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Unit 2 Test Review on Chapters 15 and 16

by: Justin Sequerra

Unit 2 Test Review on Chapters 15 and 16 CHEM 1312

Marketplace > University of Texas at Dallas > Chemistry > CHEM 1312 > Unit 2 Test Review on Chapters 15 and 16
Justin Sequerra
GPA 3.929

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This study guide goes more in-depth on Chapter 16 (Acids, Bases, and Salts) because I haven't posted notes on the chapter yet since we learned it all the week before the test. Also pay attention in...
General Chemistry II
Dr. Sibert
Study Guide
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This 9 page Study Guide was uploaded by Justin Sequerra on Wednesday February 24, 2016. The Study Guide belongs to CHEM 1312 at University of Texas at Dallas taught by Dr. Sibert in Winter 2016. Since its upload, it has received 106 views. For similar materials see General Chemistry II in Chemistry at University of Texas at Dallas.


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Date Created: 02/24/16
UNIT 2 TEST: CHAPTERS 15 AND 16 GENERAL CHEMISTRY II This general outline is meant as a supplement to the General Chemistry II (1312) course  taught at the University of Texas at Dallas and should not be taken as a standalone study  guide for the overall curriculum. However, I do hope that this broad summary of the textbook  helps you all in becoming successful undergraduate students here at UTD. ­ Justin Sequerra, “Chemistry is the study of matter. But I prefer to see it as the study of  change.” – Walter White, Breaking Bad Need­to­know Equations  1 Kw = [H3O+][OH­] = 1.0x10^­14 *Only in a aqueous solution at a constant temperature of 25 degrees Celsius* pOH = ­log[OH­] pH = ­log[H3O+] [OH­] = 10^­pOH [H3O+] = 10^­pH pH + pOH = 14.00 (at 25°C) percent ionization = ([H3O+]eq/[HA]o) x 100 ([HA]o = original weak acid concentration; [H3O+]eq = equilibrium concentration of hydronium) Ka x Kb = Kw (for any conjugate acid­base pairs Kp=Kc(RT)^Δn  Kp=Equilibrium constant for pressure Kc=Equilibrium constant for concentrations R=Gas constant (.08206 L*atm/K*mol) T=Temperature (°K) Δn=moles of gaseous products – moles of gaseous reactants NOTE: When there is no change in the number of moles (Δn=0), Kp=Kc 2 ACIDS, BASES, AND SAL TS 16.1 BRONSTED ACIDS AND BASES  Bronsted Acid: Donates a PROTON  Bronsted Base: Accepts a PROTON  Bronsted Acid  Donates  Conjugate BASE  Bronsted Base  Accepts  Conjugate ACID  Bronsted Acid + Conjugate Base = Conjugate Pair = Bronsted Base + Conjugate Acid Examples: 16.2 MOLECULAR STRUCTURE AND ACID STRENGTH  Molecular structure affects how strong an acid may be  4 Factors that Affect Strength of an Acid: BOND STRENGTH, POLARITY, OXIDATION  NUMBER, and EXTENT OF ELECTRON DELOCALIZATION  Weak bonds are easier to break, therefore making the acid stronger since the acid will  more easily ionize.  Polar bonds will easily break up into their respective ions since there is more of a  difference in charge (positive and negative) = stronger acid  BOND STRENGTH DECREASES AS ONE GOES DOWN THE COLUMN (periodic table)  In polyatomic acids, the more electronegative (high oxidation number) the central atom  (metal) is, the stronger the acid is. This is because the central atom will attract more  electrons, making the O­­­H bond more polar.  The greater the oxidation number (or the MORE OXYGENS) increases the strength of a  particular acid, where the ability of the central atom (metal) to draw in electrons from the  oxygens (electron delocalization) increases as the amount of oxygens increase (makes  the O—H bond more polar). (HClO4>HClO3>HClO2>HClO)  Resonance stabilization of conjugate base favors the ionization process, which creates a stronger acid. (the more central atoms there are, the stronger the acid)  During test, look at Bond Strength first and the Polarity second 16.3 THE ACID-BASE PROPERTIES OF WATER  Water = amphoteric (can act as an acid or a base)  H2O(l)  OH­(aq) + H3O+(aq) = autoionization of water; K =  Kw = 1.0  ONLY at  25°C  Kw = K = equilibrium constant (but for water) = also termed ion­product constant  [H3O+] = [OH­] Solution is neutral  [H3O+] > [OH­] Solution is acidic  [H3O+] < [OH­] Solution is basic 16.4 THE PH AND POH SCALES  pOH scale is opposite to the pH scale in terms of labeling: 7.00 = Neutral; <7.00 = Basic;  >7.00 = Acidic 16.5 STRONG ACIDS AND BASES  Strong acids/strong bases completely ionize/disassociate in an aqueous solution  Meaning: Strong Acids and Bases do not undergo equilibrium because they go straight  to completion Figures: Memorize These!!!!! (Recognize that there is only one arrow in the figure’s  reactions instead of two since there is no equilibrium)  Note: Only the first ionization of H2SO4 is complete  Since these reactions go to completion, the pH calculation is fairly easy (can be done in 2 4 ways) Given the concentration of the strong base (which is equal to [OH­]): 1. Use of [H3O+][OH­] = 1.0x10^­14 and then, pH = ­log[H3O+] equations 2. pOH = ­log[OH­] and then, pH + pOH = 14  I prefer the last one since I believe it is a lot faster than the first one  Be careful!!! The strong base used in the question may be a Group 2A metal hydroxide,  so you must multiply its concentration by 2 to get the [OH­]; Example: Ba(OH)2 where  Ba(OH)2 disassociates into Ba2­ and 2OH­. 16.6 WEAK ACIDS AND ACID IONIZATION CONSTANTS  Weak acids (as well as bases) do not completely ionize, but rather they only ionize  partially  [H3O+] = [H+]; Be comfortable with this idea  Ka (Acid Ionization Constant) indicates to what extent the weak acid actually ionizes  Meaning: Unlike strong acids, weak acids develop into an equilibrium state since they do not ionize completely  Large Ka value: strong acid (more ionized products)  Small Ka value: weak acid (less ionized products)  We can then use ICE tables to find pH (from product [H3O+]), equilibrium concentrations, and Ka  Since these are weak acids, DO NOT USE THE QUADRATIC EQUATION, but instead use  the Successive Approximation Method taught in class since it is faster!  Can determine Ka from initial acid concentration and pH at equilibrium since pH can be used to find [H3O], which is the same concentration as [A­]  Percent Ionization illustrates that an acid is relatively strong if it ionizes more (it’s degree of ionization)  Less the original concentration of acid, the more percent ionization (since the reaction will favor the right since there are more gas moles there). 16.7 WEAK BASES AND BASE IONIZATION CONSTANTS 5  Just like weak acids, weak bases partially disassociate and therefore can achieve  equilibrium  Same notes as acids, but be careful with calculations and know which concentration (H+  or OH­) you are supposed to calculate  May switch between Ka and Kb with equation Ka x Kb = 1x10^­14, or pH + pOh = 14 16.8 CONJUGATE ACID-BASE PAIRS  Weak Bronsted Base  Strong Conjugate Acid =(same thing as) Weak Bronsted Acid  Weak Bronsted Acid  Strong Conjugate Base =(same thing as) Weak Bronsted Base  Strong Bronsted Base  Weak Conjugate Acid = Does not react at all with water  Strong Bronsted Acid  Weak Conjugate Base = Does not react at all with water  Since strong Bronsted acids and bases don’t have an equilibrium, it makes sense why  their conjugate bases do not react with water 16.9 DIPROTIC AND POLYPROTIC ACIDS  Diprotic/polyprotic acids have more protons (H+s) to donate, so they go through  successive (stepwise) ionizations, where each ionization has its own Ka (Polyprotic  acids: H2SO4, H2CO3,H2S, etc.)  The first ionization always has the highest Ka, and it gets smaller thereafter  Determining pH: Only need to look at the first ionization of the acid  Determining Equilibrium Concentrations: Must consider successive ionizations 16.10 ACID-BASE PROPERTIES OF SALT SOLUTIONS  Salt Hydrolysis: The reaction of a salt’s dissociated ions with water to produce either  hydroxide or hydronium ions.  When given a salt problem, dissociate it and look at each ion it produces  From there, react both separate ions to water and find the conjugate acid for each ion  (the base plus the hydrogen atom)  The conjugate acids often tells us if the solution is either basic, acidic, or neutral  If the conjugate acid of an anion of the salt is a weak acid, then we can assume that the  solution is basic, since it produces OH­ as well as the weak acid A­ (aq) + H2O(l)  HA(aq) + OH(aq)  If the conjugate acid of the cation of the salt is a weak base, then the solution will be  acidic since it creates hydronium ions.  And if the cation of the dissolved salt is a highly charged/small metal ion (Al3+, Fe3+,  6 etc.) then it will produce an acidic solution (polarize the O—H bond more)  Those that produce a neutral salt solution are the group 1A and 2A metals (b/c they are large, low charge ions that do not polarize the O—H bond to the extent of the transition metals) and the conjugate base of a strong acid (such as Cl­) b/c they always ionize completely in the presence of water.  Salts that have both cations and anions that can hydrolyze with water (react with water)  can be determined basic, acidic, or neutral by comparing the Kb and the Ka of the  reaction ­ Kb > Ka, solution is basic ­ Kb < Ka, solution is acidic ­ Kb = Ka, solution is neutral 16.11 ACID-BASE PROPERTIES OF OXIDES AND HYDROXIDES  Metal Oxides (Na2O, CaO, etc) are mostly basic  Nonmetal Oxides (CO2, SO3, etc.) are mostly acidic  Metal Hydroxides = Basic or amphoteric 16.12 LEWIS ACIDS AND BASES  Lewis definitions are more general than the Bronsted definitions  Lewis Acid: Can accept a pair of electrons  Lewis Base: Can donate a pair of electrons  Lewis acid usually don’t have a lot of electrons and don’t usually have a Hydrogen atom  Lewis bases have lone pairs of electrons that they usually want to share (an anion  mostly)  Every Lewis Acid­Base reactions result in a coordinate covalent bond  Lewis Acid is almost = as Bronsted Acid in reactions; Bronsted Acid is NOT Lewis acid  in any formation/decomposition reactions (A + B  AB or vice versa)  Same thing with Lewis acids and Bronsted bases CHAPTER 15 REVIEW  Equilibrium is achieved when Q=K, when forward rate = reverse rate, and when the  concentrations of the species in the reaction are CONSTANT   K (equilibrium constant) is dependent on temperature  K > 10^2 Products are favored (more product); K < 10^­2 Reactants are favored (more  reactant)  Q (reaction quotient) always changes throughout a reaction, while K stays constant  Solids and liquids DO NOT appear in the equilibrium expression (b/c concentrations do  not generally change) Figure: How to manipulate equilibrium constant  Kp and Kc are only equal when the number of moles on the product side = number of moles on reactant side  For ICE tables, make sure you find the reaction quotient (Q) before you do the quadratic  equation so that you know whether or not the reactants/products are either  consumed/produced. (Whether x is negative on reactant or product side in the “change”  row).  If you can use perfect square or  successive approximation instead of the quadratic equation, DO IT (be careful though)  Check your answers (if you have time) by plugging in equilibrium concentrations into the equilibrium expression to see if the number that comes out is the K given     BALANCE YOUR EQUATIONS!!!!!!!!! Previous Study Outline Practice Question Answers 1. Kc = 7.1; Kc = [BrCl]^2                         [Cl2][Br2]  2. 8.1 x 10^­6 3. A) 2.3 x 10^24 B) 6.6 x 10^­13 8 C) 2.8 x 10^­15 4. H2 = 0.37 atm; I2 = 0.37 atm; HI = 2.75 atm 5. A) Neither B) Right C) Left D) Neither


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