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Chemistry II

by: Hulda Donnelly Sr.

Chemistry II LB 172

Hulda Donnelly Sr.
GPA 3.78

Samantha Cass

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Samantha Cass
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This 79 page Class Notes was uploaded by Hulda Donnelly Sr. on Saturday September 19, 2015. The Class Notes belongs to LB 172 at Michigan State University taught by Samantha Cass in Fall. Since its upload, it has received 17 views. For similar materials see /class/207239/lb-172-michigan-state-university in OTHER at Michigan State University.


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Date Created: 09/19/15
Acid Base Introduction beginning review Bronsted AcidBase chemistry involves the exchange of what Follow up question Does the acid donate or receive the prawn Follow up questions What part of the molecule causes it t0 be the acid or base What is the name of this molecule Follow up questions What part of the malecule causes it to be the acid or base What is the name of this molecule Follow up questions What part of the malecule causes it to be the acid or base What is the name of this molecule Follow up questions What part of the malecule causes it to be the acid or base What is the name of this molecule HCl H2804 HBr HNO3 HI HClO4 Follow up questions Is acetic acid strong or weak Take notes H3O is an acid in the reverse reaction CH3COO is a base in the reverse reaction CH3COO is called the conjugate base of CH3COOH CHaCOOH is called the conjugate acid of CH3COO Take notes H20 H30 are a conjugate pair or conjugate acidbase pair The acid and base in a conjugate acidbase pair differ by only H TWQ ccmjugate pairs will exist in any Bronstead acid base reaction Soluble MOH salts alkali Ba H Rquot R0 N1112 H H20 gt H2 OH 01 R39 H20 gt RH OH39 React to completion More explanation to come on why we chose these strong bases later NH3 H20 39NH4 0H Notes Weak bases establish an equilibrium with water Weak basas organic amina compounds conjugate of carboxylic acids H20 H20 H30 OH Notes Watar is amphjprotic can act as both an acid and base Kw H300H1 True for all aqueous KW 2 1 X 1039 selutions pH and pQH notes If the H304 then the solution is said to be If the H304 then the solution is said to be PH 90 If the H304 gt then the solution is said to be If the H304 gt then the solution is said to be If the H304 lt1 then the solution is said to be If the H304 lt1 then the solution is said to be Chemists use a value called pH because note pH lo gH3C and pOH 39 lQOH A Scientific Notation sucks B Chemists are laZy C That damn Arrhenius guy note pH logH3O and pOH 10gOH 1430f 2 IICli because it all iam zes O18 note pH log30 and pOII lagOH OI11f 2BaOI I2 because it all ionizes Haw 2quot Kw OH L 1378 V 9 v 14 I I g I u ng39l I quotL IJHquot fro 10Lto20L tis the pH of the diluted soluticn you can do this without your calcdatarsthink about it V 9 v 14 I I g I u ng39l I quotL IJHquot fro 10Lto20L tis the pH of the diluted soluticn you can do this without your calcdatarsthink about it I solutmn I solutmn M mamas Which of the following reactions is the base ionization constant Kb for ammonia NH3aq H200 NH4aq H200 NH4aq OH aq NH3aq OHaq NH3aq H3Oaq NH4aq OH aq 7 NH3a D H30aq NH3aq H200 NH239aq H200 NHHW H200 Which of the following reactions is the base ionization constant Kb for ammonia 39 B NH4aq H200 NH3aq H3Oaq C NH4aq OHaq NH3aq H200 I NH3aq OH aq NH2 aq H200 E NH33C1 H303ltD NHIWJ H200 Which of the following is the strongest acid A acetic acid Ka 18 x 10395 B benzoic acid Ka 63 x 10395 C dihydrogen phosphate Ka 62 x 10398 D formic acid Ka 18 x 10394 E hydrocyanic acid Ka 40 x 103910 Which of the following is the strongest acid A acetic acid Ka 18 x 10395 B benzoic acid Ka 63 x 10395 C dihydrogen phosphate K 62 x 10398 a E hydrocyanic acid Ka 40 x 103910 Which of the following weak acids has the strongest conjugate base A acetic acid Ka 18 x 10395 B benzoic acid Ka 63 x 10395 C dihydrogen phosphate Ka 62 x 10398 D formic acid Ka 18 x 10394 E hydrocyanic acid Ka 40 x 103910 Which of the following weak acids has the strongest conjugate base A acetic acid Ka 18 x 10395 B benzoic acid Ka 63 x 10395 C dihydrogen phosphate Ka 62 x 10398 D formic acid Ka 18 x 10394 39Write out equilibrium for acid than equilibrium expressmn H pKa 42 Ka benzoic acid 63 x 10395 Other possible questions What is HBO F OH PhCOO pOH NotepH of previous saluting 0012 M was 3 06 Kb ammonia 18 x 10 5 W te equilibrium nd OH L then get to pH pH 110 Til i i0 te Mt3 w ETH39AHJL 1H4 159K iodate 150K lithium iodate names from the strong base lithium hydroxide and the weak acid iodic acid 915 m 7 arc sin I L potassium oxalate potassium oxalate Potassium oxalate comes from the strong base patassium hydroxide and weak acid exalic acid perchlorate 39rquot quot3l quot39tsEIRW 1L h quotr JIM anunonhnn pen onue This salt comes from the weak base ammonia and strong acid padded acid big at 1 TU j Methylammonium phosphate 15 CHsNH33PO4 39 Methylammo nium phosphate is Salt is from a weak acid and a weak base Depends on strengths 915 m 7 sin I L sodi m carbonate 159K quot 1quotquot MW L 11 wi sodium carbonate Salt from strong base and weak acid bicarbonate limitations 21 Equot 39 n 7 l n 3 y 17 l 91 y I V V r a v f r 7 p l quot 39339 i V 7 nquot 7 quot 7 398 1 7h PH 0f the Solution Orwhat is the pH of a 15M solution of NH4NO3P pH 46 Kbammonia18x 105 What is the H30 in a 15 M solution of H2C03 K31 HZCO3 18 x 105 H30 000079 Exam 2 Review Guide Molecular Orbitals and Metals Chapter 11 Band them a model for bonding in atomic solids that comes from the molecular orbital theory in which atomic orbitals combine and become delocalized over the entire crystal Liz Li3 Li4 Antibonding orbitalsno electrons Bonding orbitals lled with electrons J J ii 00000000000 00 COO 0000 Because there are so many molecular orbitals the energy spacings between them are in nitely small discrete energy levels disappear and MO can be treated as a continuous band of energy Molecular orbitals will span the entire molecule or piece of metal note the thickened lines indicate the empty orbitals conduction band Metals Conduct electricity well No band gap Electrons can enter a metal on one side and then move through the solid in one single MO As metals cool down they become better conductors This is because as the nuclei slow down the electrons will be more successful at going through MOs since there is less resistance when the electrons are knocked off course EX Insulators Very large band gap Do not conduct electricity EX 4 band from antibonding orbitals aka conducting band 4 band gap band from bonding orbitals aka valence band Semi conductors Small band gap band gap can be overcome by applying a voltage An increase in temperature will increase conductivity because the thermal excitation of energy will cause electrons to jump over the band gap into the conduction band D I semiconductors When voltage is applied to LEDs electrons jump over the band gap to the conducting band When these electrons drop back down to the valence band energy is released in the form of light The color of light that the LED emits varies with wavelength which varies with energy which depends on the band gap Pros of LEDs I tend to emit only one color 2 energy ef cient 3 don t heat up like uorescent bulbs Cons of LEDs when used in streetlights they don t work as well because they don t generate enough heat to melt the snowice LEDs Transition Metals Can conduct electricity if energy is added to promote electrons from the valence to the conducting band Small band gap Electrons can absorb nearly any energy of light for conductivity In transition metals the d orbital has to be considered valence band is lled approximately 30 because there are 3 out of 10 electrons D Network Solids conducting band HHH lt bmdgav Valence atomic orbitals I K valence band Takes a lot of energy to excite an electron larger band gap Q Why aren t molecular compounds likely to be conductors A Electrons would have to jump large spaces from molecule to molecule since they re only held together by IMFs Solutions Chapter 12 Q What makes up a solution A l solute the minority component of a solution dissolved 2 solvent the majority component of a solution dissolver Solubiligg how much solute will dissolve Q How will a decrease in the pressure of a dissolved gas affect the solubility of that gas A the solubility will decrease ie pop can example if the temperature decreases and the pressure therefore decreases there are less molecules to hit the surface of the pop and dissolve in the liquid H em 3 Law as the pressure of a gas increases solubility increases pressure does not significantly affect the solubility of solids or liquids S g kHPg Pop can example if the pressure is increased there is more of a chance that C02 g will dissolve in the solvent v At equilibrium the rates of the 2 following reactions 0 0 will be the same 00 o o 4 Solvent C02 g gt C02 aq O O 0 C02 aq gt C02 g 0 O O 939 Dissolved gas molecules The Bends When a diver goes down the pressure increases so solubility also increases When a diver comes back up the pressure goes down so the solubility of the gas decreases gas comes out of solution when pressure goes down It is important that divers don t come up to the surface too quickly because if too much nitrogen is released into the bloodstream at once the capillaries will burst Q Why does deep diving require a He and 02 mix A The helium gas can help reduce the possibility of oxygen toxicity and reduce the narcotic effects Nitrogen Q When water boils what is in the bubbles A water vapor Q How do bubbles form A Nucleation sites are needed places where bubbles can start forming Bubbles usually form in the same spot around the edgesexample champagne Saltsugar added to pop acts as nucleation sites because they are irregularly shaped Other examples include putting mentos in diet coke and slamming down a can of beer Temperature and solubility If there is an increase in temperature there is a decrease in solubility and vice versa Examples 1 Boiling water causes dissolved gas particles to come out of the solution and therefore there is a decrease in solubility 2 A pop can on a hot summer day will expand and explode if it gets hot enough because the dissolved gases are coming out of the liquid A T T S gl because gas molecules have more energy and are more easily able to escape the solvent T T S lsT because as temperature increases the differences in energy between IMFs is less important The solubility of one liquid in another requires similar IMFs Q If we have a salt and it dissolves in H20 will heat be released A It depends on the relative IMFs in solution and within the solidliquid Q Which IMFs are greater A If AH gt 0 IMFs in solution lt IMFs in solidliquid Because energy is lower so the IMFs are stronger If AH lt 0 IMFs in solution gt IMFs in solid liquid Colligatz39ve Progertz39es properties of solution that depend only on the concentration of solute not chemical nature 1 vapor pressure boiling point boiling pt elevation freezing point depression osmotic pressure bUJN VVV l Vapor pressure pressure of a gas over a liquid at a given temperature gas particles will eventually reach an equilibrium between the air and liquid if solute is added to the solvent the particles will be located at the surface therefore decreasing the rate at which the gas particles can leave to achieve equilibrium P solvent Xsolvent X Posolvent Actual Pressure is related to the of solvent molecules at the surface Xsolvem mole fraction 139 dissociation constant Van Hoff factor Posolvent normal vapor pressure 139 of particles formed when the solute dissolves of pure solvent nsolvent X solvent nsolvent nsolute X i Q What is the dissociation constant ivalue for NaCl A 2 NaCls gt Naaq Claq Q What is the dissociation constant for CaCl A 3 CaClzs gt Ca2aq 2Claq Application of Vapor Pressure Example What is the vapor pressure of a solution at 68 degrees Celsius containing 54g of CaClz in 3000g of water Water s VP at 68 degrees Celsius is 2142 mm Hg FW CaClz 1109ng1 P solvent Xsolvent X P solvent 54g CaClz X 1 mol CaClz 004869 mol CaClz 1109g CaClz 3000gH20 X 1mol H20 16652 mol H20 18016g H20 16652 mol H20 X 2142 mm Hg 197mmHg ortorr 16652 mol H20 04869molCaClz 3 2 Boiling Point Elevation the effect of a solute that causes a solution to have a higher boiling point than the pure solvent What impacts ATBP 1 concentration of solute 2 i dissociation constant Concentr tion is usually measured in molarity mol of solute L of solution T Vl concentration of solutionl Thus we cannot use molarity for these calculationsneed a concentration that is temperature independent molality m mols solute kg solvent Application of BP elevation TBP KBP X m X i EXample What is the boiling point of a solution made of 428gCaClz and 430gH20 Kb 05121 0C m 428gCaClz X 1 mol 03856 mol CaClz 03856mol CaClz043 kg H20 08967m g 430gH20 X 1 kg 043 kg H20 1000g 05212 0C m0897m3 138 0C TfiTl39 138 quotC gt Tfi 100 138 OC gt Tf 10138 0C 6 3Freezing Point Degression The effect of a solute that causes a solution to have a lower melting point than the pure solvent Q When salt NaCl is poured on ice the liquid melts and the solution gets cold Where is the energy coming from to melt the solid A Kinetic energy from the surroundingsthe thermal energy from the surrounding environment melts the snow decreasing the freezing point Q Why is brinesalt better for melting ice than NaCl A Because brine salt has a faster reaction time dissolves ice faster than NaCl because the reaction using NaCl would require breaking down salt crystals which would be very time consuming Example What is the molality of ethylene glycol necessary to raise the boiling point of water by 20 0C KBP 05121 0C m AT Kbp X m X 139 Solve for m 39m ethylene glycol z39 1ethylene glycol is an organic molecule held together by covalent bonds so it won t dissolve in Solution How many grams of ethylene glycol must be added to 3000g of water to make this molality MW ethylene glycol 62gmol m molkg 39m ethylene glycol X mol 03 kg HzO X mol 117 mol ethylene glycol X 62 g ethylene glyco 7254 g ethylene glycol 1 mol ethylene glycol EXample What is the freezing point of a solution made by combining 18 grams of MgNO32 in 2000mL of HzO FW MgNO32 148gmol Kfp 186 0C m 18g MgNO32 X 1 mol 0122 mol MggNO3 061M 148g 02 L H20 T Kfp X m Xi A Tfi 0 OC 186 0C m061M3 T1 34 0C 4 Osmotic Pressure the pressure required to stop the osmotic ow left gt right diffusion is slower because solute particle blocks the molecules right b left diffusion is faster because there is no solute blocking the water path semipermeable membrane 7 If you apply pressure to the solution you can use the process of reverse osmosis ie reverse osmosis puri cation is used in water fountains H MRTi gt dissociation constant Osmotic pressure 39 Molarity gas temperature constant Applying osmotic pressure to the medicalfield An IV liquid needs to be a solution usually saline to effectively work If the IV is just pure water then the Molarity of the patient s blood will go down because the amount of solvent water is increased As Molarity increases so does osmotic pressure Water from the IV will move into blood cells to establish equal osmotic pressures between bodily uids and blood cells Too much water in the cell can cause it to lyse explode If an IV is too concentrated with solute then the Molarity of the patient s blood increases Water in the cells will move out into the bloodstream to create equilibrium Iftoo much water moves out of the cell then the cell may undergo crenation shrivel up and die Applying osmotic pressure to college life If you drink you increase the ethanol levels in your blood therefore changing the osmotic pressure The cells in your body will release water into your body to balance out the amount of alcohol This leads to hangovers to prevent this plenty of water should be drank so as to create equilibrium Example What is the osmotic pressure of an aqueous solution at 27 degrees Celcius if 100 L of solution contains 2423 mg of NaCl in torr A 2923mgNaCl X lg X lmolNaCl 50E4molNaCl 1000mg 5845gNaCl 50E4molNaCl X 008206atmLmolK X 30015K X 2 00246 atm X 760 torr 187 torr L 1 atm Chemical Kinetics chapter 13 Going inside of the a1row of chemical reactions Q What impacts the rate of a reaction A l the amount of reactants and products 2 time 2N205 gt 4NOz 02 Rate llgA Oz 1141A N02 gl2QA N205 At At At rate decreases as time progresses less reactant available smaller statistical chance of reactant molecules colliding Collision Theory of Reactivity l reactant molecules must collide 2 reactant molecules must collide with suf cient force dependent on speci c reaction 3 reactant molecules must collide in correct orientation need the correct overlap between molecular orbitals on colliding reactant molecules Factors that a ect rate 1 concentration as decreases rate generally decreases so there are fewer collisions in the rst place 2 temperature T as T less frequent collisions as T less force rate decreases 3 catalyst another chemical added to a reaction that speeds up reaction without itself being destroyed or is regenerated if destroyed enzyme biochemical catalyst 7 speeds up reaction without changing T or E catalysts prealign the reactant molecule to permit their orbitals to overlap correctly for a reaction Taking a closer look at the effect of concentration on rate aA bB cC overall reaction rate is proportional to A Ax rate is proportional to B Ax orders exponents must be experimentally determined because or a they can t be determined by stoichiometric coef cients They are y or b usually small integers or zero Fractional or negative orders can happen but they are rare orders reveal information about the mechanism of the reaction they tell you the number of each type of molecule in the most critical collision in mechanism rate determining step Combine proportions rate is proportional to A AxBAy Rate kAAxBAy rate law states the dependence on concentration for the rate of a reaction Orders are experimentally determined in 2 ways 1 the isolation method 2 the graphical method the isolation method vary and only one reactant concentration keep all other reactant concentrations constant examine the effect of variance on the rate of reaction Example Using the following chemical equation and table nd the rate equation and then the rate of k 2N0g 2H2g gt N2g 2H20g First step set up a rate ratio RR and a concentration ratio CR setting them equal to each other to nd X or y Rate2k gt1020X H2 20y Rate3k1 NO30X H23Ay Rate 1 k NO1X Hz10y Rate 2 k NO2X Hz20y 50E5 0010 AX 00 A 10E4 00 0 AK 0004 A 13E5 0005 0 20y 50E5 0 10 00020y 42AXX2 220yyl RatekNO02Hz Second step Plug in one of the rates from the table and the corresponding Molarity of NO and Hz to solve for the rate of k Using the information from Trial 1 13E5k0005020002 k260 1s 2 The graphical method Using reaction orders and their rate laws 1 measure A as the reaction proceeds 2 measure the elapsed time and 3 make 3 plots lnA vs I l A vs I A vs tthe straightest plot line will give you your order Reaction orders Zero order Rate kA00 k Integrated zero order law At kt A0 half life t v2 A02k First order Rate kA01 Integrated rst order rateyIJAy TkkilnA0 4 initial concentration Concentration at eloquen time rate time elapsed constant half life tiz 0693k Example If N205 0500M what is N205 after 48 minutes have elapsed Step by Step Reaction Mechanisms Elementary Steps and Reaction Rates When we normally see a reaction we see the reaction at face value AB gt CD What we don t always consider is that the reaction can be broken into multiple elementary steps that will achieve the overall desired result This can be proven by summing the elementary reactions to equal the overall reaction A B C 2D 2D B C A 23 9 2C For each of these steps there is a reaction rate law de ned by Rate kReactant1xReactant2y As long as it is an elementary step the coef cients of each of the reactants can be used as the exponents in the reaction rate Step 2 Rate k A B f Otherwise coef cients can be predicted by experimental trial Each of these reactions has an enthalpy of reaction associated with it The enthalpy of a reaction AHW is equivalent to the change in energy of breaking the bonds of reactants and forming bonds of product This can be solved with bond energies or enthalpies of formation The elementary enthalpies can be added together to get an overall enthalpy However we don t know if the elementary steps suggested are what is actually occurring For example we can have two different mechanisms that will get us the same products Mechanism 1 2N0 NO OONO Fast OONO NO 02 Slow Mechanism 239 N02 amp 1202 NO Slow 1202 N02 amp 02 NO Fast But the pathway each takes is entirely different To determine which mechanism is correct we must look at the slowest step of the mechanism The slowest step of the reaction is going to require the most energy for the reaction to occur Since the overall reaction can not take place without all of the steps the entire reaction will be determined by the slowest step and so will the rate law Next we must compare the rate laws of each mechanism Mechanism 1 kOONO Mechanism 2 kN02 What we nd is that OONO is an intermediate a compound that is produced inthe previous reaction but is used as a reactant in the next and although rate laws can be determined by intermediates they are hard to measure and harder to change in a controlled fashion since they will normally be very low concentrations Because the 2nd step is the slowest in mechanism 1 we can assume that the 1st step is at equilibrium this will be fully explained in Graphing Mechanisms At equilibrium we can assume that the forward and reverse reaction rates will be equal Therefore kfwd vOzf mN000N0 k5 dN022 OONO kmNO Plug it into the original rate law Rate k k5 dN022 kmvNO In some cases solving for the intermediate will produce an equation with a product in the denominator as in mechanism 1 Because of this we can assume that it is not the rate law since this inhibition product will skew the slope on the graph So then mechanism 2 should be the correct mechanism Reaction Coordinates Mechanisms can be displayed on a graph as shown below Initial Initial Energy Energy En En er er gy gy Reactants Products Reactants Products On the first graph we see that the nal energy is lower than the initial energy suggesting an exothermic reaction We also see two hills where each hill corresponds to a reaction step The valley between the two hills represents an intermediate product that will be used in the next hill s reaction ie OONO as in the mechanisms before Intermediates will normally be the result of an endothermic reaction where the creation of it will increase the energy in the system In order for the reactions to take place the molecules must overcome the energy needed to break the bonds between the atoms this is depicted by the height of the hills This is known as activation energy E Arrhenius equation k AeEaRT which shows that a greater Ea will increase the k Note how the rst hill has a much larger height than the second as shown by the double a1rows As we discussed before the slow step of a mechanism determines the reaction law The first hill is the slow step because it takes more energy to complete the reaction therefore the mechanism is dependant on whether it can overcome the activation energy Since the slow step is the first step the reaction has produced enough kinetic energy to overcome all other reactions afterwards As you can see by the dashed line this is kinetic energy reached by the first reaction The second hill does not come close to this height which means that second reaction does not take nearly as much energy for the reaction to occur Since enough kinetic energy is behind it the second reaction will occur without trouble If the 2quotd step is a slow step as in the second graph then the first step will be in equilibrium This is because the activation energy for the fast step is not enough overcome the slow step s activation energy Until there is enough energy to overcome the second step the rst step will establish an equilibrium The intermediate is also at a higher energy level than the initial energy Most reactions occur because the products are lower in energy anal therefore more stable Equilibrium Equilibrium Constant Keg Things about equilibrium 1 No macroscopic changes are obvious 2 Rate of the forward and reverse reactions are equal A f B H C D k1 A B kzlcl D k1 C D k2 A B Keq k1 Products k2 Reactants REMEMBER Products are always over reactants When solving for Keq remember that pure liquids and solids do not matter This is because pure solids and liquids have a particular density for a given molecule at a specific state For gases and solutions the molarity does matter since different types of molecules can easily mix in with each other solute and solvent Since the concentration of gases and solutions can change by adding more moles or more volume their concentration at a particular point in the reaction is important This formula is also equal to the reaction quotient Q but the key difference between Keq and Q is that Keq only applies to reactions at equilibrium Therefore if a reaction is not at equilibrium we can use the Keq and Q difference to determine if the reaction is shifted favors products or reactants If the Q is very high it is likely the reaction is product favored high number high numerator high concentration of product If it is very low the opposite occurs If the Q is between 001 and 1000 it s hard to tell whether the reaction favors one side or the other Ifwe know that the reaction has a Keq and we know what the Q value is we can also determine which way the reaction will shift in order obtain equilibrium The Q is going to want to get closer to the Keq So if the Q is higher than the Keq the reaction will shift to the left and make more reactants since that will decrease the Q value Question 2 A solution contains 15 M of acetic acid and 2 M of CH3COO and H30 If the Keq is 18x10 5 is the reaction in equilibrium If not is it shifting to the right 01 the left CH3COOHaq H20l lt gt 143076151 CH3COO39aq Q H3OCH3COO39 CHgCOOH Q 0202 26x10392 26x10392 gt 18x105 Because Q gt Keq the reaction will shift to the left to create more reactants so that Q can decrease to the value of Kgq note water is not included in the equation because water is a pure liquid KP analLeChatelz39er Kp pressure equilibrium constant is a variety of Keq The Keq that was used in Question 2 is technically Ka acid equilibrium When you re not dealing with acids or bases Keq is commonly known as Kc concentration equilibrium However Kp is a constant of pressures rather than concentrations It has a similar formula to Kc except that the concentrations are replaced with the pressures of each gas You can switch between Kp and Kc with the given formula Kp Kc 082057 T T is in kelvin and An is the difference between the of moles in product and of moles in reactant LeChatelz39er39s Principlequota system in equilibrium will react in order to relive applied stress If the overall volume of a system is decreased the reaction will shift to the side that will produce fewer moles This is because there is less room in the container for the gas molecules Ifthe pressure increases the opposite will occur and the reaction will shift toward more gas molecules Other changes that can be made to equilibrium reaction include heat and water Heat can be added to the reaction as a product or a reactant depending on whether the reaction is endothermic or exothermic Water will dilute aqueous solutions and decrease the concentration of solutions therefore uXing the Q value Question 3 The given reaction takes place within a 2L container If the volume is decreased to 1L which way will the reaction shift Hug My H 2mg Neither Both sides contain the same number of moles Therefore no matter which way the reaction shifted the same amount of moles would be produced ICE Tables With values such as Keq and a general idea as to which way the reaction will shift we can use ICE tables Initial Change Equilibrium to determine the concentrations of the products and reactants at equilibrium The amount of change is determined by the balanced equation since it is the coef cient of the compound Question 4 What will be the concentration of H30 in a 15 M solution of acetic acid Ka 18x10 5 CH3COOHaq H20l lt gtH3oaq CH3COO39aq AceticAcid H20 1130 021130 Initial 15 0 0 0 Change x 0 x x Equilibrium 15 x 0 x x Keg 1130 CH3COO39 CHgCOOH 18x105 xx 15 x Because the value of x is normally very small compared to the concentration of the reactants the x in the denominator can be disregarded from the equation IF YOU USE THISMETHOD ON THE EXAM be sure to write on the problem that you are making this assumption or talk to your professor before hand 18x10515 W2 7x105 x 52x103M Acids and Bases Definition and Conjugates Acids and bases can be de ned by two different ideas Bronstead Lowry Acid increases H in solution Base increases OH39 in solution primarily refers to solutions Lewis Acid accepts electron pair Base donates electron pair relates to solids liquids etc Strong Acid HCl HClO4 H2804 HBr HI HNO3 Strong Base LiOH NaOH KOH RbOH BaOH2 Hquot NHz39 RO39 Rquot R means carbon chain To every base there is a conjugate acid in the product and every acid has a conjugate base A conjugate is a compound that has one more or one less H than the initial acidbase The strength of the conjugate is determined by the strength of the original base or acid If the initial is a strong base or acid it is thought to be so acidicbasic that the conjugate will be neutral But for weak acid or bases they produce weak conjugates With this principle we can look at salts and determine whether they would be acidic or basic accept or donate hydrogens NH4C1 gt NH4 Cl39 We know that Cl39 is a conjugate base because it carries a negative charge therefore it normally is an acid NH4 is a conjugate acid because of its positive charge The initial acidbase of these compounds are NH3 and HCl Are either of these strong bases or acids HCl is a strong acid so Cl39 is a neutral conjugate base and will not produce OH ions when dissolved with water NH3 however is a weak base therefore a weak conjugate acid so it will create H30 in the solution and be known as acidic The salt would break down in this fashion in water NH4C1 s gt NH4 aq Cl39 aq NH aq H200 lt gt NH3aq H30 aq pH One way to measure the strength of acids and bases is through the pH scale This measures how much H30 ions are present in the solution on a scale of 0 to 14 note Though most acids will not extend past 0 a negative pH is possible as is a pH above 14 pH logH30 pOH is the same as pH except with the concentration of OH ions Remember that 14 pOH pH If you39re given pH you can find molar concentration with the following equation Hgot 1039pH When solving for the pH of a solution you must take into account whether the base or acid is strong or weak Strong bases and acids will dissociate completely in solution so you can solve for the concentration in the same way that you would with regular stoichiometry But if the acid or base is weak you need to solve for the H30 concentration with an ICE table pH is an important measure to gage the health of an environment For example some species of fishes and plant life can only survive in water that is close to 7 If the pH uxes there could be damage to these population and it could be an indicator ofpollution Ka and Kb Along with pH acidsbases can be compared with Ka or Kb values equilibrium constants for when the compound is acting as an acid or base Weak acids will have a Ka value or an equilibrium constant when they are increasing the amount of H30 in the solution Weak bases on the other hand will have Kb constants for when the compound increases the OH39 concentration in a solution Conjugate acids and bases have a particular Ka or Kb As with pH and pOH Ka and Kb are inversely proportioned because they have to equal leO3914 Ka Kb Question 5 NH4C1 is dissolved in water to form a 32M solution If the Ka for NH4 equals 52x10 6 what is the Kb value What is the pOH 0f the solution Bonus what39s concentration of OH ions in the solution 1x10 K Kb 52x10quot Kb 1x10 K 19x109 52x105 As shown in the De nition and Conjugate section NH4Cl is an acidic salt so it will increase the concentration of H30 Because NH4 is a weak acid we need an ICE table NH4 H20 1130 NH3 Initial 32 0 0 0 Change x 0 x x Equilibrium 32 x 0 x x K H3ONH3 NHf 52x10quot xx 32 x 52x10632 1 7x105 x 41x103M With the H3O we can nde pH iog41x103 2 39 14 pH pOH 239 pOH 14 239 pOH 1161 UsepOH tofind OH OH 10P0H 10 2455x103912M PolyproticAcids As the name describes polyprotic acids have more H than monoprotic acids This means that the acid has to undergo a few equilibrium steps in order to completely lose all of its hydrogen For example HZCO3 needs to two equilibrium steps H2C03aq H20l lt gt HC0339aq H30aq HC0339aQ t H200 lt gt CO3Z39aQ H30aQ Because H2C03 goes through these two steps H2C03 has two Ka values one for HZCO3 and one for HCO339 However after the rst deprotonation equilibrium the Ka values drop in powers from 45XlO397 to 4 7X1011 This means that the concentration of H30 after the rst equilibrium is insigni cant compared to the rst Therefore if you want to determine pH from a polyprotic acid you just need to solve for the rst equilibrium Bu ers Buffersolution that is resistant to changes in pH when an acid or a base is added Buffers are important to humans since we need buffers in our blood stream to keep the pH at 74 As noted before constant pH is essential for aquatic wildlife Minerals deposited in the rocks of river can be dissolved by the river water and help buf er the water Testing alkalinity measures the bu quotering power of the river In order for buffers to keep a constant pH a base and an acid must be present in the solution so if a base is added the acid can neutralize it and vise versa However acid and bases react together to form salts so how can we prevent this from happening Answer use a weak acidbase and its conjugate The compounds will be at equilibrium with each other and even when a base or an acid is added to the solution and uxes the equation the concentrations of the acidbase and its conjugate will even out and retain the buffer pH To determine the pH of a buffer you can use a handy equation known as the HendersonHasselbach equation pH pKa logA39HA where pKa is the negative log of the Ka A39 is the moles of conjugate base and HA is the moles of the original acid The most effective buffers are those where the pH Z pKa because this means that the concentrations of the acid and conjugate are both relatively high This means that the buffer can effectively resist both acid and base addition effectively Bases also have a similar equation pOH pr logHBB One big problem with the HendersonHasselbach equation is that it assumes that X is insigni cant when being added or subtracted from initial concentrations If the ICE table is used instead you can spot this more easily Titrations A titration normally includes the addition of a strong acid or base depending on the solution to a solution containing a weak acid or base in order to nd an equivalency point The equivalency point is when the moles of the strong acidbase are equal to the moles of weak acidbase Because acids and bases are being added together the solution and additive will hydrolyise to form water and a conjugate When no strong acidbase is added the pH of the solution is equal to the equilibrium of the weak acidbase If some strong acidbase is added but not greater in moles than the weak acidbase the conjugate and the leftover weak acidbase will determine the pH this is because the strong acidbase was completely used up Should enough moles of the strong acidbase be added to make the weak acidbase the limiting reactant then pH will be determined by the leftover strong acidbase Although the solution will have the maximum amount of conjugate because all ofthe weak acidbase was used up the equilibrium it forms will be supercided by the strong acidbase Remember strong acidbases dissociate completely and this will affect the pH more so than the weak acidbase Question 6 5 mol of HC2H302 and 09 mol of KC21 1302 are dissolved into a 1L solution that has a pH of 4 If 20mL 0f 1 M KOH is added to the solution what will the pH be Ka of acetic acid 18x10 5 BONUS What if 5L 0f the KOH was added to the solution what would the pH be use an ICE table Acid and Base will go through hydrolysis HC2H302aq KOH gt K aq C2H30239aq H20l 1mol 1L 20mL 02molKOH L 1000mL KOH is limiting and reaction is a one to one ratio so moles of C 2H 30239 02 New concentration of HC2H302 is 498 5 02 Henderson Hasselbach pH pKa logA39HA pH log18x10395 log902498 pH 474 733 40066 For the BONUS 1 mol 5L 5 mol KOH L Now there is an equal amount of KOH and HC2H302 which means that the reaction will use up both of the reactants completely since both are the limiting reagents This also means that there is 5 mol ongHgOg39 in the solution Because there is no HC2H302 left all ofthe pH is going to be dependant on C 2H 30239 acting as a base Therefore we need to find the Kb value for C 2H 30239 1x10 Ka Kb 18x105 Kb 1x10 Kb 556x10quot 18x103 C2H30z aq H20 gt HC2H302 aq Exam 3 Review Guide Ifyou have any questions about any of the material or feel as ifI have missed anything please feel free to email me or call me Email fadelraemsuedu Phone 13132057485 Chapter 13 Chemical Kinetics o A Reaction Mechanism is a series ofindividual steps in an overall reaction 0 The sum of the individual steps gives the overall equation 0 Compile the individual steps by adding together or canceling out to find overall reaction I Compounds on similar side of equation add up I Compounds on different sides cancel out I EX N02 CO 9 C02 NO overall equatlon The 2N02 and N02 cancel to form N02 and the NOg39s cancel out 1st step 2N02 9 N03 NO 2 101 step CON03 9 N02 C02 0 Elementary Steps I Elementary steps are simple reactions that include one or more similar or different compounds to form product I The amount and type ofproduct dictates rate law 0 A 9 product Rate kA1 o A A 9 product Rate kA2 o A B 9 product Rate kA1B1 0 Rate Determining Step I Within each reaction mechanism there is one intermediate step which is the slow step I This slow step or the quotRate Determining Step dictates what the rate law of the overall reaction will be 0 EX 2NO2H29N22H20 1st step 2ND 69 N202 Fast 2ndstep N202H2 9N20H20 Slow 3rdstep N20H29N2H20 Fast In this case the 2 101 step would be the Rate Determining Step Therefore Rate kReactants of Slow Step Rate kNzOz H2 However N202 is an intermediate and is consumed and used up in the reaction Therefore we cannot determine its concentration In order to replace an intermediate kfNO2 krN202 Set kr intermediate kfreactant in overall equation kfkrNO2 N202 Now replace N202 with new concentration Rate k kfkrNO2 H2 or Rate k N02 H2 k kkfkr 0 Energy Diagrams 0 Energy Diagrams are diagrams illustrating the overall behavior of a reaction 0 Energy Diagrams tell us 1 Amount of Elementary Steps of humps in the diagram 2 Amount of Intermediates ofwells in the diagram 3 Activation Energy requirements The Transition State 4 Rate Determining Step the hump that requires most Energy 5 If reaction is Exothermic or Endothermic If reaction ends at a lower Energy point than it began it is Exothermic If reaction ends at a higher Energy point than it began it is Endothermic I Ex Figure 1316 on page 596 in textbook 0 1 Count the humps in the diagram this will tell you how many steps are in the reaction 2 Count the number of wells in the diagram this will tell you how many intermediates there are 0 3 In order to find how much Ea Activation Energy is required begin at the well just after the previous hump if you are at the first hump begin where the line starts and draw a straight line across connecting to the next line over The are from the drawn line and the point on top of the hump just over the line determines Ea The greater the space the greater the Ea 4 In order to find the RDS find the highest hump reactions maximum This hump elementary step is the RDS So if the hump is between points A and B and there is another smaller hump between points B and C then the hump between A and B would be the RDS 5 The reaction line should begin and end at certain points If the beginning point is higher in Energy than the end point the reaction is Endothermic If the beginning point is lower in Energy than the end point the reaction is Exothermic o Catalysis o Catalysts lower the activation energy of a reaction which speeds up the overall reaction I Ex Production of Sulfuric Acid 0 O O 1 Step S 02 9 S02 Fast 2nd Step S02 12 02 9 S03 Slow Use Catalyst Here 3rd Step S03 H20 9 H2S04 Fast Chapter 14 Chemical Equilibrium 0 Dynamic Equilibrium is when the rate of the forward reaction equals the rate of the reverse reaction 0 Take the formation of H1 for example H2g12g 69 2H1 g As H2 and 12 are reacting their concentrations are decreasing decreasing the rate of the forward reaction Simultaneously as H1 is begins to form its concentration is increasing increasing the rate of the reverse reaction Eventually these rates will reach equilibrium 0 Therefore at equilibrium it is not the concentrations ofproducts and reactants that are equal Instead it is the rate of the forward reaction that is equal to the rate of the reverse reaction Equilibrium Constant 39Keq39 o The E ofthe conce 0 Large 0 Ifa ch quilibrium Constant or 39Keq is the ratio at equilibrium of the concentrations products raised to their stoichiometric coefficients divided by the ntrations of the reactants raised to their stoichiometric coefficients Keq C15 D101 Aa Bb vs Small Keq A large Keq Kgt1 implies that the numerator of the equation which specifies the concentration of products is larger than the denominator which specifies the concentration of reactants of the reaction Therefore the forward reaction is favored Conversely a small Keq Klt1 implies that the denominator is larger than the numerator thus favoring the reverse reaction emical equation is modified in some way then the equilibrium constant for the equation must be changed to re ect the modification Ifyou reverse an equation the equilibrium constant must be inverted Ifyou multiply the coefficients in the equation by a factor you must raise the equilibrium constant to the same factor Ifyou are adding two or more chemical equations together to obtain an overall equation multiply the corresponding Keq values together to obtain the overall eq 0 SOLIDS AND LIQUIDS ARE NEVER CALCULATED IN THE EQUILIBRIUM EXPRESSION Kp vs KC Compounds with s or I infront of them should never be put in the Keq equation because they do not change within a reaction Compounds with aq or g must always be placed into the Keq equation 0 Kc is the equilibrium constant ofa reaction in terms of concentration 0 Kpist I Reaction Qu Kc IlAV The concentration ofindividual gases is represented by AI1AV Therefore Kc CCDd AaBb keeping KC in terms of concentration he equilibrium constant ofa reaction in terms ofpartial pressures PAVnART for any gaseous compound A PAnARTV or PAnAVRT Partial pressure of any gaseous compound and Kp PA if PA was calculated at equilibrium After substituting A in for nAV you get PAA RT or APART Inputting the new equation for the A in terms of partial pressure into the original Kc you get Kc PCRTCPDRTd PARTAPBRTb After simplifying the new Kc equation you are left with an equation that incorporates Kc and Kp which uses the equilibrium constant in terms ofboth concentration and partial pressures which could then be used to solve for either Kp or Kc Kp Kc RT 5d39ab otient 39Q o The Reaction Quotient or 39Q looks identical to the Keq expression or Kc expression However the reaction quotient expression is only used when the concentrations of the compounds are NOT in equilibrium R 08206 LatmmolK T Kelvin Celsius 273 Q CCDd AP IBb You can Compare 39Q and 39Keq to determine which way the reaction will favor 0 Before we discussed when Keq is large there is more product and when Keq is small there is more reactant 0 However when QltKeq the reaction will move toward the products to the right because there are more reactants than products and the reaction will attempt to restore equilibrium Conversely when QgtKeq the reaction will move toward the reactants to the left because there are more products than reactants and the reaction will attempt to restore equilibrium 0 EX The following reaction takes place in a 10 L vessel 214 K Nz 3ll2 69 2NH3 KC237X10393 The partial pressures were measured N2 12 atm H2 155 atm NH3 184 atm What is the QC of the reaction Which direction will the reaction go in order to reach equilibria 0 Given N2 12 atm N2 12 atm08206X214K 06M ii2 155 atm H2 155 atm08206X214K 88M NHg 184 atm NH3 184atm08206X214K 105M T 214 K Kc 237x103 QC 1052 88306 234 PMRT Qc gt Kc or Q gt Keq meaning the reaction will go Le Chatelier s Principle to the left 0 Le Chatelier s Principle states When a chemical system at equilibrium is disturbed the system shifts in a direction that minimizes the disturbance o In other words there are several changes that a system may experience which could throw the equilibrium of that system off When such changes are experienced the system will shift right to favor products or left to favor reactants Therefore if a change in the system increases the concentration ofproducts the system will shift to the left to create more reactants to combat the change in concentration of product and vice versa 0 Some of the quotstressorsquot that can affect a systems equilibrium include I Concentration o If the concentration of product increases the reaction will shift to the left to create more reactants and vice versa I PressureVolume 0 An increase in pressure is directly related to an increase in concentration so an increase in the pressure ofproducts increase in concentration ofproducts results in the system shifting to the left to create more reactants to restore equilibrium and vice versa 0 An increase in Volume is indirectly related to an increase in concentration so if the volume ofa system is increased and the concentration ofproducts decreases the system will shift to create more product to restore equilibrium o ONLY TRUE FOR GASSES I Temperature 0 An increase in heat results in a system shift toward the products 0 A decrease in heat decrease in temperature results in a system shift toward the reactants I Catalysts o Catalysts do not affect the reaction s equilibrium 0 Catalysts only increase the rate of the overall reaction increasing both the forward and reverse reactions favoring no side 0 ICE Tables 0 ICE tables are tables that include the Initial concentration Change in concentration and Equilibrium concentration 0 concentrations at equilibrium 0 Solving for Kc using ICE tables I EX Ifinitial concentrations are given ICE tables can help solve for individual 100M H2 and 100M 12 are in a system At equilibrium H2 020M and 12 020M What is the concentration ofHI at equilibrium Whatis the KC of this system H2 12 69 Initial 100M 100M 0 Change 100X 100X 2X Equilibrium 020M 020M 160M X is calculated out by nding the difference between the initial and equilibrium concentrations of either H2 or 12 Once X is solved for plug it in to 2X to Solve for the H1 at equilibrium 0 What if the concentration of reactantsproducts at equilibrium are not given I Plug initial concentrations into ICE table I Find the change X or X depending on ifQ is greater than or less than Keq KC 2H1 lHIlm Given H2 100M lz 100M H2eq 020M IZeq 020M K HI2 H2 12 1602 020020 64 o If Q is less than Keq the change for products will be X and the change for reactants will be X 0 Conversely if Q is greater than Keq the change for products will be X and the change for reactants will be X I EX Solve for the concentrations at Equilibrium COg H200 69 C02 g H2 g 100M DC 0 0 X DC X X Kc 406 C02II2 C0 XX 100X XZ 100X After using quadratic formula X 416 0977 The X value cannot be negative so the 416 value is thrown out Therefore X 098 C0 100 X 100 098 002M C02 X 098M Hz X 098M Kc 406 C0Initial 100M TipC02initial 0 As smlg ayeigapproach this problem you notice H20 is in 1 form meaning we DC don39t care about it Also in order to solve for the concentrations at EQ X must rst be solved for X can then be plugged into EQ to nd concentrations E 100X DC X X 0 To check your X value and your concentrations at equilibrium plug the concentrations into the KC eXpression and see if you get the same KC as the one given to you at the beginning of the problem 0 What if the initial concentrations given to you are not equal I EX Solve for the concentrations at Equilibrium H2 12 69 2H1 Given K 543 I 006M 004M 02 2M thnmal 006M C quotX quotX 2X IZInilial 004M E 006X 004X 0222X HI1nmal 022M Kc 543 HI2 Hz 12 0222X2 006X 004 X 4X2088X4 84 X10394 X2010X24 X10395 Using quadratic formula X 011 0015 and since X cannot be negative X 0015 HI 0222x 025M 12 004x 0025M Hz 006x 0045M 0 Solving Equilibrium Problems Steps to Solving Equilibrium Problems 0 1 Write Balanced Equation Write Keq Expression 0 2 Determine Direction of Reaction I Solve for Q I If QgtK products are X reactants are X I If QltK products are X reactants are X o 3 Fill in ICE Table 0 4 Use equilibrium values AmmalX and plug into Keq expression 0 5 Set Keq eXpression equal to given Kvalue and solve for 39X o 6 Use 39X value to determine concentration of reactantsproducts at Equilibrium o NEGATINGX 0 One of the most promising and convenient shortcuts learned in this section is the 39negate X shortcut 0 Most ofthe time the change in Xin the denominator is so small compared to the overall equation that we could just negate it I This is the case when reactions have very small Ka meaning the overall reaction is mostly reactants I The change in reactants is small To negate X simple remove it from the denominator and go about solving the equation without the quadratic formula saving you tons of time In order to determine if removing X from denominator is valid divide X by the initial reactant concentration then multiply it by 100 The answer must be less than 500 Ifit is then you can negate X Simple tip I always believed that by the time you test X and do all that you could have completed quadratic formula This shortcut if not tested does have HUGE risks so be cautious about using it 0 O 0 Chapter 15 Acids and Bases Definitions ofAcids and Bases o Arrhenius Definition Acid A substance that produces H ions in aqueous solution Base A substance that produces 0H39 ions in aqueous solution HClaq 9 Haq Cl39aq 0 Under this definition HCl is an acid because it produces H ions in solution 0 BronstedLowry Definition Acid Proton or H ion donor Base Proton or H ion acceptor HClaq H20l 9 H30aq Cl39aq 0 Under this definition HCl is the acid because it donates a proton to the H20 0 Lewis Definition Acid Electron Pair Acceptor Base Electron Pair Donator Definitions ofAcids and Bases continued 0 Common Acids HCl Hydrochloric Acid H2S04 Sulfuric Acid HF Hydro uoric Acid HN03 Nitric Acid 0 Common Bases Na0H Sodium hydroxide NHg Ammonia KOH Potassium hydroxide NaHC03 Sodium bicarbonate o In many situations you will find that a compound you once though was an acid actually acts as a base These compounds are called amphoteric substances Examples of amphoteric substances include water amines alcohol 0 Conjugate AcidsBases In a reaction an acid will donate the H to the base The result is products which are called conjugate acidsbases Ex NH3aq H20l 69 NH4aq 0H39aq Base Acid Conj Acid Conj Base In this example NHg accepts the donated H from H20 forming NH4 and OH This is an equilibrium reaction so it can go forward and reverse If the reaction was ipped and NH4 and 0H39 were on the reactant side and NH4 and H20 were on the product side the NH4 would act as an acid donating an H to the 0H39 hence the names conjugate acid and conjugate base The conjugate acid of a weak base is a strong acid The conjugate base ofa weak acid is a strong base The conjugate acid of a strong base is a weak acid The conjugate base ofa strong acid is a weak base Acid Strength and Ionization 0 Strong Acid is an acid that completely ionizes when dissolved in H20 I Common Strong Acids o HCl Hydrochloric Acid 0 HBr Hydrobromic Acid 0 HI Hydroiodic Acid I Virtually there will be no intact HCl once it is dissolved in H20 I Strong acids will never be present in an equilibrium equation because they are completely ionized 0 Weak Acid is an acid that does not completely ionize in solution I Common WeakAcids 0 HF Hydro uoric Acid 0 H2S03 Sulurous Acid I Virtually only some weak acid molecules ionize so there are intact weak acid molecules to reverse the reaction I Therefore weak acids are present in equilibrium equations because they can be used in the reverse reaction 0 Acid Ionization Constant K3 I K3 is the equilibrium constant for the ionization reaction ofa weak acid The relative strengths of these weak acids are represented by K3 0 In general the smaller the value of K3 the weaker the acid and vice versa 0 pKa logKa I Ka ProductsReactants 0 As before compounds with s or I are never included in this expression I In a Ka expression a weak acid will donate H to H20 to form a conjugate base and H30 I Ifa problem involves an equation that produces H30 as a product but the Kb is given you must calculate for the K3 I will show how to do this after the Base section 0 Base Strength and Ionization o A Strong Base is a base that completely dissociates in solution I Na0Haqj 9 Naaq 0H39aq I Most strong bases contain 0H and separate into a cation and OH I Common Strong Bases o Li0H Lithium Hydroxide o Na0H Sodium Hydroxide o Ca0H2 Calcium Hydroxide o A Weak Base is a base that doesn t completely dissociate in solution I NH3aq H20l 69 NH4aq 0H39aq I Unlike strong bases that contain 0H39 and dissociate in water the most common weak bases produce 0H39 by accepting a proton from water ionizing water to form OH I Common Weak Bases o NH3 Ammonia o CH3NH2 Methylamine 0 Base Ionization Constant Kb I Kb is the equilibrium constant for the ionization reaction ofa weak base The relative strengths of these weak bases are represented by Kb o In general the smaller the value of Kb the weaker the base and vice versa 0 pr logKb I In a Kb expression a weak base will take a H from H20 to form an OH39 as a product I Ifa problem involves an equation that produces OH39 but the K3 is given you must calculate for the Kb 0 Using Kw to solve for either Kb or Ka depending on which one is given 0 O O 0 KW is the ion product constant for water dissociation constant for water KW H30OH39 HOH39 Ka X Kb KW 10x103914 25 degrees Celsius Dr Cass will assume each reaction in which she intends us to use KW to be at this temp so assume 10x103914 KW always So ifa problem has an equation that produces H30 you know you must use Ka However if the Kb is given you must calculate for the K3 I KW Ka X Kb I Ka Kb KW and you know KW is always 10x103914 so just plug whatever Kb value Dr Cass gives you to solve for Ka Ifa problem has an equation that produces OH39 you know you must use Kb However if the K3 is given you must calculate for the Kb approach this problem exactly how you would approach the previous one except youre solving for Kb instead of K3 I Ex A solution has OH39 32x10394 What is the H30 W H30 OH39 10x103914 H30 32x10394 H30 3125x103911 0 pH Scale pOH and Acidity O O 0 If somethingis added to a solution and the concentration of H becomes larger than OH39 the solution is Acidic If something is added to a solution and the concentration of OH39 becomes larger than H the solution is Basic I Ex Is the previous problem acidic or basic H 3125x103911 OH 32x10394 H lt OH39 Solution is Basic pH is a compact way to specify the acidity of a solution I pH logH30 H30 is the same as H I pH 14 pOH I pHlt7 Solution is acidic I pH7 Solution is neutral I pHgt7 Solution is basic 0 pOH is a compact way to specify the basicity ofa solution I pOH logOH39 I pOH 14 pH 0 pOH pH 14 o Calculating the pH and pOH from strong acids and bases 0 When you are dealing with strong acids and strong bases it is easy to solve for the pH or pOH H30 X 00146 pH logH30 2836 0 EX A solution has 200M HCl What is the pH pH 10gH pH log200 pH 699 o EX A solution has 05 0M HI What is the pOH pOH logOH 14 pH pH logH log050 130 pOH 14 130 12699 0 In the two above examples solutions contained strong acidsbases and asked to find the pH or pOH In these situations when the concentration ofa strong acid or base is given and elemention connected to the H in the acid or OH39 in the base you know is whats called a spectator ion This means that it is not significant so the OH in a solution of NaOH is equal to the NaOH Determination of HgO and OH39 in weak acidbase solutions 0 BACK TO ICE TABLES o EX What is the pH of120M solution of Acetic Acid Tip logHgO knowing this you know you want to solve for H30 at equilibrium pH the concentration of CH3C02H H20 69 CH3C0239 H30 Given Ka 18X10395 120M DC 0 0 CH3C02H 12M X DC X x 120MX DC X X K3 18X10395 CH3C02H30 CH3C02H K3 18X10395 X2 12X Using quadratic formula X 001461 00148 but X cannot be negative so X 00146 In the above problem it asked you to solve for the pH of the solution As soon as the problem asks for the pH you must understand it is asking to first find the H30 and then take the log of that concentration to satisfy the equation pH logH Conversely as soon as the question asks for the pOH of the solution you must note that it is asking for the concentration of OH first then you must plug in OH into pOH logOH Using ICE tables allows you to find the concentrations of OH and H at equilibrium so you could use them to solve for pH or pOH Again finding OH and H using ICE tables only holds true for weak bases and weak acids that do not completely dissociate in solution The equation must have a 69 arrow to symbolize equilibrium For strong acids and strong bases see previous section on how to calculate pH and pOH Effect of Salts on pH 0 Salts in which neither the cation nor the anion acts as an acid or a base form pH neutral solutions 0 Salts in which the cation does ot act as an acid and the anion acts as a base form basic solutions Salts in which the cation acts as an acid and the anion does not act as a base for acidic solutions 0 Salts in which the cation acts as an acid and the anion acts as a base form solutions in which the pH depends on the relative strengths of the acid and the base 0 o This table represents the effects of salts on pH Conjugate base of strong Conjugate base ofweak acid Cation on on o Polyprotic acids amp Determination of pH in these solutions 0 Polyprotic acids ionize in successive steps each step with its own Ka value I EX The ionization of sulfurous acid 1st step H2S03aq 69 Haq HS0339aq K31 16X10392 2 101 step HS03aq 69 Haq S03239aq K32 64X10398 o In this example the ionization takes place in 2 separate steps The K3 is smaller for the second step than the first which will be the case for all polyprotic acids I EX Find the pH of01M Sulfuric Acid solution 1st step H2SO4aq H20l 9 H30aq HSO439aq Strong 2 1 step HSO4aq H20l 69 H30aq SO4239aq K32 12X10392 The first step is a strong acid completely ionizing in water so the H30 is 1M Again H20 is in 1 form so we do not count it in our ICE table H504 69 l39lgO 504239 I 010M 010M 0 C X X X E 10MX 010MX X Kaz 12X10392 H30SO4 HSO439 H30 010M X 12X10392 010X X2 010 X 010M X2 022X 00012 0145M After quadratic formula X 027 0045 and pH logH30 9 184 since X cannot be negative X 0045 Chapter 16 Aqueous Ionic Equilibrium o Buffer resists pH change by neutralizing added acid or added base 0 Buffers contain TONS ofweak acid and conjugate base or weak base and conjugate acid 0 Why do we care about buffers I Blood is buffered and is highly resistant to pH change This must be the case because ipr in the blood was to drop the bloods 02 distribution efficiency diminishes which is bad I Stomach acid must remain acidic duh in order to properly break down foods in order to digest properly 0 Determination opr of Buffer systems 0 EX Whatis the pH ofa buffer solution containing 2M acetic acid and 2M sodium acetate Ka 18X10395 cngcozmaq H200 69 cngcoziaq H30aq I 200M DC 200M 0 C X DC X X E 200X DC 200X X Ka 18X10395 CH3C0239H30 CH3C02H pH logH30 18X10395 200X X2 200X log18x105 After using quadratic formula X 18X10395 47 0 Determination opr on Buffer systems continued 0 In order to completely understand how the pH ofa system changes depending on what acidbase is added to it lets take a look at what the addition of NaOH to the previous example would do 0 EX Whatis the new pH of the system after 15M NaOH is added For a problem like this 2 ICE tables must be set up one to calculate the concentrations of CH3C02H and CH3C02 upon addition of NaOH and the second to calculate both concentrations once in equilibrium The conc found CH3C02Haq Na0Haq 9 H200 CH3C027an inthe rstICE I 200M 15M DC 200M table for C 15 15 DC 15 A gftllileb CH3C02H and F 05M 0 DC 35M used 11 s 6 Eli 02 age because it is the en use as initial conc in CH3C02Haq H200 69 cngcozrmq 04an a strorlllg 133e the second ICE 1 05M DC 35M 0 SO rea y is the only part of table C X DC X X NaOH that will E 05 DC 35 X X X take part in the K3 18X1075 CI 02 1430 CH3CO2H reaction 18X10395 X35X05X after calculation X 25X10 396 pH log25x106 559


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