Advanced Placement Presenter Copy Chemistry Acid Base Equilibrium 2014
47.90 91.22 178.49 (261) 50.94 92.91 180.95 (262) 52.00 9.94 18.85 (26) 54.98 (98) 186.21 (262) 55.85 1.1 190.2 (265) 58.9 2.91 192.2 (266) Li Na K Rb Cs Fr e Ne Ar Kr Xe Rn Ce Th Pr Pa Nd U Pm Np Sm Pu Eu Am Gd Cm Tb Bk Dy Cf o Es Er Fm Tm Md Yb No Lu Lr 1 11 19 7 55 87 2 18 6 54 86 Be Mg Ca Sr Ba Ra B Al Ga In Tl C Si Ge Sn Pb N P As Sb Bi O S Se Te Po F Cl Br I At Sc Y La Ac Ti Zr f Rf V Nb Ta Db Cr Mo W Sg Mn Tc Re Bh Fe Ru Os s Co Rh Ir Mt Ni Pd Pt Cu Ag Au Zn Cd g 1.0079 6.941 22.99 9. 85.47 12.91 (22) 4.0026 20.179 9.948 8.80 11.29 (222) 140.12 22.04 140.91 21.04 144.24 28.0 (145) 27.05 150.4 (244) 151.97 (24) 157.25 (247) 158.9 (247) 162.50 (251) 164.9 (252) 167.26 (257) 168.9 (258) 17.04 (259) 174.97 (260) 9.012 24.0 40.08 87.62 17. 226.02.811 26.98 69.72 114.82 204.8 12.011 28.09 72.59 118.71 207.2 14.007 0.974 74.92 121.75 208.98 16.00 2.06 78.96 127.60 (209) 19.00 5.45 79.90 126.91 (2) 44.96 88.91 18.91 227.0 58.69 6.42 195.08 (269) 6.55 7.87 196.97 (272) 65.9 112.41 200.59 (277) 4 12 20 8 56 88 5 1 1 49 81 6 14 2 50 82 7 15 51 8 8 16 4 52 84 9 17 5 5 85 21 9 57 89 58 90 *Lanthanide Series: Actinide Series: * 59 91 60 92 61 9 62 94 6 95 64 96 65 97 66 98 67 99 68 0 69 1 70 2 71 22 40 72 4 2 41 7 5 24 42 74 6 25 4 75 7 26 44 76 8 27 45 77 9 28 46 78 1 29 47 79 111 0 48 80 112 Periodic Table of the Elements Not yet named
Throughout the test the following symbols have the definitions specified unless otherwise noted. L, ml liter(s), milliliter(s) mm g millimeters of mercury g gram(s) J, kj joule(s), kilojoule(s) nm nanometer(s) V volt(s) atm atmosphere(s) mol mole(s) ATOMIC STRUCTURE E hν c λν E energy ν frequency λ wavelength Planck s constant, h 6.626 4 J s Speed of light, c 2.998 8 ms 1 Avogadro s number 6.022 2 mol 1 Electron charge, e 1.602 19 coulomb EQUILIBRIUM K c K p K a c d [C] [D] a b, where a A + b B c C + d D [A] [B] c d ( PC )( PD ) a b ( PA) ( PB) + - [ ][A ] [A] - + [O ][B ] [B] K w [ + ][O ] 1.0 14 at 25 C Equilibrium Constants K c (molar concentrations) K p (gas pressures) K a (weak acid) (weak base) K w (water) K a p log[ + ],, po log[o ] 14 p + po p pk a + log [A - ] [A] pk a logk a, p log KINETICS ln[a] t ln[a] 0 kt 1-1 kt [ A] [ A] 0 t t ½ 0.69 k k rate constant t time t ½ half-life
GASES, LIQUIDS, AND SOLUTIONS PV nrt P A P total X A, where X A P total P A + P B + P C +... n m M K C + 27 D m V KE per molecule 1 2 mv2 moles A total moles Molarity, M moles of solute per liter of solution A abc P V T n m M D KE Ã A a b c pressure volume temperature number of moles mass molar mass density kinetic energy velocity absorbance molarabsorptivity path length concentration -1-1 Gas constant, R 8.14 J mol K 0.08206 L atm mol K -1-1 62.6 L torr mol K 1 atm 760 mm g 760 torr STP 0.00 C and 1.000 atm -1-1 TERMOCEMISTRY/ ELECTROCEMISTRY q mcdt DS ÂS products -ÂS reactants D ÂD products -ÂD reactants f DG ÂDG products -ÂDG reactants DG D - TDS -RT ln K -nfe q I t f f f q heat m mass c specific heat capacity T temperature S standard entropy standard enthalpy G standard free energy n number of moles E I q t Faraday s constant, F standard reduction potential current (amperes) charge (coulombs) time (seconds) 96,485 coulombs per mole of electrons 1 joule 1volt 1coulomb
ACID BASE EQUILIBRIUM K a,, and Salt ydrolysis What I Absolutely ave to Know to Survive the AP Exam The following might indicate the question deals with acid base equilibrium: p, po, [ O + ], [O ], strong and weak, salt hydrolysis, solubility product, K a,, acid or base dissociation constant, percent ionized, Bronsted-Lowry, Arrhenius, hydronium ion, etc Arrhenius Acids: ydrogen ion, +, donors Bases: ydroxide ion, O donors Bronsted-Lowry Acids: proton donors (lose + ) Bases: proton acceptors (gain + ) Acids and Bases: By Definition! Cl + + Cl NaO Na + + O NO + 2 O O + + NO N + 2 O N 4 + + O ydrogen Ion, ydronium Ion, Which is it!!!? ydronium + riding piggy-back on a water molecule; water is polar and the + charge of the naked proton is greatly attracted to Mickey's chin (i.e. the oxygen atom) O + Anthony + Tony Often used interchangeably in problems; if O + is used be sure water is in the equation! Bronsted Lowry and Conjugate Acids and Bases: What a Pair! Acid and conjugate base pairs differ by the presence of one + ion. C 2 O 2 + 2 O O + + C 2 O 2 C 2 O 2 is the acid; thus C 2 O 2 is its conjugate base (what remains after the + has been donated to the 2 O molecule) 2 O behaves as a base in this reaction. The hydronium ion is its conjugate acid (what is formed after the 2 O accepts the + ion) N + 2 O N 4 + + O N is the base; thus N 4 + is it s conjugate acid (what is formed after the N accepts the + ion) 2 O behaves as an acid in this reaction. The hydroxide ion is its conjugate base (what remains after the + has been donated to N ) Understanding conjugate acid/base pairs is very important in understanding acid-base chemistry; this concepts allows for the understanding of many complex situations (buffers, titrations, etc ) Important Notes Amphiprotic/amphoteric--molecules or ions that can behave as EITER acids or bases; water, some anions of weak acids, etc fit this bill. Monoprotic acids donating one + Diprotic acids donating two + Polyprotic acids donating + + Regardless, always remember: ACIDS ONLY DONATE ONE PROTON AT A TIME!!! Copyright 2014 National Math + Science Initiative, Dallas, Texas. All Rights Reserved. Visit us online at www.nms.org Page 1
Ionization: That s What it s All About! Relative Strengths A strong acid or base ionizes completely in aqueous solution (0% ionized [or very darn close to it!]) The equilibrium position lies far, far to the right (favors products) Since a strong acid/base dissociates into the ions, the concentration of the O + /O ion is equal to the original concentration of the acid/base respectively. They are strong electrolytes. Do Not confuse concentration (M or mol/l) with strength! Strong Acids ydrohalic acids: Cl, Br, I; Nitric: NO ; Sulfuric: 2 SO 4 ; Perchloric: ClO 4 Oxyacids! More oxygen atoms present, the stronger the acid WITIN that group. The + that is donated is bonded to an oxygen atom. The oxygen atoms are highly electronegative and are pulling the bonded pair of electrons AWAY from the site where the + is bonded, polarizing it, which makes it easier [i.e. requires less energy] to remove thus the stronger the acid! O O Br < O Br O < O Br Strong Bases Group IA and IIA (1 and 2) metal hydroxides; be cautious as the poor solubility of Be(O) 2 and Mg(O) 2 limits the effectiveness of these 2 strong bases. IT S A 2 for 1 SALE with the Group 2 (IIA) ions), i.e. 0.M Ca(O) 2 is 0.20M O O Where the fun begins A weak acid or base does not completely ionize (usually < %) They are weak electrolytes. The equilibrium position lies far to the left (favors reactants) [ + ] is much less than the acid concentration thus to calculate this amount and the resulting p you must return to the world of EQUILIBRIUM Chemistry! The vast majority of acid/bases are weak. Remember, ionization not concentration!!!! Acids and Bases ionize one proton (or + ) at a time! For weak acid reactions: A + 2 O O + + A K a [ O+ ][A - ] [A] For weak base reactions: B + 2 O B + + O [B+ ][O - ] [B] where K a is typically much less than 1 where is typically much less than 1 Copyright 2014 National Math + Science Initiative, Dallas, Texas. All Rights Reserved. Visit us online at www.nms.org Page 2
Ionization: That s What it s All About! con t. Working it Out! Calculating the p of weak acids simply requires some quick problem solving MONUMENTAL CONCEPT On the AP Exam the weak acid will be significantly weak so that the initial [A] o is mathematically the same as the equilibrium concentration i.e. there is no need to subtracting the amount of weak acid ionized as it is mathematically insignificant. If given the concentration of the weak acid, A and the ionization constant for the acid, K a, plug both into the equilibrium expression and solve. Both O + and A are equal (remember it's 1:1 always) so call them x and solve. K a [ O+ ][A - ] [A] Thus all you need to know is Never forget K a K a [x][x] [M o ] where x [ O+ ] or [x][x] [M o ] where x [O - ] K 1 b 14 Very important when you are given the ionization constant for the acid but need the conjugate base and vice-versa. Percent Ionization Often will be asked to determine how ionized the weak acid or base is % [x] [M o ] 0 where x [ O+ ]or[o ] Salts and p: It s all about ydrolysis A salt is the PRODUCT of an acid base reaction; SOME salts affect the p of the solution. So an easy way to tell, you ask? Ask yourself, which acid and which base are needed to make that salt?.were they strong or weak? 1. A salt such as NaNO does not affect the p of the solution; if it could hydrolyze water the following would happen: Na + + 2 2 O NaO + O + NO + 2 O NO + O Neither of the salt s ions can hydrolyze water because it would result in the formation of a strong acid (NO ) or a strong base (NaO), which would ionize 0% back into the ions. 2. K 2 S would make the solution more basic SB (KO) & WA ( 2 S) NOTE: Only the ion that forms a weak acid or base through water hydrolysis can actaully hydrolyze water. In this case: S 2 + 2 O 2 S + O The increase in O ions means the anion of this salt makes the solution more basic. N 4 Cl would make the solution more acidic since WB (N 4 O or N ) & SA (Cl) In this case: N 4 + + 2 O N + O + The increase in O + ion means the solution is more acidic Copyright 2014 National Math + Science Initiative, Dallas, Texas. All Rights Reserved. Visit us online at www.nms.org Page
Salts and p: It s all about ydrolysis con t Working it Out! Calculating the p of a salt solution A salt in solution is either behaving as a weak acid or a weak base. Remember, this means they do not ionize much. So treat them just like any other weak acid or base. The MAJOR DIFFERENCE; no K a or value will be provided. The salt is the conjugate acid or base to some weak acid or base that is provided - that K will be given you need to convert it. Example, the salt NaC 2 O 2 ; the K a for C 2 O 2 is typically provided. The needs to be calculated. K a 1 14 Very important! Then solve as shown below to find the p [x][x] [M o ] where x [O - ] Copyright 2014 National Math + Science Initiative, Dallas, Texas. All Rights Reserved. Visit us online at www.nms.org Page 4
Acid Base Equilibrium Cheat Sheet Relationships Equilibrium Expression K a [ O+ ][A - ] [A] K a [x][x] where x [ + O ] [M o] [x] + % 0 where x [O ]or[o ] [M o] [B+ ][O - ] [B] [x][x] where x [O - K b ] [M ] o K 14 a Kb 1 p log[ + ] p + po 14 po log[o ] K w [ O + ][O ] 1.0 14 @ 25 C Acidic p < 7 [ O + ] > [O ] Neutral p 7 [ O + ] [O ] Basic p > 7 [ O + ] < [O ] Conjugate acids and bases K sp [M + ] x [A ] y MA(s) xm + (aq)+ ya (aq) Connections Equilibrium Buffers and Titrations Precipitation and Qualitative Analysis K a or with salt p Potential Pitfalls Bonding and Lewis Structures justify oxyacid strengths Weak acids and bases be sure you know what you are using, acid or base K a or ; solving a problem gives [O ] thus you are finding the po! Weak is about IONIZATION not CONCENTRATION Copyright 2014 National Math + Science Initiative, Dallas, Texas. All Rights Reserved. Visit us online at www.nms.org Page 5
NMSI SUPER PROBLEM N (aq) + 2 O( ) N + 4 (aq) + O (aq) 1.80-5 1. Ammonia reacts with water as indicated in the reaction above. (a) Write the equilibrium constant expression for the reaction represented above. + [N 4 ][O ] [N ] 1 point is earned for the correct expression (b) Calculate the p of a 0.150 M solution of N x2 M where x [O ] 1.80x 5 x2 0.150 1.64x x log x po 2.784 p 11.216 1 point is earned for the correct set up and for calculating the concentration of hydroxide ions 1 point is earned for the correct p (c) Determine the percent ionization of the weak base N. [x] % 0 [M ] o 1.64 % 0 1.09% [0.150] 1 point is earned for the correct percent ionization (d) Calculate the hydronium ion, O +, concentration in the above solution. Be sure to include units with your answer. [ O ][O ] 1.00 + 14 1.0 1.00 [ O ] 6. [O ] 1.64 OR 14 14 + 12 [ O ] + p [ O ] 6.08 + 11.216 12 M M 1 point is earned for the concentration of hydronium ions with units of mol L 1 or M Copyright 2014 National Math + Science Initiative, Dallas, Texas. All Rights Reserved. Visit us online at www.nms.org Page 6
When a specified amount of ammonium nitrate (N 4 NO ) is dissolved in water, the ammonium ions hydrolyze the water according to the partial reaction shown below. The resulting solution has a p of 4.827. (e) Complete the reaction above by drawing the complete Lewis structures for both products of the hydrolysis reaction. N + O + 1 point is earned for each correct Lewis structure. 2 points possible (f) Determine the (i) molarity (M) of the ammonium ions in this solution [ O + ] p 4.827 1.49 5 K a K w 1.00 14 5.56 5 1.80 5.56 [1.49 5 ][1.49 5 ] [M] M 0.400 1 point is earned for calculating the K a for ammonium ions 1 point is earned for calculating the concentration of ions in the solution. (ii) number of moles ammonium ions in 250 ml of the above solution. M 0.400 0.400M x mol 0.250 L x 0.0 mol 1 point is earned for calculating the molar concentration and the number of moles of ammonium ions Copyright 2014 National Math + Science Initiative, Dallas, Texas. All Rights Reserved. Visit us online at www.nms.org Page 7