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Chemistry Final Review.docx

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Chemistry
Course
01:160:159
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Professor Marvasti
Semester
Fall

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Chemistry notes Important terms *Mass of element in a sample Periods- horizontal rows on a periodic table Groups- vertical rows on a periodic table Ionic compound – electron transfer from a metal to a non-metal Covalent compound- electron sharing between two non-metals Hydrates- have a specific number of water molecules associated with each formula unit ( shown with a *#H O2 Aqueous solutions are solutions in water Combustion analysis- add O  2 O +C2 2 Limiting reactant- substance that stops the reactions from proceeding Molarity- concentration of a solution (mole/liters) Polar molecule : a molecule that is electronegativity charged Precipitate reaction- when two soluble ionic compounds combined to form an insoluble product Acid-base reaction- acid reacts with a base to form a neutral substance Titration- one solution of known concentration is used to determine the concentration of another solution through a monitored reaction - End point- tiny excess of OH ion changes Oxidation-reduction reaction(redox)- net movement of electrons from one reactant to the other, goes from less electronegative to more electronegative rules 1.For atoms in their elemental form, the oxidation number is 0 2.For ions, the oxidation number is equal to their charge 3.For single hydrogen, the number is usually +1 but in some cases it is -1 4.For oxygen, the number is usually -2 5.The sum of the oxidation number (ONs) of all the atoms in the molecule or ion is equal to its total charge. Standard temperature and pressure(STP)- 273.15K and 1 atm Standard molar volume = 22.4 L Universal gas constant= .082058 atm*L Mol*K Mole fraction(X) the fraction of each element in a compound Kinetic-molecular theory- describes behavior of gas at macroscopic levels Root mean squared speed (rms) a molecule moving at this speed has the average kinetic energy Effusion- gas escaping through a tiny hole Diffusion- gas moving through another gas Equations Mass of compound in sample * mass of element in compound Mass of compound Molecular mass= sum of atomic masses 23 Avogadro’s number= 6.022*10 = 1 mole Mass % of element x = moles of X in formula * molar mass of X (g/mole) Mass(g) of 1 mole of compound %Yield=actual yield *100 theoretical yield PV=nRT n= m=PV M RT Pa=X a P total Urms= √3RT/M R= 8.314 (J/k*mol) Rate of effusion Ratea=√M b Rate b √M a Vander Waals equation = (Pn a) (V-nb) =nRT V2 a and b are Vander Waals constants A relates to the number of electrons B relates to molecular volume Laws *law of multiple proportions if element A and B react to form tow compounds, the different masses of B than combine with a fixed mass of A can be expressed as a ratio of small whole numbers *law of mass conservation the total mass of a substance does not change during a chemical reaction. The number of substance may change. The total amount of substance is constant *law of definite composition no matter what its source a particular compound is composed of the same elements in the same parts by mass. *coulombs law the energy of attraction is directly proportional to the product of the charges and inversely proportionate to the distance between them *ideal gas law PV=nRT *Boyles law at constant temperature the volume occupied by a fixed amount of gas is inversely proportionate to the applied pressure *Charles law at constant pressure, the volume occupied by a fixed amount of gas is directly proportionate to its absolute temperature *Dalton’s law of partial pressure in a mixture of unreacting gas the total pressure I the sum of the partial pressure of the individual gases P total 1P 2P +3 … 4 *Grahams law of effusion the rate of effusion of a gas is inversely proportional to the square root of molar mass Chapter 16 kinetics Factors that influence reaction rate 1. Concentration – molecules must collide to react a. Reaction rate is proportional to concentration of reaction 2. physical state – molecules must mix to collide a. the more finely divided a solid or liquid reaction , the greater its surface area per unit volume, the more contact it makes with the other reactant, the faster it occurs 3. Temperature- molecules must collide with enough energy to react a. At higher temperature more collisions occur in a given time. b. Raising the temperature raises the reaction rate Express your reaction Rate of motion = Change in position = x -x = Δx 2 1 Change in time 2 1t Δt Reaction rate- change in concentration; reactant decreases while product increase at set rate Rate = - Δ[A] [ ] represent concentration in moles per liter Δ t Average , instantaneous, and initial reaction rates Rate itself varies with time as the reaction proceeds The instantaneous rate decreases as during the course of the reaction The slope of straight line from two points on an x/y graph is the average Initial rate- the instantaneous rate when the reaction occurs aA+ bB cC+ dD Rate= - 1 Δ[A] = - 1 Δ[B] = 1 Δ[C] = 1 Δ[D] c Δt b Δt c Δt d Δt 16.3 Rate law and its components rate law expresses rate as a function of reactant concentrations, product concentrations, and temperature. Rate depends only on reactant concentration and temperature Rate= k[A] [B] n k is reaction constant M and n are the reaction orders aka how the rate is affected by reactant concentration The component of the rate law- rate, reaction, order and rate constant must be found through experiment Reaction order terms First order Rate = k[A] Second order Rate = k[A] 2 Zero order- 0oesn‘t depend on A Rate = k{A} Example 2NO + 2H  N2+ O 2 2 Rate = k[NO] [H ]2 Second order in NO and first order in H overall it‘s a third order 2 Another example: CHCl + C3  C2l +HCl 4 Rate= k[CHCl ][Cl ] 1/2 3 2 Reaction orders experimentally Run a series of experiments staring each with a different set of reactant concentration and obtaining an initial rate for each case Integrated rate law m Rate -Δ{A} =k[A] Δt First order is Ln[A] =0t [A]t second order is 1 -1 =kt [A]t[A] 0 zero order is [A}t-[A]0= -kt half-life of rate law half-life of first order is constant ln 2 k = half life second order half life t1/21 . k[A] 0 zero order half life t1/2[A] 0 2k Effects of temperature on reaction K increases as temperature increase Use Arrhenius equation K=Ae =Ea/RT A is a constant Ea is activation energy- minimum energy needed for a reaction R is universal gas constant T in temperature in K Higher T larger k increase in rate Rewrite the equation Ln k 2 E (a - 1 ) K 1 R(T 2 T 1 16.6 Explanation of effect of concentration and temperature There are two theories that explain 1. collision theory-views the reaction rates as the result of parties colliding with certain frequencies and minimal energy. 2. transition state theory- offers close up view of how energy of collision converts from reactant to product collision theory measured in collisions per time explains why reactant concentrations are multiplied together in the rate law Why concentrations are multiplied Because the different paths that a particle can take to get to the end result Temperature affects the rate because collisions must have enough energy to overcome the minimum amount needed in a reaction called activation energy Two types of activation energy 1. E aorward- energy difference between activated state and reactant. 2. E aeverse- difference between activated energy and product the smaller the E the larger the value of k, and the faster a reaction a larger Ea(or lower T)  smaller k decrease rate effective collision- the molecules must collide so that the reacting atoms make contact. Aka must have enough energy and particular molecular orientation TRANSITION STATE THEORY Explains why activation energy is needed and what the molecules look like. - if the potential energy is less than the activation energy, molecules recoil, like billiard balls. - The kinetic energy pushes them together with enough force to overcome repulsions and react Transition state aka atiatied complex: a state that is not product or reactant; very frail; highest potential energy 16.7 Reaction mechanisms: steps in the overall reaction reaction mechanism- a sequence of single reaction steps that sum to an overall reaction A+B E +D Reaction intermediate- a substance that is formed and used up during the overall reaction Elementary steps- a single molecular event, molecules decompose or collide Unimolecular reaction A  B +C Bimolecular reaction A +B  C We use the equations coefficient as the reaction order in rate law for elementary steps Rate determining step: a step in a reaction slower than every other that affects overall rate Slow step determines rate We can‘t prose from just data that particular mechanism represents actual chemical change 16.8 Catalysis: speeding up a chemical reaction catalysis: a substance that increases the rate without being consumed in the reaction catalysis  lower activation energyrate constant higher homogeneous catalysis: exist in solution with reaction mixture heterogeneous catalysis: speed up reaction that occur in separate stages enzyme- a catalyst found in the human body, it is a protein active site- a small region whose shape results from those of side chains of amino acids, every enzyme has one 2 models of enzymes 1. lock and key – key(substance) lock is active site 2. induced fit – hand entering a glove all enzymes function by binding to the reactions transition state and thus stabilizing it ozone reactions uv O 2 2O O + O 2O (o3one) O + O 3 2O (o2one breakdown) Chapter 17 Equilibrium: the extent of chemical reactions Focuses on 17.1 Equilibrium state & equilibrium constant summary:  kinetics and equilibrium are distinct aspects of a chemical reaction, thus the rate extent of a reaction are related.  When the forward and reverse reactions occur at the same rate, the system has reached dynamic equilibrium.  The equilibrium constant (K) is a number based on a particular ratio of products and reactant concentrations: K is small for reaction that reach equilibrium with a high concentration of reactant(s) and large for reactions that reach equilibrium with a low concentration of reaction 17.2 reaction quotient & equilibrium constant 17.3 expressing equilibrium w/ pressure  The reaction quotient and the equilibrium constant are most often expressed in terms of concentration(Q & Kc). foc gases, they can also be expressed in terms of partial pressure(Q & Kp) p  The value of K apd K arecrelated by the derivation that relies on the Δngas ideal gas law K pK (cT) 17.4 reaction direction: comparing Q and K  We compare the value of Q and K to determine the direction I which a reaction will proceed towards equilibrium  If Qc K core reactant is formed  If Qc=K chere is no net charge 17.5 how to solve equilibrium problem  In equilibrium problems, we typically use quantities( concentrations or pressure) of reactants and products to find K, or we use K to find quantities  Reaction tables summarize the initial quantities, how they change and the equilibrium quantities  To simplify calculation, we assume that if K is small and the initial quantity of reactant is large, the unknown change in reactant (x) can be neglected. If this assumption is not justified( that is, if the error that results is greater than 5%), we use the quadratic formula to find x  To determine reaction direction, we compare the values of Q and K 17.6 reaction conditions : Le Chatelier‘s principle  Le Chatelier principle states that if a system at equilibrium is disturbed, the system undergoes a net reaction that reduces the disturbance and allows equilibrium to be retained  Changes in concentration cause a net reaction away from the added component or toward the removed component  For a reaction that involves a change in the number of moles of gas, an increase in pressure(decrease in volume) causes a net reaction toward fewer moles of gas, an increase in pressure(decrease in volume) cause a net reaction towards fewer moles of gas and a decrease in pressure causes the opposite change.  Although the equilibrium concentration of components changes as a result of concentration and volume, K does not change. A temperature change, however, does change K: higher T increase K for an endothermic reaction (positive ΔH rxnnd decreases K for an exothermic reaction ( negative ΔH rxn  A catalyst causes a system to reach the equilibrium point more quickly  Ammonia production is favored by high pressure, low temperature, and continual removal product. To make the process economical, an intermediate temperature and a catalyst are used 17.1 equilibrium state and the equilibrium constant equilibrium: reactant and product concentrations stop changing because the forward reverse rates have become equal rate = rate fwd rev no further change is observed at equilibrium because changes in one direction are equal to changing in the opposite direction 2 K fwdN2O 4 eq [rev] 2 eq K =K fwd= [NO 2 eq = product K rev= [N2O 4 eq reactant K is the equilibrium constant- number equal to particular ratio of equilibrium concentration of product and reactant at a particular temperature 1. small K – reaction yields very little product ; it looks like there is ‗no reaction‘ 2. Large K – reaction reaches equilibrium with very little reactant remaining; ― goes to completion‖ 3. Intermediate K – significant amount of both reactant and product are present 17.2 reaction quotient and equilibrium constant History: In 1864 two Norwegian chemist, Cato Goldberg & Peter Waage found that at a given temperature a chemical system reaches a state in which a particular ratio of reactant to product concentration has a constant value Aka law of chemical equilibrium or law of mass action Reaction quotient (Q) –aka mass-action expression Q= [product] [reactant] @ equilibrium Q=K note 1. ratio of initial concentration varies widely but always gives the same ratio of equilibrium concentration 2. individual equilibrium concentrations are different in each case, but the ratio of these concentrations is constant reaction quotient (Q) is a ratio made up from product concentration terms multiplied together and divided by reactant concentration terms multiplied together. aA +bB cC + dD a , b ,cc, d are stoichiometry coefficients Q c[C] [D] [A] [B]b To construct the reaction quotient you must write the balance equation first Q & K are unitless numbers because it is a ratio If the overall reaction is the sum of two or more reactions, the overall reaction quotient is the product of the reaction quotients Q overallQ 1Q * 2 *….3 Koverall K1*K 2K *3 The form of the reaction depends on the direction the balanced equation is written. Q c(fwd] 1/Q c(rev) K c(fwd) 1/ Kc(rev) If you multiply a reaction by a coefficient the Q must be raises to that power If all coefficients of the balanced equation are multiplied by some factor, that factor becomes the exponent for relating the reaction quotients and the equilibrium constant. N (aA +bB cC +dD) Q‘=Q =([C] [D] )d n K‘=K n ([A] [B] )n heterogeneous equilibrium- components are in different phases a pure solid and liquid always has the same concentration, same density We eliminate the terms for pure liquids and solids from reaction quotient… they do not change 17.3 expressing equilibria with pressure terms: relation between K & Kc p we start with the ideal gas law PV=nRT if T is constant then pressure is directly proportional to molar concentration ) reaction quotient based on partial pressures (Q p 2NO(g) + O (g22NO (g) 2 Q p P 2NO2 . P 2 *P NO O2 K p equilibrium constant based upon pressure K p/= K c But you can use change in moles of gasΔn to find Δn K p K (cT) 17.4 reaction direction: comparing Q &K more products make Q larger more reactant makes Q smaller 3 types of relative size of Q and K 1. if Q< K reactant products 2. if Q>K, reactant  product 3. if Q=K, reactantproducts 17.5 How to solve equilibrium problems Review: -reactant and product concentrations are constant over time -the forward reaction rate equals the reverse reaction rate -the reaction quotient equals the equilibrium constant Q=K Principles - we are given equilibrium quantities and solve for K - we are given K and initial quantities and some for equilibrium quantities USE QUANTITES TO FIND EQUILBRUM CONSTANT Example: flask 1.5 Liters H 2g) + I2(g) 2HI(g) We have At equilibrium 1.8 mol H 2.8 mol I an2 .520 mol HI. We find K chrough concentrations and substituting them into reaction quotient. Q = [HI] 2 . c [H2][2 ] We first convert amount (mol) to concentration (mol/Liters) using the 1.5 liter flask. [H 2=1.8 mol = 1.2 M 1.5 L We get [I 2 =1.2 mol, & [HI] =.347M Substitute values for Q c K c (0.347) 2 . = 8.36 *10-2 (1.2)(1.2) Using a reaction table CO +2C(graphite)  22O(g) Q p P CO P CO2 X atm CO 2x atm CO 2 Pressure at equilibrium P CO2(initial)P CO2 Setup a table Pressure (atm) CO 2 + C(graphite)  2CO(g) Initial 0.458 ------- 0 Change -x -------- +2x Equilibrium 0.458-x -------- 2x Use the quadratic formula to solve if in doubt If a reaction has a relatively small K and a relatively large initial reactant concentration, the concentration change (x) can often be neglected, x does not equal 0 [A]initialA]reacting [A}eq=[A]initial 17.6 reaction conditions and the equilibrium state: Le Chatelier‘s principle Le Chatelier Principle: when a chemical system at equilibrium is disturbed, it retains equilibrium by undergoing a net reaction that reduces the effects of the disturbance Q does not equal K ― Disturbance‖ Three common are change in 1. concentration of component a. if the concentration goes up, system reacts to consume some i. equilibrium position moves right when reactant added b. if the concentration does down, systems reacts to reduce some i. equilibrium position move to the left when reactant reduced The equilibrium system reacts to consume some of added substance or produce some of the removed substance New table Concentration(M) PCl 3g) + Cl2(g)  PCl5(g) Original equilibrium .2 .125 .6 Disturbance 0 +.075 0 New initial .2 .2 .6 Change -x -x +x New equilibrium .2-x .2-x .6+x K ctays the same 2. change in pressure(volume) a. change in concentration of gas b. adding an inert gas (gas does not take part) i. no effect c. changing the volume of vessel i. decrease volume raises concentration ii. increase volume decreases concentration iii. does not alter c iv. more volume = more product v. more pressure leads to side with less moles gas 3. change in temperature a. only way to alter K b. temperature rises, K rises and ΔH rxnxis positive c. temperature down, K down ΔH rxnnegative A catalyst shortens the time it takes to reach equilibrium but has no effect on the equilibrium position. Haber Process N 2g) + 3H (2)  2NH (g) 3 ΔH rxn=-91.8Kj 1. Decrease concentration of ammonia. NH is the 3roduct, so removing it will shift the equilibrium position towards producing more 2. Decreasing Volume(increase in pressure). Because 4 moles of gas react to form two moles of gas, decrease volume will shift the equilibrium position toward fewer moles of gas. 3. Decrease temperature. Because the formation of ammonia is exothermic, decreasing the temperature will shift the equilibrium position towards formation of product, thereby increasing K c Chapter 18 Acid base equilibria 18.1 acid bases in water  In aqueous solution, water binds the proton released from an acid to form the hydrated species represented by H O (aq3 + +  In Arrhenius definition , ac-ds contain H and yield H O 3n water, bases contain OH and yield OH in water, and acid-base reaction(neutralization) is the reaction of H and OH to form H O 2  Acid strength depends on [H O] 3elated to [HA] in aqueous solution. Strong acids dissociate completely and weak acids slightly  The extent of dissociation is expressed by the acid-dissociation constant K . weak acids range from 10 to 10 -12 a  Many acids and bases can be classified qualitatively as strong or weak based on their formulas 18.2 auto ionization of water and pH scale  Pure water has a low conductivity because it auto ionizes to a small extent. This process is described by an equilibrium reaction whose -14 equilibrium constant is the ion-product for water, K (1w ). Thus [H O] 3 and [OH] are inversely related: in acidic solution, [H O3 is greater than[OH]; the reverse is true in basic solutions; and the two are equal in a neutral solution  To express small values of [H O]m3re simply, we use the pH scale (pH = -log[H 3]). A high pH represents a low [H O].3In acidic solution, pH <7.00; I basic solutions, pH>7.00; and in neutral solutions pH=7.00. Similarly pOH= -log[OH], and pK = -logK. The sum of pH and pOH equals pK (14) w 18.3 proton transfer and he Bronsted-Lowry acid- base definition  The bronsted-lowry acid-base definition does not require that bases contain OH or that acid-base reaction occur I aqueous solutions  An acid is a species that donates a proton and a base is one that accepts it  An acid and a base act together in proton transfer. When an acid donates a proton, it becomes the conjugate base; when a base accepts a proton it becomes the conjugate acid. I an acid-base reaction, acids and bases form their conjugates. A stronger acid has a weaker conjugate base and vice versa  An acid-base reaction proceeds in the net direction I which a stronger acid and base form a weaker base and acid 18.4 solving problems involving weak-acid equilibra  Two common types of weak- acid equilibrium problems involve K from a a concentration and find a concentration from K a  We summarize the information in a reaction table (ICE) and we simplify the arithmetic by assuming [H O 3 from H2Ois so small relative to [3 O]from HA that it can be neglected and weak acids dissociate so little that [HA] =[HA] init equilibrium  The fraction of weak acid molecules that dissociates is greater in a more dilute solution, even though the total [H 3] is less  Polyprotic acids have more than one ionizable proton, but we assume that the first dissociation provides virtually all3H O 18.5 weak bases and their relation to weak acids  The extent a weak base accepts a proton from water to from OH is expressed by base dissociation constant K b  Bronsted-lowry base include NH and 3mines and the anions of weak acids. All produce basic solution by accepting H from water, which yields OH and makes [H O]<3OH}  A solution of HA is acidic because [HA] >>[A], so [H O]3>[OH].  A solution of A is basic because [A]>>[HA] so [OH]>>[H O] 3  By multiplying the expression for K of HA and K of A we obtain K . this a b w relationship allows us to calculate either K af BH, the cation conjugate acid of weak base B, or K ofbA, the anion conjugate base for weak acid HA 18.6 molecular properties and acid strength  The strength of an acid depends on the ease with which the ionizable proton is released  For nonmetal hydrides, acid strength increases across the period, with the electronegativity of the nonmetal (E, and don a group, with the length of E—H bond  For oxoacids with the same number of O atoms, acid strength increases with electronegativity of E; for oxoacids with the same E, acid strength increases number of O  Small, highly charged metal ions are acidic in water because they withdraw electron density from the O-H bond of bound H O mol2cules, releasing an H ion to the solution 18.7 acid-base properties of salt solutions  Salts that always yield a neutral solution consist of ions that do not react with water  Salt that always yield an acidic solution contain unreactive anion and a cation that releases a proton to water  Salts that always yield a basic solution contain an unreactive cation and an anion that accepts protons from water  If both cation and anion react with water, the ion that reacts to the greater extent (higher K) determines the acidity or basicity of the salt solution  If the anion is amphiprotic(first anion a polyprotic acid) the strength of the anion as an acid (K )aor a base (K ) betermines the acidity of the salt solution 18.8 electron-pair donation and the Lewis acid-base definition  The Lewis acid base definition focuses on the donation or acceptance of an electron pair to form a new covalent bond in an adduct, the product of an acid-base reaction. Lewis bases donate the electron pair, the Lewis acids accept it. Many species that do not contain are Lewis acids.  Molecules with polar double bonds act as Lewis acids, as do those with electron deficient atoms.  Metal ions act as Lewis Acids when they dissolve in water, which acts as a Lewis base to from an adduct, a hydrated cation  Many metal ions function as Lewis acids in biomolecules 18.1 acids and bases in water Acids and bases react, each cancels the properties of the other in a process called neutralization. Water is a product of all reactions between strong acids and strong bases: HCl(aq) + NaOH(aq)  NaCl(aq) + H O(l) 2 An acid dissociates in water to make a hydronium molecule HA + H O2 A + H O 3 Arrhenius acid-bade definition- acids and bases are classified in terms of their formula and their behavior in water - an acid is a substance that has H in its formula and dissociates into water + to yield H 3 - a Base is a substance that has OH in its formula and dissociates into water to yield OH - Acids contain covalently bonded H atoms that ionize when their molecules dissolve in water. The H ion and OH ion form the base combine h O 2 The heat of reaction is abut -57Kj per mole of water formed Strong acids dissociate completely into water. Weak acids dissociate slightly into water Specific equilibriu+ (K-a- weak acid in water K cH 2]= K =aH O 3[A ] [HA] + stronger acid higher[H O 3larger K a smaller K a lower dissociation of HA weaker acid Acids Strong acids Hydrohalic acids HCl, HBr, HI Oxoacids- 2 more O‘s than H‘s HNO ,3H S2 , 4ClO 4 Weak acids- a lot Hydrohalic acid HF H is not bonded to O or Halogens HCN, H S2 Oxoacids- number of O‘s exceeds by 1 over H‘s HClO, HNO , 2 PO3 4 Carboxylic acids- (RCOOH) CH COOH, C H COOH 3 6 5 Bases 2- - Strong bases- water soluble compound containing O or OH Group 1A(Li, Na, K, Rb, Cs) M 2, MOH Group 2A(Ca, Sr, Ba) MO, M(OH) 2 Weak bases- nitrogen with lone pair Ammonia NH 3 Amines (RNH , R2NH,2R N) 3 CH CH NH , (CH ) NH, (C H ) N 3 2 2 3 2 3 7 3 18.2 autoionization of water and the pH Scale Water is an extremely weak electrolyte. Water dissociates in equilibrium process called autoionization (self- ionizing) Ion-product constant for water (K ) wx 2 + - -14 K cH 2] = K = wH O ]3OH ] = 1.0*10 (at 25degrees C) Pure waters concentration 55.5M One H O3and one OH ion appear for each H O molec2le that dissociates 1. A change in{H O3 causes an inverse change in [OH] Higher [H O] lower [OH] 3 2.Both ions are present in all aqueous systems In an acidic solution [H 3]>[OH] In a basic solution [H O]7.00 hydroxide ion concentration can be expressed as pOH pOH = -log[OH] acidic solutions have a higher pOH than basic solution pK= -logK a lower pK corresponds to a higher K relationship between pH, pOH, pK -14 w K w[H O3[OH]=1.0*10 -logK w (-log[H O]3 + (-log [OH]) = 1*10 -14 pK w pH + pOH= 14 @25degrees C table 18.1 Relationship between K andapK a Acid Name (formula) - [email protected] -2 pK a Hydrogen sulfate ion (HSO ) 4 1.0*10 1.99 Nitrous acid (HNO ) 2 7.1*10 -4 3.15 -5 Acetic acid (CH C3OH) -98*10 4.74 Hypobromous acid (HBrO) 2.3*10 8.64 Phenol (C H OH) 1.0*10 -10 10.00 6 5 In the lab we measure acidity with a pH indicator or a pH meter. 18.3 proton transfer and the Bronsted-lowry acid base definition The Bronsted- Lowry acid-base definition took out the limitations of Arrhenius definition + 1. an acid is a proton donor, and species that donates a H ion a. there must be a H in the formula 2. a base is proton acceptor, any species that accepts a H ion + a. a base must contains a lone pair of electron to bind the H b. the only requirement for an acid-base reaction is that one species donates a proton another species accepts it;an acid-base reaction is a proton-transfer process. Typical acidic and basic solutions 1. acid donates a proton to water- a. HCl + H O  Cl +H O + 2 3 2. base accept a proton from water a. NH +3H O 2NH 4++ OH - H 2+NH  3S + NH 4 Conjugate acid-base pair- HS is conjugate base of acid H S 2 NH is the conjugate acid of the base NH 4 3 Every acid has a conjugate base, every base has a conjugate acid Table 18.2 the conjugate pairs in some acid-base reactions _____________________conjugate pair__ | | acid + Base  Base + Acid | | ------------------conjugate pair-------- - + reaction1 HF + H2O  F + H3O reaction 2 HCOOH+ CN -  HCOO - + HCN + + 2- - reaction 3 NH 4 -+ CO -  NH 3 -2 + HCO reaction 4 H 2O 4 + OH  HPO 4 + H2O reaction 5 H 2O 4 + N2H5+  HSO 4- + N2H 62 reaction 6 HPO 2-+ SO 2-  PO 3- + HSO - 4 3 3 3 A reaction proceeds to the greater extent in the direction in which a stronger acid and stronger base forma weaker acid and weaker base Strong  weak Kc>1 if strong is on left A weaker acid has a stronger conjugate base An acid-base reaction proceeds to the right if the acid reacts with a base that is lower on the list 18.4 solving problems involving weak-acid equilibria 2 general problems 1. given equilibrium concentration, find K a 2. given K aad some concentration information, find the other equilibrium constants - problem solving approach o start with what is given in the problem, then apply steps 1. write the balanced equation an K expression; these will tell you a what to find 2. define x as the unknown change in concentration during reaction. Through the use of certain assumptions, also equals [H O ] and [A ] - 3 at equilibrium 3. construct a reaction table that incorporates the unknown 4. make any assumptions that simplify calculation a. [H 3] from autoionization is so much smaller [H O] of3 dissociation that it can be ignored b. weak acid has a small K a 5. substitute the value of K ana solve for x 6. check the assumption are justified, test it percent HA dissociated =[HA] dissoc100 [HA] init as the original acid concentration decreases, the percent dissociation of the acid increases. do not confuse the concentration of HA dissociated with eh percent of HA dissociated polyprotic acid- acids with more than one ionizable proton. Each time a proton is taken away it gets a new K . a deareases as protons go K a1 >Ka2 a3 18.5 Weak bases and their relation to weak acids a base must have a lone pair base-dissociation constant (K )-nb base is dissociated during the process B(aq) + H O2 BH (aq) +OH(aq) K = [BH ][OH ] - b [B] ammonia is the simplest nitrogen-containing compound that acts as a weak base in water -5 NH (3q) + H O 2 NH (aq) +O4(aq) kb=1.76*10 K b[NH ]4OH] [NH ] 3 Amine group: was a NH but H3atom gets replaced by N group RNH , 2 NH2 R N 3 Has a lone pair of electrons to bind proton Rules 1. the acidity of HA(aq) a. the equilibrium position moves to the left. 2. basicity of A (aq) a. the relative concentration of HA and A determine the acidity or basicity of the solution 3. In a HA solution, [HA]>>[A ] & [H O]3 from HA>[OH] from H2O so the solution is acidic 4. In a A solution, [A]>>[HA]&[OH] from A> [H 3] from the H2Oso the solution is basic HA + H O2 H O +A 3 A + H O2 Ha +OH 2H O  H O + OH 2 3 the sum of the two dissociated reactions is the autoionization of water. K a K = b w 18.6 Molecular properties and acid strength The strength of an acid depends on its ability to donate a proton. Factors for nonmetal hydrides 1. cross a period, nonmetal hydride acid strength increase, because electronegativity increases. 2. Down a group, nonmetal hydride acid strength increases, as E becomes longer the E-H bond increases, decreasing the bong strength a. H O HNO 3 2 acids of hydrated metal ions hydrated ions transfer H to water. n+ M(NO ) 3 n +xH O(2) M(H O) 2 x (aq) +nNO (3q) If M is small and highly charged 18.7 Acid-base properties of salt solutions Salts that yield neutral solutions A salt consist of the anion of a strong acid and the cation of a strong base yields a neutral solution because the ions do not react with water - HNO (l3 +H O2NO (aq) 3 H O(aq) 3 The anion of a strong acid is a weak base in water, a strong acid anion is hydrated but nothing more NaOH(s) h2o Na + Oh All strong bases behave this way Salt containing ions such as NaCl or Ba(NO ) yie3 2neutral solutions because no reaction takes place between the ions and water. Salts the yield acidic solutions -A salt consisting of the anion of a strong acid and the cation of a weak base yield an acidic solution because the cation acts as a weak acid and the anion does not reaH2O NH C4(s) NH +4l {dissolution and hydration} NH (aq) + H O(l)  NH + H O { dissociation of weak acid} 4 2 3 3 Small highly charged metal ions make up another group of cations that yield H O 3 in solution H2O Fe(NO )3 36H O 2 Fe(H O2 (6q) +3NO (dis3ociates and hydrates) FE(H O) (AQ) + H O  Fe(H O) OH +H O (dissociate of weak acid) 2 6 2 2 5 3 Salts that yield basic solutions -a salt consist of the anion of a weak acid and the cation of a strong base yields a basic solution in water because the anion acts as a weak base and the cation does not react. Salts of weak acid and weak base ions Acidity depends on K aad K b NH 4H O 2 NH + H O 3 3 CN+H O  HCN +OH 2 -14 -10 K af NH =4K w . = 1.0*10 = 5.7*10 K bf HCN 1.76*10 -5 -14 -5 K bf CN = K w . = 1.0*10 = 1.6*10 K af HCN 6.2*10 -10 Solution is basics because K >b a Table 18.3 behavior of salts in water Salt solution pH Nature of ions Ion that react with examples water ex: Neutral: 7.0 Cation of strong None Nacl,Kbr,Ba(NO ) 3 2 acid Anion of strong base Acidic: <7.0 Cation of weak Cation: NH Cl, NH NO , base NH + H O 4 4 3 4 2 CH 3H Br3 Anion of strong NH 3 H O3 acid Acidic: <7.0 Small, highly Cation: AL(NO )3 3CrBr ,3 charged cation Al(H2O) 6H O2 FeCl Anion of strong  Al(H O) OH 3 2 5 acid Acidic/ basic <7.0 if Cation of weak NH 4H O2 NH 4lO ,2NH CN4 Ka(cation b(anion) base NH +H 3 3 Pb(CH C3O) 2 >7.0 If Anion of weak CN+ H O2 Kb(anion)Ka(cation) acid HCN + OH Acidic/ basic: <7.0 if Cation of strong Anion: NaH P2 , 4HCO , K 3 a(anion) b(cation) base HSO +3 O2 NaHSO 3 >7.0 if Anion of polyprotic SO + 3 O 3 Kb(cation)a(anion) acid HSO +3H O2 H2SO +3H 18.8 electron-pair donation and the Lewis acid-base definition definition: - a base is any species that donates an electron pair - an acid is any species that accepts an electron pair compared to Bronsted- Lowry Lewis expands on classes of acid the product of a Lewis acid-base reaction is called Adduct – a single species that contains a new covalent bond B: + H  B-H + \ covalent bond Lewis acid has electron deficiency - one surrounded by less than 8 electrons Chapter 19 ionic equilibria in aqueous systems 19.1 equilibria of acid- base buffer systems  a buffered solution exhibits a much smaller change in pH when H O 3 or OH is added that does an unbuffered solution  a buffer consists of relatively high concentrations of the components of a conjugate weak acid-base pair. The buffer component concentration ratio determines the pH and the ratio and pH are related by the Henderson- hasselbalch equation. As H O or 3 OH is added, one buffer component concentration ratio, and consequently the free[H O]3(and pH), change only slightly  a concentration buffer undergoes small changes in pH than a dilute buffer. When the buffer pH equals the pK of ahe acid component, the buffer has its highest capacity  a buffer has an effective range of pK +a pH unit.  To prepare a buffer choose the conjugate acid-base pair, calculate the ratio of buffer components, determine buffer concentration, and adjust the final solution to the desires pH 19.2 acid-base titration curves  An acid-base (pH) indicator is a weak acid that has differently colored acidic and basic forms and changes color over about 2pH unit  In a strong acid-strong base titration, the pH starts out low, rises slowly, then shoots up near the equivalence point (pH=7)  In a weak acid- strong base titration, the pH starts out higher, than the strong acid-strong base titration, rises slowly in the buffer region (pH=pK aa the midpoint), then rises more quickly at the equivalence point (pH>7)  A weak base-strong acid titration curve has a shape that is the inverse of the weak acid-strong base curve, with the pH decreasing to the equivalence point(pH<7) 19.3 equilibria of slightly soluble ionic compounds  As an approximation, the dissolved portion of a slightly soluble salt dissociates completely into ions  In a saturated solution, the ions are in equilibrium with the solid and the product of the ion concentrations, each raise to the power of its subscript in the compound‘s formula, has a constant value (Q sp= sp  The value of K spn be obtained from the solubility , and vice versa  Adding a common ion lowers an ionic compound‘s solubility  Adding H O (lowering pH) increases a compounds solubility if the 3 anion of the compound is that of a weak acid  If Qsp K spr an ionic compound, a precipitate forms when two solution, each contains one of the compounds ions, are mixed  Lakes bounded by limestone-rich soil form buffer systems that prevent harmful acidification by acid rain 19.4 equilibria involving complex ions  A complex ion consists of a central metal ion covalently bonded to two or more negatively charges or neutral ligands. Its formation is described by a formation constant( K ) f  A hydrated metal ion is a complex ion with water molecules as ligands, other ligands can displace the water in a stepwise process. In most cases, the K value of each step is large, so the fully f substituted complex ion forms almost completely in the presence of excess ligand  Adding a solution containing a ligand increases the solubility of an ionic precipitate if the cation forms a complex ion with the ligand 19.1 equilibria of acid base buffer systems buffer- something that lessens the impact of an external force. Acid-base buffer- a solution that lessens the impact of pH from the addiction of acid or base. Example: a shelter when it rains The components of a buffer are the conjugate acid-base pair of a weak acid (or base). How a buffer works Common-ion effect- it‘s how buffers work Example : CH CO3H(aq) +H O  H2O(aq) + CH3COO(aq) 3 From Le Chatelier principle equations shifts left if CH CO3 is added CH COO is common ion in this case, it occurs when a given ion is added to an 3 equilibrium mixture that already contains that ion, and shifts the position away from forming more of it. Common ion suppresses the dissociation of CH COOH, making the solution less 3 acidic. %Dissociation = dissoc/init *100 [CH CO3H] - a Buffer consist of high concentrations of acidic (HA) and basic (A ) components when H O3or OH are added small amounts of one buffer component to convert into the other. As long as the amount is small the added ions have little effect on the pH because they are consumed by one or the other buffer components. K a [CH C3O][H O] 3 [H O] = K3*[CHCOaH] [CH 3OOH] [ CHCOO] [H O] of solution depends directly on the buffer-component ratio 3 [CH 3OOH] [CH 3OO] The conversion of one component into the other produces a small change in the buffer- component concentration ratio and consequently a small change in [H O] 3 and in pH. Henderson equation Henderson-Hasselbalch equation pH= pK +alog[base]/[acid] can find [H O] very quick 3 allows us to prepare a buffer of desired pH Buffer capacity- is a measure of the ability to resist pH change and depends on both the absolute and relative component concentration The more concentrated the components of a buffer, the greater the buffer capacity. The pH of a buffer is distinct from its buffer capacity. For the given addition of acid or bases, the buffer-component concentration ratio changes less when the concentrations are similar than when they are different. *a buffer has the highest capacity when the component concentrations are equal  [A]/[HA] =1 A buffer whose pH is equal to or near the pK ofaits acid component has the highest buffer capacity buffer range- the pH range over which the buffer acts effectively, and its related to the relative component concentration. Buffers have a usable range with +- 1 pH unit of pK ofaacid component Preparing a buffer Step 1. Decide on the conjugate acid-base pair. Based off desired pH. pH is close to pK a step 2. Find the ratio [A-]/[HA] that gives the desired pH, using the Henderson Hasselbalch equation pH = pK +aog[A]/[HA] step 3. Choose the buffer concentration and calculate the amount to mix . find the amount of other component using the buffer- component concentration ratio step 4. Mix the amounts together and adjust the buffer pH to the desired value. Add small amounts of strong acids or strong base while monitoring the solution with a pH meter 19.2 acid-base titration curves we track the pH of titration with an acid –base titration curve a plot of pH vs. volume of titrant added monitoring pH with an indicator what is an acid base indicator? A weak organic acid (Denoted as Hln) that has different color than its conjugate base (IN) HIn + H 2 H O + 3n K a [H O3[In]/[HIn] [H 3]/K =a[HIn]/[In] An indicator changes color in units of 2pH Strong acids strong base titration curve Features of the curve 1. pH starts out low, shows high [H O]3 goes up gradually as base is added 2. sudden, steep pH rise , dues to moles of OH equaling mole H O 3 3. after steep there is an increase but slowly until equilibrium equivalence point- number of moles OH is equal to number of mole H O 3 at this point the solution consist of the anion of a strong acid and the cation of a strong base ions do not react with water at this point so the solution is neutral. Before the titration begins , we add an indicator to acid to signal when equivalence point. The end point occurs when the indicator changes color An indicator with an end point close to the equilibrium point **** The visible change in color of indicator(end point) indicates point where moles of base equal mole of acid originally.****** calculating the pH of a system 1. original solution of strong HA. pH=-log[H O]3 2. before the equivalence point. Initial amount of H O – change= which equals 3 amount of OH added. Concentration [H O] 3 3. at the equilibrium point. Weak acid- strong base titration features of the curve 1.the initial pH is higher , because weak acid dissociates slightly 2. a gradual rise portion of curve called the buffer region, appears before the steep rise to the equivalence point, the pH equal the pKa at half the original HPr reacted 3. the pH at the equivalence point is greater than 7.00 4. beyond the equivalent point the pH increases slowly as excess OH is added 19.3 equilibria of slightly soluble ionic compounds ion-product expression(Q ) and the solubility-product constant (K ) sp sp equilibrium exists between solid solute and aqueous ions. PbSO (s4 Pb 2+ (aq)+SO (4q) 2+ 2- Q c [Pb ][SO ]/[4bSO ]soli4 2+ 2- Ion-product expression Q = QspPbSOc] =[Pb 4[SO ] 4 When [PbSO ] re4ches equilibrium then we get K  solubspity equilibrium constant Depends on the temperature not the concentration Also equals the subscript of each ion in the compounds formula 2+ - Cu(OH) 2Cu + 2OH Ksp [Cu][OH] 2 Exceptions include insoluble metal sulfides . Sulfur ion is so basic it is not soluble in water MnS Mn +S S + H O  HS + OH 2 MnS + H O2 Mn + HS + OH Ksp[Mn][HS][OH] The value of K ispicates how far to the right the dissolution proceeds at equilibrium (saturation) Calculations with K sp Most are in grams of solute dissolved in 100 grams H O =102ml of solution Then we convert grams per 100ml solution into molar solubility: The amount of mol of solute dissolved per liter of solution. Using K vspues to compare solubilities If compounds have the same total number of ions. Then the higher the K the sp greater the solubility Effects of a common ion on solubility The presence of a common ion decreases the solubility of a slightly soluble ionic compound….. using le Chateliers principle Effects of pH on solubility If the compound contains the anion of a weak acid or CO , addi3ion of H O(from3 a strong acid) increases its solubility. CaCO (s) Ca(aq) +CO (aq) 3 3 Add some strong acid , reacts with CO to f3rm weak HCO 3 CO (3q) + H O(3q) HCO (aq) + H O(l) 2 More CaCO dis3olves  CaCO  Ca + CO - ->HCO - -> H CO  CO +H O+Ca 3 3 H3O 3 H3O 2 3 2 2 Predicting the formation of a precipitate: Q spvs. K sp Q sp whsp the system is saturated Q sp prspipitate forms until solution is saturated( need more water) Q sp Sosption is unsaturated and no precipitate forms(can add more solid) ACID RAIN Substances involved 1. sulfurous acid(H 2O )-3coal burns to form sulfurous acid a. H O2+2 SO 2 H 3O +H 2 4 2 2. sulfuric acid(H2SO )4 forms through atmosphere oxidation of SO 2 3. nitric acid- from car engines a. CO + 2H O2H O +HCO 3 3 Lakes bounded by limestone-rich soils form buffer system that prevent harmful acidification by acid rain 19.4 equilibria involving complex complex ion consist of central metal ion covalently bonded o two or more anions or molecules called ligands OH, Cl, CN, all complex ions are Lewis adducts the metal ion acts as a Lewis acid formation of complex ion M(H O) (aq) + 4NH (aq)  M(NH ) (aq) + 4H O(l) 2
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