2014 Acid Base Equilibrium

Chemistry Acid Base Equilibrium Advanced Placement

2014 Periodic Table of the Elements

1 2 H He 1.0079 4.0026 3 4 5 6 7 8 9 10 Li Be B C N O F Ne 6.941 9.012 10.811 12.011 14.007 16.00 19.00 20.179 11 12 13 14 15 16 17 18 Na Mg Al Si P S Cl Ar 22.99 24.30 26.98 28.09 30.974 32.06 35.453 39.948 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr 39.10 40.08 44.96 47.90 50.94 52.00 54.938 55.85 58.93 58.69 63.55 65.39 69.72 72.59 74.92 78.96 79.90 83.80 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe 85.47 87.62 88.91 91.22 92.91 93.94 (98) 101.1 102.91 106.42 107.87 112.41 114.82 118.71 121.75 127.60 126.91 131.29 55 56 57 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 Cs Ba * La Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn 132.91 137.33 138.91 178.49 180.95 183.85 186.21 190.2 192.2 195.08 196.97 200.59 204.38 207.2 208.98 (209) (210) (222) 87 88 89 104 105 106 107 108 109 110 111 112 † Fr Ra Ac Rf Db Sg Bh Hs Mt § § § §Not yet named (223) 226.02 227.03 (261) (262) (263) (262) (265) (266) (269) (272) (277)

58 59 60 61 62 63 64 65 66 67 68 69 70 71 *Lanthanide Series: Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu 140.12 140.91 144.24 (145) 150.4 151.97 157.25 158.93 162.50 164.93 167.26 168.93 173.04 174.97 90 91 92 93 94 95 96 97 98 99 100 101 102 103 †Actinide Series: Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr 232.04 231.04 238.03 237.05 (244) (243) (247) (247) (251) (252) (257) (258) (259) (260) ADVANCEDADVANCED PLACEMENTPLACEMENT CHEMISTRYCHEMISTRY EQUATIONSEQUATIONS ANDAND CONSTANTSCONSTANTS

ThroughoutThroughout thethe testtest thethe followingfollowing symbolssymbols havehave ththee definitionsdefinitions specifiedspecified unlessunless otherwiseotherwise noted.noted.

L,L, mLmL = = liter(s), liter(s), milliliter(s)milliliter(s) mmmm HgHg = = millimeters millimeters ofof mercurymercury gg = = gram(s)gram(s) J,J, kJkJ = = joule(s), joule(s), kilojoule(s)kilojoule(s) nmnm = = nanometer(s)nanometer(s) VV = = volt(s)volt(s) atmatm = = atmosphere(s)atmosphere(s) molmol = = mole(s)mole(s)

ATOMICATOMIC STRUCTURESTRUCTURE EE == energyenergy EE == hhνν νν == frequencyfrequency cc == λνλν λλ == wavelengthwavelength

Planck’sPlanck’s constant,constant, hh == 6.6266.626 × × 1010−−3434 JJ s s SpeedSpeed ofof light,light, cc == 2.9982.998 ×× 101088 msms−−11 Avogadro’sAvogadro’s nnumberumber == 6.0226.022 ×× 10102323 molmol−−11 ElectronElectron charge,charge, e e == −−1.1.602602 ××1010−−1919 coulombcoulomb

EQUILIBRIUMEQUILIBRIUM [C][C]cdcd [D] [D] KKcc == ,, wherewhere aa AA ++ bb BB RR cc CC ++ dd DD EquilibriumEquilibrium Constants Constants [A][A]abab [B] [B] cdcd KKcc (molar(molar concentrations)concentrations) ((PPPPCC )( )(DD ) ) KKpp == abab KKpp (gas(gas pressures)pressures) ((PPPPABAB )( )( ) ) KKaa (weak(weak acid)acid) [H[H+-+- ][A ][A ] ] KK == K (weak base) aa [HA][HA] Kbb (weak base) KK (water)(water) [OH-+-+ ][HB ] ww KK == [OH ][HB ] bb [B][B] ++ −− −−1414 KKww == [H[H ][OH][OH ]] == 1.01.0 ××1010 atat 2525°°CC

== KKaa ×× KKbb pHpH == −−loglog[H[H++]] , , pOHpOH == −−log[OHlog[OH−−]] 1414 == pHpH ++ pOHpOH -- pHpH == ppKK ++ loglog[A[A ] ] aa [HA][HA]

ppKKaa == −−loglogKKaa,, ppKKbb == −−loglogKKbb

KINETICSKINETICS ln[A]ln[A] −− ln[A]ln[A] == −−ktkt tt 00 kk = = raterate constantconstant 1111 tt == timetime -- = = ktkt AAAA [][]tt [][]00 tt½½ == half-lifehalf-life tt == 0.6930.693 ½½ kk GASES, LIQUIDS, AND SOLUTIONS P = pressure V = volume PV = nRT T = temperature moles A n = number of moles PA = Ptotal × XA, where XA = total moles m = mass M = molar mass Ptotal = PA + PB + PC + . . . D = density n = m KE = kinetic energy M Ã = velocity K = °C + 273 A = absorbance a = molarabsorptivity m D = b = path length V c = concentration KE per molecule = 1 mv2 2 Gas constant, R = 8.314 J mol--11 K Molarity, M = moles of solute per liter of solution --11 = 0.08206 L atm mol K --11 A = abc = 62.36 L torr mol K 1atm= 760 mm Hg = 760 torr STP= 0.00D C and 1.000 atm

THERMOCHEMISTRY/ ELECTROCHEMISTRY q = heat m = mass q= mcD T c = specific heat capacity T temperature DSSDD= ÂÂ products- S D reactants = SD = standard entropy DD D DDHH= ÂÂproducts- D Hf reactants f HD = standard enthalpy DD D GD standard free energy DDGG= ÂÂf products- D Gf reactants = n = number of moles D DDGD = HDD - TS D E = standard reduction potential I = current (amperes) = -RTln K q = charge (coulombs) t = time (seconds) = -nFED

q Faraday’s constant , F = 96,485 coulombs per mole I ϭ t of electrons 1 joule 1volt = 1coulomb ACID – BASE EQUILIBRIUM Ka, Kb, and Salt Hydrolysis

What I Absolutely Have to Know to Survive the AP Exam The following might indicate the question deals with acid – base equilibrium: + − pH, pOH, [H3O ], [OH ], strong and weak, salt hydrolysis, solubility product, Ka, Kb, acid or base dissociation constant, percent ionized, Bronsted-Lowry, Arrhenius, hydronium ion, etc…

Acids and Bases: By Definition! Arrhenius • Acids: Hydrogen ion, H+, donors • Bases: Hydroxide ion, OH− donors HCl → H+ + Cl− NaOH → Na+ + OH− Bronsted-Lowry • Acids: proton donors (lose H+) Bases: proton acceptors (gain H+) • + − HNO3 + H2O → H3O + NO3 + − NH3 + H2O ⇌ NH4 + OH

Hydrogen Ion, Hydronium Ion, Which is it!!!? Hydronium − H+ 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) + • H3O “Anthony” • H+ “Tony” + • Often used interchangeably in problems; if H3O 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 H+ ion. + − HC2H3O2 + H2O → H3O + C2H3O2 − + • HC2H3O2 is the acid; thus C2H3O2 is its conjugate base (what remains after the H has been donated to the H2O molecule) • H2O behaves as a base in this reaction. The hydronium ion is its conjugate acid (what is formed after the H2O accepts the H+ ion) + − NH3 + H2O → NH4 + OH + + • NH3 is the base; thus NH4 is it’s conjugate acid (what is formed after the NH3 accepts the H ion) + • H2O behaves as an acid in this reaction. The hydroxide ion is its conjugate base (what remains after the H has been donated to NH3)

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 EITHER acids or bases; water, some anions of weak acids, etc… fit this bill. Monoprotic − acids donating one H+ Diprotic − acids donating two H+ Polyprotic − acids donating 3+ H+

Regardless, always remember: ACIDS ONLY DONATE ONE PROTON AT A TIME!!!

Ionization: That’s What it’s All About! Relative Strengths A strong acid or base ionizes completely in aqueous solution (100% 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 H3O /OH 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 • Hydrohalic acids: HCl, HBr, HI; Nitric: HNO3; Sulfuric: H2SO4; Perchloric: HClO4 • Oxyacids! More oxygen atoms present, the stronger the acid WITHIN that group. The H+ 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 H+ is bonded, “polarizing it”, which makes it easier [i.e. requires less energy] to remove − thus the stronger the acid! O

HO Br < HO Br O < HO Br

O • Strong Bases • Group IA and IIA (1 and 2) metal hydroxides; be cautious as the poor solubility of Be(OH)2 and Mg(OH)2 limits the effectiveness of these 2 strong bases. − • IT’S A 2 for 1 SALE with the Group 2 (IIA) ions), i.e. 0.10M Ca(OH)2 is 0.20M OH

Where the fun begins… A weak acid or base does not completely ionize (usually < 10%) • They are weak electrolytes. • The equilibrium position lies far to the left (favors reactants)… • [H+] is much less than the acid concentration − thus to calculate this amount and the resulting pH 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 H+) at a time! + − • For weak acid reactions: HA + H2O → H3O + A [H O+ ][A- ] K = 3 where K is typically much less than 1 a [HA] a + − • For weak base reactions: B + H2O → HB + OH [HB+ ][OH- ] K = where K is typically much less than 1 b [B] b

Ionization: That’s What it’s All About! con’t. Working it Out! Calculating the pH 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 [HA]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, HA and the ionization constant for the acid, Ka, plug both into the + − equilibrium expression and solve. Both H3O and A are equal (remember it's 1:1 always) so call them x and solve. [H O+ ][A- ] K = 3 a [HA] • Thus all you need to know is…

[x][x] + [x][x] - K a = where x = [H 3O ] or K b = where x = [OH ] [M o ] [M o ] −14 • Never forget KKab×=110× • 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] + − % = ×100 where x = [H 3O ]or[OH ] [M o ]

Salts and pH: It’s all about Hydrolysis A salt is the PRODUCT of an acid base reaction; SOME salts affect the pH 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 NaNO3 does not affect the pH of the solution; if it could hydrolyze water the following would happen: + + Na + 2 H2O → NaOH + H3O − − NO3 + H2O → HNO3 + OH − Neither of the salt’s ions can hydrolyze water because it would result in the formation of a strong acid (HNO3) or a strong base (NaOH), which would ionize 100% back into the ions. 2. K2S would make the solution more basic—SB (KOH) & WA (H2S) − NOTE: Only the ion that forms a weak acid or base through water hydrolysis can actaully hydrolyze water. In this case: 2− − S + H2O → H2S + OH The increase in OH− ions means the anion of this salt makes the solution more basic… 3. NH4Cl would make the solution more acidic—since WB (NH4OH – or NH3) & SA (HCl) In this case: + + NH4 + H2O → NH3 + H3O + The increase in H3O ion means the solution is more acidic…

Salts and pH: It’s all about Hydrolysis con’t Working it Out! Calculating the pH 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 Ka or Kb 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 NaC2H3O2; the Ka for HC2H3O2 is typically provided. The Kb needs to be calculated. −14 • K a × K b = 1×10 Very important! • Then solve as shown below to find the pH…

[x][x] - K b = where x = [OH ] [M o ]

Acid – Base Equilibrium Cheat Sheet

Relationships

Equilibrium Expression [H O+ ][A- ] [HB+ ][OH- ] K = 3 K = a [HA] b [B]

[x][x] + [x][x] - K a = where x = [H3 O ] K b = where x = [OH ] [Mo ] [Mo ]

[x] + − %=× 100 where x = [H3 O ]or[OH ] [Mo ]

−14 + KKab×=110× pH = −log[H ] pH + pOH = 14 pOH = −log[OH−]

+ − −14 + − Kw = [H3O ][OH ] = 1.0×10 @ 25°C Acidic pH < 7 [H3O ] > [OH ]

+ − + − Neutral pH = 7 [H3O ] = [OH ] Basic pH > 7 [H3O ] < [OH ]

+ x − y + − Conjugate acids and bases Ksp = [M ] [A ] MA(s) ⇌ xM (aq)+ yA (aq)

Connections

Equilibrium Buffers and Titrations

Precipitation and Qualitative Analysis Bonding and Lewis Structures − justify oxyacid strengths

Potential Pitfalls

Ka or Kb with salt pH

Weak acids and bases − be sure you know what you are using, acid or base – Ka or Kb; solving a Kb problem gives Weak is about IONIZATION not CONCENTRATION [OH−] thus you are finding the pOH!

NMSI SUPER PROBLEM

+ − -5 NH3(aq) + H2O() ⇌ NH4 (aq) + OH (aq) Kb = 1.80 × 10

1. Ammonia reacts with water as indicated in the reaction above.

(a) Write the equilibrium constant expression for the reaction represented above.

(b) Calculate the pH of a 0.150 M solution of NH3

+ (d) Calculate the hydronium ion, H3O , concentration in the above solution. Be sure to include units with your answer.

When a specified amount of ammonium nitrate (NH4NO3) is dissolved in water, the ammonium ions hydrolyze the water according to the partial reaction shown below. The resulting solution has a pH of 4.827.



(e) Complete the reaction above by drawing the complete Lewis structures for both products of the hydrolysis reaction.

(f) Determine the (i) molarity (M) of the ammonium ions in this solution

(ii) number of moles ammonium ions in 250 mL of the above solution.