Chapter 15,16,17 Study Guide PDF

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Chapter 15,16,17 Study Guide Buffers  Mixture of a weak acid and its conjugate base, both in substantial concentrations o “Substantial concentrations” – within a factor of ten  When a SA or SB is added to buffered solution, it is best to deal with the stoichiometry of the resulting reaction first (moli, molc, molf). Be careful of limiting reagent. Then consider the equilibrium calculations (ICE table from molf data).  Henderson-Hasselbach Equation o Used to solve for new pH of a buffer solution [ conj .base] o (or can just do a new ICE table) pH= p K a +log [ weak acid ]  Two ways to make a buffer 1. a) mix an excess of weak acid with strong base (i.e. H3PO4 and NaOH) b) mix an excess of weak base with strong acid 2. a) mix weak acid and salt containing conjugate base (i.e. CH3COOH and CH3COONa) b) mix weak base and salt containing conjugate acid  Always make sure to carefully examine what have left at end of ICE or molc table Buffer Capacity  Capacity – Indicated by the molarity of the buffer components o Ideally want [HA] = [A-] in high concentration o Do not want [HA]:[A-] to be too big or small bad buffer, reacts w/ H+ or OHBuffer Range  Range – 1 pH unit on either side of pKa [ conj .base]  if [HA] = [A-], pH = pKa pH= p K a +log [ weak acid ]  Ex) If pKa of HA = 5 then range is 4-6, translates into a [H3O+] change by factor 10 Titration  Uses burette (calibrated for volume) that contains solution of known concentration  Titration of __(in flask)______ with ___(from burette)_____  Inflection point of titration curve (mL NaOH added vs pH) is the equivalence point  Do molc chart first. Look at what have left. If excess SA or SB, do a direct pH calculation. If excess WA or WB, set up a hydrolysis ICE table and calculate that pH.  Titration Curve I: SA with SB or SB with SA o If titrate SA with SB, equivalence pt always pH = 7  Ex) mol NaOH = mol HCl o Initially, pH does not change very much because HCl and NaOH are reacting, creating products that do not affect pH (NaCl and water) o Know: VNaOH added from burette, M of NaOH, VHCl in flask o Measure: pH in flask

o Want: Molarity of HCl in flask  Titration Curve II: WA with SB o Initial pH higher than with a SB, because weak Titration of polyprotic WA 

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acid has higher pH than strong o Buffer region because have a WA and its conjugate base  Buffer ends when use up the acid o Equivalence point pH > 7  Because HA + OH- reacts to form A- as a major species  That A- reacts with water  OHTitration Curve III: WB with SA o Similar buffer region because WB and its conjugate acid  Buffer ends when use up the base o Equivalence point pH < 7  Because A- + H+ reacts to form HA  That HA reacts with water  H+ Titration Calculation o Always look what you have left over after molc table o If have [HA] and [A-] its in the buffer  use Henderson-Hasselbach o If just have [HA] or just have [A-]  react with water and make ICE table o If have weak acid HA left over, but have ANY excess strong acid H3O+  convert to molarity then do direct pH calculation (same if any excess SA, OH-) Titration Graph Interpretations – Comparing two titrations curves o Which liquid in flask has higher [ ] if titrated with same base or acid?  Look at Equivalence Point  which ever takes more titrant to reach equivalence point has the higher concentration initially o Which is stronger acid/base initially?  Look at initial pHs  tells which liquid in flask has what pH and thus which is stronger (can tell which has stronger/weaker conjugate too) Indicators o Choosing one: choose an indicator with a pKa close to the pH at the equivalence point of titration o Strong acids and bases have a bigger range for indicators. Weak acids and bases have a small range to choose an indicator

Acids and Bases Arrhenius Bronsted-Lowry Lewis

Acid H3O donor in aqueous soln H+ donor Electron pair acceptor +

Base OH donor in aqueous soln H+ acceptor Electron pair donor -

ex) metal ions ex) ligands Lewis Acid – electron pair acceptors  Atoms within a molecule that have an empty orbital  Atoms within a molecule that are electron deficient (AlCl3)  Can often contain pi bonds that get broken (CO2) Solubility Product Constant, Ksp  AxBy (s)  xA+(aq) + yB- (aq) Ksp = [A+]x[B-]y  Does not include AxBy because solids are not included in equilibrium constants  Ksp < 1 (small) if only slightly soluble or insoluble  Ksp > 1 (big) if fully soluble  For ionic compounds (salts) that are sparingly soluble or relatively soluble K sp 0 o ΔSuniverse = ΔSsystem + ΔSsurroundings

o ΔSsurroundings = -ΔHsys/T (opposite sign of ΔH because if heat is released, KE increases, so more microstates are available)  Calculating ΔSo o 3rd Law of Thermodynamics: entropy of a pure, perfect crystal is 0 J/k at 0 K  Anything that isn’t pure crystal is ___ J/K more entropy o ΔSrxn/sys = ΣSomolar products – ΣSomolar reactants o Somolar measured in J/molK so make sure account for stoich of eq when calc o Standard state symbol indicates [X-(aq)]= 1M, P=1atm, T=25 degrees Celsius Ways to Determine Spontaneity 1. ΔSuniv > O 2. ΔG  free energy a. ΔG = ΔH – TΔS b. ΔGf° c. ΔGf° = - RT ln K Gibbs Free Energy, ΔG  Spontaneity is related to ΔH and ΔS o ΔG = ΔH – TΔS (units in all kJ, kJ, K, kJ/K or J, J, K, J/K)  ΔH ΔS ΔG Spontaneity Δ + Spontaneous at all temperatures + + Spontaneous at high temperatures + + + Non-Spontaneous at low temperatures Spontaneous at low temperatures _ + Non-Spontaneous at high temperatures + + Non-spontaneous at all temperatures G < 0 (spontaneous) ΔG > 0 (non-spontaneous ΔG = 0 (equilibrium) Driving Force of a Reaction  If ΔH = + , then the system is entropy-driven to make –ΔG spontaneous reaction  If ΔS = - , then the system is enthalpy driven to make –ΔG spontaneous reaction The Effect of Temperature: ^ not true at all temperatures**  ΔH does not change with changes in temperature  TΔS increases when T increases  ΔG = 0  equilibrium when TΔS=ΔH  All reactions at temperatures higher than equilibrium point are spontaneous  All reactions at temps lower than equilibrium point are non-spontaneous  Temperature which there is a switch from spont to non-spont o T = ΔH / ΔS 

Standard Free Energy of Formation  ΔGf° = Σ ΔGf° products – Σ ΔGf° reactants  ΔGf° : free energy change for the formation of 1 mol of substance from its elements in their standard states  ΔGf° for elements = 0 kJ/mol  Is stoichiometric-dependent (need to multiply by number of moles from eq in calc) ΔG and Equilibrium  A  B K = [B]/[A] (or B  A) (equilibrium shifts all depend on Q and K) Situation QK Q=K

Shift in Eq. Shift right Shift left No shift

Q/K 1 =1

Ln (Q/K) + 0

ΔG - (spont) + (non-spont?) 0 (equil)

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o Shift is non-spont? No, only as written. Spont for reverse reaction ΔG° = - RT ln Keq (derived from ΔG =RT ln (Q/K) if Q is standard state of 1M solns) Can plug in above into a form of ΔG =RT ln (Q/K) to get ΔG = ΔG° + RT ln Q Graph of free energy vs reaction progression o Equil at minimum. Closer to Rct if ΔG° = +. Closer to Prdt if ΔG° = o ΔG is found by subtracting final ΔG from initial o K1 if ΔG° =  Graph is mirror image if plot of reverse reaction...