O 2 H + lost

17 Neutralization (in the Lewis acid context) occurs when a Lewis acid reacts and forms a covalent bond with a Lewis base. The bond is called a coordinate bond. acid base adduct The adduct contains a new covalent bond. Identify the Lewis acids and Lewis bases in the following reactions: (a) H + + OH H 2 O (b) Cl + BCl 3 BCl 4 (c) K + + 6H 2 O K(H 2 O) + 6 adduct M 2+ H 2 O(l) M(H 2 O) 4 2+ (aq) Identify the Lewis acids and Lewis bases in the following reactions: (a) H + + OH H 2 O (b) Cl + BCl 3 BCl 4 (c) K + + 6H 2 O K(H 2 O) + 6 Look for electron pair acceptors (acids) and electron pair donors (bases). acceptor (a) H + + OH donor H 2 O (c) K + + 6H 2 O K(H 2 O) 6 + acceptor donor donor (b) Cl + BCl 3 BCl 4 acceptor