Chemistry Buffers and Titrations Advanced Placement 2014 2 Ar Kr 10 18 36 54 86 Xe He Ne Rn (222) 83.80 4.0026 20.179 39.948 131.29 I 9 F Cl At Br 17 35 53 85 Lr 71 Lu 103 (210) 19.00 79.90 (260) 35.453 126.91 174.97 8 S O Te 16 34 52 84 Se Po 70 Yb No 102 (209) 16.00 32.06 78.96 (259) 127.60 173.04 7 P N Bi 15 33 51 83 As Sb 69 Md Tm 101 74.92 (258) 14.007 30.974 121.75 208.98 168.93 6 C Si 8.71 14 32 50 82 Sn Pb Er Ge 68 Fm 100 28.09 72.59 207.2 (257) 12.011 11 167.26 11 §Not yet named 5 B Tl In Al 4.82 13 31 49 81 67 99 Es Ga Ho (252) 26.98 69.72 10.8 11 204.38 164.93 2 § 30 48 80 Zn Cd Hg Cf 66 98 Dy 11 (277) 65.39 (251) 112.41 200.59 162.50 § 29 47 79 Ag Au Cu 65 97 Tb Bk 111 (272) 63.55 (247) 107.87 196.97 158.93 0 § Ni Pt 28 46 78 Pd 64 96 Gd 11 Cm (269) (247) 58.69 106.42 195.08 157.25 Ir 27 45 77 Mt Co Rh 63 95 Eu 109 Am ble of the Elements (266) 58.93 192.2 (243) 102.91 151.97 Ta 26 44 76 Fe Hs Ru Os 62 94 Pu 108 Sm (265) 55.85 101.1 190.2 (244) 150.4 Tc 25 43 75 61 93 Bh Re Np Mn Pm 107 (98) Periodic (262) (145) 54.938 186.21 237.05 U W Cr 24 42 74 Sg 60 92 Mo Nd 106 (263) 52.00 93.94 183.85 144.24 238.03 V Pr Ta 23 41 73 59 91 Pa Nb Db 105 (262) 50.94 92.91 180.95 140.91 231.04 Ti Zr Hf Rf 22 40 72 58 90 Th Ce 104 (261) 47.90 91.22 178.49 140.12 232.04 Y 21 39 57 89 La Sc Ac * † 44.96 88.91 138.91 227.03 4 Sr 12 20 38 56 88 Be Ba Ca Ra Mg 9.012 24.30 40.08 87.62 137.33 226.02 †Actinide Series: 1 3 K H Li Fr 11 19 37 55 87 *Lanthanide Series: Cs Na Rb (223) 6.941 22.99 39.10 85.47 1.0079 132.91 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 == millimetersmillimeters 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 numbernumber == 6.0226.022 ×× 10102323 molmol−−11 ElectronElectron charge,charge, e e == −−11.602.602 ××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 == −−log[Hlog[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 BUFFERS AND TITRATIONS As Easy As It Ever Gets What I Absolutely Have to Know to Survive the AP Exam The following might indicate the question deals with buffers and/or titrations: buffer solution, common ion, conjugate base/acid, equivalence point, ½ equivalence point, end-point, weak acid/base “reacted with a strong..”, titrant, buret, analyte, etc… Buffers: All about those “common ions” IS: A buffer is a solution of a weak acid or base and its salt [which is its conjugate]. DOES: A buffer resists a change in pH. HOW IT WORKS: Since a buffer consists of both an acid or base and its conjugate, which differ by an H+, both a weak acid and a weak base are present in all buffer solutions. In order to understand buffers and buffer problems we must be proficient with the following… • MUST know the conjugate A/B concept to be successful at buffer problems. This means understanding that HA has A− as it’s conjugate and that NaA is not only a salt, but that any soluble salt releases the common ion, A− • pH = −log [H+] • Ka × Kb = Kw • Any titration involving a weak A/B, from the first drop to the last, before equivalence, is a buffer problem. If HA is titrated with NaOH, as soon as the first drop splashes into the container, I’ve added A− ions to the solution and the presence of both HA and A− constitutes a buffer solution. Buffers: Calculating pH Solving Buffer Problems with the one and only, handy dandy, all I’m ever going to need equation… + [Acid] ⎡⎤HO = K ⎣⎦3 a [Base] Where: Ka – is the acid dissociation constant [Acid] – is the concentration of the substance behaving as the weak acid in the buffered solution § Either the weak acid or the salt of the weak base [Base] – is the concentration of the substance behaving as the weak base in the buffered solution § Either the weak base or the salt of the weak acid To Solve: § MUST recognize who’s who in the buffered and then solve with the equation § If the buffer is a weak acid, HA, and its conjugate salt, A− then HA is the acid and A− is the base. § If the buffer is a weak base, B, and its conjugate salt, BH+, then B is the base and BH+ is the acid. § You still use the Ka; thus you must calculate it first using the Kb provided for the weak base § If concentrations are given plug and chug § If concentrations of separate solutions are given with volumes and the two are added together you must recalculate the “new concentrations” due to dilution of additive volumes. Copyright © 2014 National Math + Science Initiative, Dallas, Texas. All Rights Reserved. Visit us online at www.nms.org Page 1 Buffers and Titrations Buffers: Calculating pH Con’t Example: −4 Calculate the pH of a solution containing 0.75 M lactic acid, HC3H5O3 (Ka = 1.4×10 ), and 0.25 M sodium lactate, NaC3H5O3. § Lactic acid is the weak ACID and lactate ion is its conjugate BASE +[]Acid −−44[]0.75 ⎡⎤HO ==K (1.410)× = 4.210× ⎣⎦3 a [Base] [0.25] pH=− log(4.2 × 10−4 )= 3.38 Example: −5 A buffered solution contains 0.250 M NH3 (Kb = 1.80×10 ) and 0.400 M NH4Cl. Calculate the pH of this solution. + § NH3 is the weak BASE and NH4 is its conjugate ACID § When you are given a weak base and its salt you will often need to calculate Ka then plug in – don’t get in a hurry and mistakenly use the Kb! −14 KKab×=1.00× 10 1.00× 10−14 K ==5.56× 10−10 a 1.8× 10−5 +[]Acid −−10 []0.400 10 ⎡⎤HO ==K (5.56× 10 )= 8.89× 10 ⎣⎦3 a [Base] [0.250] pH=− log(8.89 × 10−10 )= 9.051 Example: Calculate the pH of a solution prepared by mixing 20.0 mL of 0.300 M acetic acid, HC2H3O2, with 20.0 mL of 0.350 M -5 NaC2H3O2. Ka for acetic acid is 1.80×10 § Acetic acid is the weak ACID and acetate ion is its conjugate BASE § Recalculate molarities OR better yet just calculate the number of moles (volumes are the same as they are all in the same solution so they cancel)….