Chem 2 Exam 2 Study Guide PDF

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Chapter 17 ● We can assume that x ([A]reacting) can be neglected if: ○ Kc is relatively small or ○ [A]initial is relatively large ● If [A]initial > 400, the assumption is justified; neglecting x introduces an error < 5% Kc ● If [A]initial < 400, the assumption is not justified; neglecting x introduces an error > 5% Kc SIMPLIFYING ASSUMPTION Le Châtelier’s Principle ● When a chemical system at equilibrium is disturbed, it retains equilibrium by undergoing a net reaction that reduces the effect of disturbance ● How is a system disturbed? ○ A system is disturbed when a change in conditions forces it temporarily out of equilibrium ● What does a “net reaction” mean? ○ The system responds to a disturbance by a shift in the equilibrium position ■ A shift to the left is a net reaction from product to reactant ● Reactants ← Products ■ A shift to the right is a net reaction from reactant to product ● Reactants → Products The Effect of a Change in Concentrations ● If the concentration of A increases, the system reacts to consume it: ○ If a reactant is added, the equilibrium position shifts to the right ○ If a product is added, the equilibrium position shifts to the left ● If the concentration of A decreases, the system reacts to replace some of it ○ If a reactant is removed, the equilibrium position shifts to the left ○ If a product is removed, the equilibrium position shifts to the right ● Only substances that appear in the expression for Q can have an effect

○ A change in concentration has no effect on the value of K

The Effect of a Change in Pressure ● Changing the concentration of a gaseous component causes the equilibrium to shift accordingly ● Adding an inert gas has no effect on the equilibrium position, as long as volume doesn’t change ○ All concentrations and partial pressures remain unchanged ● Changing the volume of a reaction vessel will cause equilibrium to shift if Δ n gas doesn’t equal 0 ○ If the moles is not equal to 0, and if the volume becomes smaller(pressure is higher), the reaction shifts so that total number of gas molecules decreases ○ If the moles is not equal to 0, and if the volume becomes larger(pressure is lower), the reaction shifts so that total number of gas molecules increases ● Changes in pressure(volume) have no effect on the value of K ● Volume and pressure are inversely related The Effect of a Change in Temperature ● To determine the effect of a change in temperature on equilibrium, heat is considered a component of the system ○ Heat is a product in an exothermic reaction (less than 0) ○ Heat is a reactant in an endothermic reaction (greater than 0) ● An increase in temperature adds heat, which favors the endothermic reaction ● A decrease in temperature removes heat, which favors the exothermic reaction Temperature and K ● The only factor that affects the value of K of a given equilibrium system is temperature ● For a reaction with greater than 0, increase in temp will cause K to increase ● For a reaction with less than 0, decrease in temp will cause K to decrease ● VAN’T HOFF EQUATION The Effect of a Catalyst ● A catalyst lowers the activation energy barrier for the reaction ○ Therefore, a catalyst will decrease the time taken to reach equilibrium ○ A catalyst does NOT affect the composition of the equilibrium mixture

Chapter 18

Arrhenius Acid-Base Definition ● This is the earliest acid-base definition, which classifies these substances in terms of their behavior in water ● An acid is a substance with H in its formula that dissociates to yield H3O+ ● A base is a substance with OH in its formula that dissociates to yield OH● When an acid reacts with a base, they undergo neutralization: ○ H+ from H3O+ reacts with OH- to form H2O Strong and Weak Acids ● A strong acid dissociates completely into ions in water: ○ HA(g or l) + H2O(l) → H3O+(aq) + A-(aq) ● A dilute solution of a strong acid contains no HA molecules ● A weak acid dissociates slightly to form ions in water: ○ HA(aq) + H2O(l) ↔ H3O+(aq) + A-(aq) ● In a dilute solution of a weak acid, most HA molecules are undissociated ○ Kc = [H3O+][A–] has a very small value [HA][H2O] ● STRONGER ACID→ HIGHER [H3O+] → LARGER Ka ○ AKA: Acid-dissociation constant Classifying the Relative Strength of Acids ● Strong acids include ○ The hydrohalic acids (HCl, HBr, and HI) ○ Oxoacids in which the number of O atoms exceeds the number of ionizable protons by 2 or more ● Weak acids include ○ The hydrohalic acid HF ○ Acids in which H is not bonded to O or to a halogen ○ Oxoacids in which the number of O atoms equals/exceeds the number of ionizable protons by 1 ○ Carboxylic acids which have the general formula RCOOH Classifying the Relative Strength of Bases ● Strong bases include ○ Water-soluble compounds containing O2- or OH- ions ○ The cations are usually those of the most active metals ■ M2O or MOH, where M= Group 1A(1) metals (Li, Na, K, Rb, Cs) ■ MO or M(OH)2, where M= Group 2A(2) metals (Ca, Sr, Ba) ● Weak bases include ○ Ammonia (NH3) ○ Amines, which have the general formula RNH2, R2NH, or R3N ■ The common structural feature is an N atom with a lone pair

Autoionization of Water ● Water dissociates very slightly into ions in a equilibrium process known as autoionization or self-ionization ○ 2H2O(l) ↔ H3O+ (aq) + OH- (aq) The Ion-Product Constant for Water (Ka) ● The concentration of water remains essentially constant and therefore it is considered a pure liquid ○ It is eliminated from the equilibrium expression and thus we get the ion-product constant for water, Kw ■ Kc ∙ (1)2 = Kw = [H3O+][OH-]= 1.0 ∙ 1014 (at 25 ∘C ) ● A change in [H3O+] causes an inverse change in [OH-], and vise versa: ○ Higher [H3O+] ⇒ lower [OH-] ○ Higher [OH-] ⇒ lower [H3O+] ● Both ions are present in all aqueous systems ○ All acidic solutions contain a low [OH-] and all basic solutions contain a low [H3O+] ○ And from that we get: < 7.00 = 7.00 > 7.00

pH, pOH, and pKa ● pH= -log [H3O+] and [H3O+]= 10-pH ○ The higher the pH, the lower the [H3O+] ○ That’s why an acidic solution has a lower pH compared to a basic solution ● pOH= -log [OH-] and [OH-]= 10-pOH ○ Acidic solutions have a higher pOH than basic solutions ● pK= -log K ○ For weak acids: pKa= -log Ka and Ka= 10-pKa ○ A low pK corresponds to a high K ■ A reaction that reaches equilibrium with mostly products present has a low pK and a high K ■ A reaction that has mostly reactants present at equilibrium has a high pK and a low K ● pKw= pH + pOH= 14.00 Bronsted-Lowry Acid Base Definition ● An acid is a proton donor⇒ any species which donates a H+ ○ An acid must contain H in its formula ○ All arrhenius acids are Bronsted-Lowry acids ● A base is a proton acceptor⇒ any species which accepts a H+ ○ A base contains a lone pair of electrons ● An acid-base reaction can be viewed from the standpoint of reactant and product ○ An acid reactant will produce a base product and the two will constitute an acid-

base conjugate pair Conjugate Acid-Base Pairs ●

In the forward reaction:

●

In the reverse reaction:

● H2S and HS- are a conjugate acid-base pair: ○ HS- is the conjugate base of the acid H2S ● NH3 and NH4+ are a conjugate acid-base pair: ○ NH4+ is the conjugate acid of the base NH3 ● Acid1 + Base2 ⇔ Base1 + Acid2 Net Direction of Reaction ● The net direction of an acid-base reaction depends on the relative strength of the acids and bases involved ● A reaction will favor the formation of the weaker acid and base

● This reaction favors the formation of the products ● The strong the acid, the weaker the conjugate base ● A reaction proceeds to the right if an acid reacts with a base lower on the list Solving Problems Involving Weak-Acid Equilibria 1. Write a balanced equation 2. Write an expression for Ka 3. Define x as the change in concentration that occurs during the reaction

4. Construct an ICE table in terms of x 5. Make assumptions that simplify the calculation 6. Substitute values into the Ka expression and solve for x 7. Check that the assumptions are justified PROBLEM SOLVING APPROACH Concentration and Extent of Dissociation ● If the initial concentration over Ka is greater than 400, then the assumption is justified and the error is less than 5% ● If the initial concentration over Ka is less than 400, then the assumption is not justified and the error is greater than 5% ● ●

As the initial acid concentration decreases, the percent dissociation of the acid increases ○ The equation is: ■ HA(aq) + H2O (l) ⇔ H3O+ (aq) + A- (aq) = [HA]dissoc ○ A decrease in the initial concentration means a decrease in the dissociation concentration causing a shift towards the products

Polyprotic Acids ● A polyprotic acid is an acid with more than one ionizable proton ○ Each dissociation step has a different value for Ka ○ The lower the number of Ka, the higher the stronger the acid or base ● Usually Ka1 > Ka2 > Ka3 ○ Neglect [H3O+] produced after first dissociation Acid Strength of Nonmetal Hydrides ● For nonmetal hydrides (E-H), acid strength depends on two factors: ○ The electronegativity of the central nonmetal (E) ○ The strength of the E-H bond ● Across a period, acid strength increases ○ Electronegativity increases so acidity of E-H does too ● Down a group, acid strength increases ○ Length of E-H increases down so strength decreases Acid Strength of Oxoacids ● All oxoacids have the acidic H bonded to an O atom ● Acid strength of oxoacids depends on: ○ The electronegativity of the central nonmetal (E) ○ The number of O atoms around E ● For oxoacids with the same number of O atoms, acid strength increases as the electronegativity of E increases ● For oxoacids with different numbers of O atoms, acid strength increases with the number of O atoms

Hydrated Metal Ions ● Some hydrated metal ions are able to transfer an H+ ion to H2O ○ These metals will form acidic solutions ● Consider a metal ion in solution, Mn+ ○ Mn+(aq) + H2O(l) → M(H2O)xn+(aq) ● If Mn+ is small and highly charged it will withdraw enough e- density from the O-H bonds of the bound H2O molecules to release H+ Weak Bases ● A Bronsted-Lowry base is a species that accepts an H+ ● For a weak base that dissolves in water; ○ B(aq) + H2O(l) ⇔ BH+(aq) + OH- (aq) ● The base-dissociation or base-ionization is given by the Kb ● No base actually dissociates in solution, but ions are produced when the base reacts with H2O ● STRONGER BASE→ HIGHER [OH-] → LARGER Kb Anions of Weak acids as Weak Bases ● The anions of weak acids often function as weak bases ○ A-(aq) + H2O(l) ⇔ HA (aq) + OH- (aq) ● A solution of HA is acidic while a solution of A- is basic ○ HF(aq) + H2O(l) ⇔ H3O+(aq) + F-(aq) ● HF is a weak acid, so this equilibrium lies to the left ○ [HF] > [F-], and [H3O+]from HF > [OH-]from H2O ■ Acidic solution ● If NaF is dissolved in H2O, it dissolves completely, and F- can act as a weak base ○ F-(aq) + H2O(l) ⇔ HF(aq) + OH-(aq) ● HF is considered a weak acid, so this equilibrium also lies to the left ○ [F-] > [HF], and [OH-]from F- > [H3O+]from H2O ■ Basic solution ● pKa + pKb = pKw Salts that Yield Neutral Solutions ● A salt that consists of the anion of a strong acid and the cation of a strong base yields a neutral solution ○ For example: ■ NaNO3 consists of a Na+ and NO3● This solution will be neutral because neither Na+ nor NO3 will react with H2O to any great extent Salts that Yield Acidic Solutions ● A salt that consists of the anion of a strong acid and a cation of a weak base yields an acidic solution ○ For example:

■ NH4Cl ● This solution will be acidic because NH4+ will react with H2O to produce H3O+ ● A salt consisting of a small, highly charged metal cation and the anion of a strong acid yields an acidic solution because the cation acts as a weak acid Salts that Yield Basic Solutions ● A salt that consists of the anion of a weak acid and the cation of a strong base yields a basic solution ○ For example: ■ CH3COONa ● This solution will be basic because CH3COO- will react with H2O to produce OH-

Salts of Weakly Acidic Cations and Weakly Basic Anions ● If a salt consists of the anion of a weak acid and the cation of a weak base, the pH of the solution will depend on the relative acid strength/base strength of ions ○ For example: ■ NH4CN ● The reaction that proceeds farther to the right determines the pH of the solution, so we need to compare the Ka of NH4+ with the Kb of CN○ Since Kb of CN- > Ka of NH4+, CN- is a stronger base than NH4+ is as an acid ■ Solution will be basic ● Amphiprotic→ can act as an acid and release a proton to water or as a base and abstract a proton from water The Leveling Effect ● All strong acids and bases are equally strong in water ● All strong acids dissociate completely to form H3O+, while all strong bases dissociate completely to form OH● In water, the strongest acid possible is H3O+, and the strongest base possible is OH○ H2O exerts a leveling effect on any strong acid or base The Lewis Acid-Base Definition ● A Lewis base is any species that donates an electron pair to form a bond ○ Must have a lone pair of electrons to donate ■ Any substance that is a Bronsted-Lowry base is also a Lewis base ● A Lewis acid is any species that accepts an electron pair to form a bond ○ Must have a vacant orbital ○ Many substances that are not Bronsted-Lowry acids are Lewis acids ■ For example: ● B: + H+ ⇔ B--H+ Electron-Deficient Molecules as Lewis Acids ● B and Al form electron deficient molecules and these atoms have an unoccupied p orbital that can accept a pair of electrons

○ BF3 + NH3 ⇔ BF3--NH3 ■ Accepts an electron pair Lewis Acids with Polar Multiple Bonds ● Molecules that contain a polar multiple bond often function as Lewis acids:

Metal Cations as Lewis Acids ● A metal cation acts as a Lewis acid when it dissolves in water to form a hydrated ion: ○ M2+ + 6H2O(l) ⇔ M(H2O)62+ (aq) ● O atoms of H2O donates a lone pair to an available orbital on metal cation

Chapter 19 The Common-ion Effect ● Acetic acid in water dissociates slightly to produce some acetate ion ○ CH3COOH(aq) + H2O(l) ⇔ CH3COO-(aq) + H3O+(aq) ● If CH3COOHNa is added, it provides a source of CH3COO- ion, and the equilibria shifts to the left ○ CH3COO- is common to both solutions ● The addition of CH3COO- reduces the % dissociation of the acid ● This effect occurs when a given ion is added to an equilibrium mixture that already contains that ion, and the position of equilibrium shifts away from forming it Acid-Base Buffers ● An acid-base buffer is a solution that lessens the impact of pH from the addition of acid or base ● An acid-base buffer usually consists of a conjugate acid-base pair where both species are present in appreciable quantities in solution ● An acid-base buffer is therefore a solution of a weak acid and its conjugate base, or a weak base and its conjugate acid How a Buffer Works ● The buffer components (HA and A-) are able to consume small amounts of added OH- and

H3O+ by a shift in equilibrium position ● The shift in equilibrium position absorbs the change in [H3O+] or [OH-] and the pH changes only slightly

Relative Concentrations of Buffer Components

● Since Ka is constant, the [H3O+] of the solution depends on the ratio of the buffer component concentrations ○ If the ratio [HA]/[A-] increases, [H3O+] increases ○ If the ratio [HA]/[A-] decreases, [H3O+] decreases The Henderson-Hasselbalch Equation ● For any weak acid, HA, the dissociation equation and Ka expression are ○ HA + H2O ⇔ H3O+ + A○ Ka= [H3O+] [A-]/[HA] ● The equation is: ○ pH= pKa + log ([base]/[acid]) Buffer Capacity and Range ● The buffer capacity is a measure of the “strength” of the buffer, its ability to maintain the pH following addition of strong acid or base ● The greater the concentrations of the buffer components the greater its capacity to resist pH changes ● The closer the component concentrations are to each other, the greater the buffer capacity ● The buffer range is the pH range over which the buffer is effective ● Buffer range is related to the ratio of buffer component concentrations ○ The closer [HA]/[A-] is to 1, the more effective the buffer ○ If one component is more than 10 times the other, buffering action is poor ■ Since log 10=1, buffers have a usable range within +/- 1 pH unit of the pKa of the acid component Preparing a Buffer

● Choose the conjugate acid-base pair ○ The pKa of the weak acid component should be close to the desired pH ● Calculate ratio of buffer component concentrations ● Determine the buffer component concentrations and calculate required volume of stock solutions and/or masses of component ● Mix the solution and correct the pH Acid-Base Titrations ● Add strong base (NaOH) with known concentration to solution of strong acid (unknown acid) ○ Start: pH of strong acid< 7 ○ Add base: ■ before equivalence, pH given by amount of strong acid in excess so pH< 7 ■ At equivalence so pH=7

Acid-Base Indicators ● Weak acid-base conjugate pair ● Very small amount added (pH not affected) ○ Hln(aq) + H2O(l) ⇔ H3O+(aq) + ln(aq)

Acid-Base Titrations ● Solution of known concentration base (or acid) slowly added to acid (or base) via buret ● Acid-base indicator→ equivalence point ● In practice, when indicator changes color for sure→ end point ○ Determined by indicator color change range ● Must choose indicator with color change range which includes equivalence pH ○ Ensures indicator endpoint reveals equivalence point ● Use of pH meter-titration curve ○ Graph of pH as function of added titrant Calculating Strong-Acid Strong-Base Titration 1. Initial solution of strong acid (pH) a. pH= -log [H3O+] 2. Before the equivalence point

a. Find the amount (mol) of H3O+ remaining i. Possible ICE table b. Calculate [H3O+] i. Amount (mol) remaining/original volume of acid + volume of added base ii. pH calculation 3. At the equivalence point 4. After the equivalence point a. Find the amount (mol) of OH- remaining i. Possible ICE table b. Calculate [OH-] i. Amount (mol) remaining/original volume of acid + volume of added base ii. pOH calculation Calculating Weak-Acid Strong-Base Titration 1. Initial acid solution a. Find [H3O+] 2. Solution of acid and added base a. pH= pKa + log ([base]/[acid]) 3. Equivalent amounts of acid and added base a. Find Kb from Ka b. Convert [OH-] to [H3O+] 4. Solution of excess base a. [OH-]= amount (mol) of excess OH-/total volume Amino Acids as Polyprotic Acids ● An amino acid contains a weak base (-NH2) and a weak acid (-COOH) in the same molecule ● Both groups are protonated at low pH and the amino acid behaves like a polyprotic acid Equilibria of Slightly Soluble Ionic Compounds ● Any “insoluble” ionic compound is actually slightly soluble in aqueous solution ○ We assume that the very small amount of such a compound that dissolves will dissociate completely ● For a slightly soluble ionic compound in water, equilibrium exists between solid solute and aqueous ions ○ PbF2(s) ⇔ Pb2+(aq) + 2F-(aq) ■ Kc=[Pb2+][ F-]2/[PbF2] ■ Ksp=Kc[PbF2] ⇔ [Pb2+][ F-]2 ● Qsp is called the ion-product expression for a slightly soluble ionic compound ○ For any slightly soluble compound MpXq which consists of ions Mn+ and Xz○ Qsp= [Mn+]p[Xz-]q ● When the solution is saturated, the system is at equilibrium and Qsp=Ksp, the solubility product constant ● The Ksp value of a salt indicates how far the dissolution proceeds at equilibrium Solubility and Ksp

● To convert solubility to Ksp ● Solubility needs to be converted into the molar solubility ● Molar solubility is converted into the molar concentration of ions at equilibrium (equilibrium equation calculation) ● Ksp is the product of equilibrium concentration of ions Effect of pH on Solubility ● If a slightly soluble ionic compound contains the anion of a weak acid, addition of H3O+ increases the compound’s solubility ● The net effect of adding H3O+ to CaCO3 is the removal of CO32- ions which causes an equilibrium shift to the right ○ More CaCO3 will dissolve Predicting the Formation of a Precipitate ● If Q > Ksp precipitate occurs until Q=Ksp ○ Solution is supersaturated ● If Q=Ksp equilibrium exists ○ Solution is just saturated ● If Q < Ksp solid disso...