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UI CHEM 1120 - Exam 2 Study Guide
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Chem 1120 1st EditionExam # 2 Study Guide Lectures: 12-25Lecture 12 (February 16)Describe Arrhenius Acids and Bases versus Bronsted-Lowry Acids and BasesArrhenius Acids = a substance that directly yields H+ ions when dissolved in water Arrhenius Bases = a substance that directly yields OH- ions when dissolved Acid = proton donor, any species which donates a H+Base = proton acceptor, any species that accepts a H+Identify the Bronsted acids and bases in the following equations:List the most common strong acids and strong bases (have these memo-rized)Most common strong acids = HCl, HBr, HI, HClO3, HClO4, HNO3, H2SO4 Most common strong bases = LiOH, NaOH, KOH, (group 1a metals with hydroxide), Ca(OH)2, Ba(OH)2, Sr(OH)2, (heavy group 2a metals with hydroxide)Remember the important equations when converting pH to pOH, etcKw = [H+][OH-] = 1.0 X 10^-14 @ 25opH = -log10[H+] pOH = -log10[OH-] pH + pOH = 14Lecture 13 (February 18)Know the differences between strong acids and strong basesStrong acids are by definition strong electrolytes and exist totally ionized in aqueous solutions; For monoprotic strong acids, [H3O+] = acidStrong bases are soluble hydroxides; alkali metals and heavier alkali earth metal hydroxides (Ca2+, Sr2+, and Ba2+); These also dissociate completely in aqueous solutionsFor all strong acids, Ka is greater than one (Ka measures acid strength, the weaker the Ka the weaker the acid); same for bases with KbCalculate the concentrations of all the species in 0.2 M H2SO4. (Ka=1.3 X 10^-2 and Kb= 6.3 X 10^-8)Lecture 14 (February 23)What are the 2 major types of weak bases? Know how to go through with Kb equilibrium calculations (same as acids just with Kb values, more conver-sions) and about the relationship between Ka and Kb/how to do the conver-sions2 major categories of weak bases:1) Neutral molecules with lone pairs of electrons, ex: NH3, CH3NH22) Conjugate bases of weak acids, ex: HCO3, F-Example: Calculate the pH of 0.10 M CH3NH2 (Kb=4.4 X 10^-4)**pKa = -log(Ka) pKb = -log(Kb) pKa + pKb = 14.00Example: The pKa of HF is 3.17. What is the value of Kb for the F- ion?Solution: 14 - 3.17 = pKb = 10.83 Kb = 10 ^ -10.83 Kb = 1.5 X 10^-11Understand Lewis Base chemistry, is the acid or base the electron acceptor?Lewis acid = electron pair acceptor, Lewis base = electron pair donor, anything that could be a Brondted Lowry base is a Lewis baseExample: Identify the Lewis acid and Lewis base for each reaction.Lecture 15 (February 25)What is the common ion effect and how would solve a common ion problem?Common Ion Effect: the extent of ionization of a weak electrolyte is de-creased by adding to the solution a strong electrolyte that has an ion com-mon with the weak electrolyte OR a shift in the equilibrium caused by the ad-dition of an ion already present in solutionWhat is the pH of a solution 0.1M in HCN and 0.2M NaCN?What is a buffer solution and what equation often proves useful in those types of problems?Buffers = solutions that resist pH change when small amounts of acid or base are addedWhat is a buffer comprised of? ——> weak acid and its conjugate base OR weak base and its conjugate acid**Henderson-Hasselbalch Equation:When 5 mL of HCl is added to a buffer containing 0.1 M NaA and 0.1 M HA (pKa = 7) what is the resulting pH? How do you know what kind of indicator you need for a buffer titration?To select the acid/base pair for a buffer of a specific pH, 1) choose an acid whose pKa is equal to the desired pH and 2) solve the Henderson-Hassel-balch equation for the [base]/[acid] ratio. Choose an indicator with a pH near the equivalence point.Lecture 16 (February 27)Review buffer capacity and titrations, what is the equivalence point, the end-point, a titration in general?Buffer Capacity = the amount of acid or base a buffer can neutralize before the pH changes significantly; More concentrated buffers have a higher buffer capacity; Buffer capacity is highest when the concentrations of weak acid and conjugate base are equaltitration = the accurate measurement of the volume of solution required to completely react with a sampleEquivalence point = the point at which stoichiometrically equivalent quanti-ties of reactants are brought togetherEndpoint = the point at which indicator changes color (very near to equiva-lence point)What are the four regions/types of problems to go along with titrating a strong acid into a strong base?1) before titrant is added (single point) 2) region up to the equivalence point2) region up to the equivalence point3) at the equivalence point (single point) 4) region after the equivalence pointRegion 1: Treat as strong acid problemRegion 2: Treat as limiting reagent problemRegion 3: Treat as complete neutralizationRegion 4: Treat as limiting reagent problemLecture 17 (March 2)Divide up the curve of weak acid-strong base (strong base into the weak acid) and know how to do calculations for each of the four partsRegion 1: before titrant, weak acid problemPoint 2: before equivalence point, buffer problem@ 5 mL 0.1 M NaOh is added0.1M X 25 mL = 2.5 mmol HAc0.1 M X 5 mL = 0.5 mmol NaOHHAc (aq) + OH- (aq) ——> H2O (l) + Ac- (aq)before 2.5mmol 0.5mmol 0mmolchange -0.5 -0.5 +0.5after 2mmol 0mmol 0.5mmolUse Henderson-Hasselbalch Equation:[base]/[acid] = (0.5mmol/30mL) / (2mmol/30mL) = 0.5/2 = 0.250pH = 4.745 + log(0.25) = 4.143Region 3: equivalence point, neutralization then weak base problem0.1M X 25 mL = 2.5 mmol NaOH0.1 X 25 mL = 2.5 mmol HAc(make ICE table) end up with 0 mmol HAc, 0 mmol OH-, 2.5 mmol Ac-[Ac-] = (2.5mmol) / (25mL+25mL) = 0.05 MAc- + water ——> HAc + OH-(make another ICE table) end up with 0.05-X Ac-, X HAc, X OH-plug into the Kb equation, Kb = Kw/Ka you end up getting x = [OH-] = 5.27 X 10^-6 which means the pOH is -log(5.27 X 10^-6) = 5.278pH = 14 - pOH = 8.721Region 3: equivalence point, neutralization then weak base problem0.1M X 25 mL = 2.5 mmol NaOH0.1 X 25 mL = 2.5 mmol HAc(make ICE table) end up with 0 mmol HAc, 0 mmol OH-, 2.5 mmol Ac-[Ac-] = (2.5mmol) / (25mL+25mL) = 0.05 MAc- + water ——> HAc + OH-(make another ICE table) end up with 0.05-X Ac-, X HAc, X OH-plug into the Kb equation, Kb = Kw/Ka you end up getting x = [OH-] = 5.27 X 10^-6 which means the pOH is -log(5.27 X 10^-6) = 5.278pH = 14 - pOH = 8.721Region 4: after equivalence point, limiting reagent/strong base


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UI CHEM 1120 - Exam 2 Study Guide

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