Chemistry Exam 2

Order from strongest to weakest acid: AsH3, HBr, KH, H2Se

HBr, H2Se, AsH3, KH

Le Chatelier's Principle: Pressure/Volume

-A decrease in volume leads to an increase in pressure (increase in the "concentration" of molecules) The system will respond by shifting to the side that has fewer molecules N2 (g) + 3 H2 (g) = 2 NH3 (g) => rxn will shift to the right (fewer molecules)

Acidity and Ka

-Acid strength is measured by the acid dissociation constant Ka -The Smaller Ka, the weaker the acid -Larger Ka = lower pH= higher acidity - [H^+] = Sqr root of (ka*HA) -[OH^-]= Sqr root of (Kb*A^-/base)

Brønsted-Lowry Concept

-Acid: H+ donor (donates proton) -Base: H+ acceptor (Accept proton) -Doesn't have to be in water A substance with transferable protons is an acid, such as HNO2 in this example: HNO2(aq) + H2O(l) ⇌ NO2−(aq) + H3O+(aq) A substance that can receive a transferable proton is a base, such as water in this example: HNO2(aq) + H2O(l) ⇌ NO2−(aq) + H3O+(aq) -Not limited to aqueous solutions and can be applied to gases, liquids, and solids. HSO3−(aq)+H2O(l)→H2SO3(aq)+OH−(aq) HSO3^- is Bronsted Lowry Base H2O is Bronsted Lowry Acid (CH3)2NH(g)+BF3(g)→(CH3)2NHBF3(s) Neither is acid or base

Polyprotic Acids

-Acids that have more than 1 ionizable H atom can donate more than 1 H^+ -Only eq with larger Ka is taken into account

Autoionization of Water

-Amphiprotic: Ability to act as acid or base depending on equation -Kw= Ionization constant of water/Ion product of water

Q: Reaction Quotient

-Constant at a specific point in a reaction -Used to predict direction -Non-equilibrium concentration -If K>Q, reaction goes right (forward reaction proceeds to form more products/ goes towards products)

Exothermic Reaction

-DeltaH is negative -Negative Enthalpy

Electronegativity Trend

-Fluorine is the most electronegative

Le Chatelier's Principle

-If a system at equilibrium is disturbed, it will shift its eq so as to counteract the disturbance. -After equilibrium is re-established, Keq is the same as before -After new eq is established, the concentration of reactants/products is different, but the ratio (key) is the same

In an exothermic reaction, if the temperature is increased, how will the reaction respond to relieve the stress?

-It will consume products to make reactants. -Leftward shift

At 2000 ∘C the equilibrium constant for the reaction 2NO(g)⇌N2(g)+O2(g) is Kc=2.4×10^3. A) If the initial concentration of NO is 0.171 M , what is the equilibrium concentration of NO? B) If the initial concentration of NO is 0.171 M , what is the equilibrium concentration of N2? C) If the initial concentration of NO is 0.171 M , what is the equilibrium concentration of O2?

A) 1.7*10^-3 B) 8.5*10^-2 M C) 8.5*10^-2 M 2400 = (x)(x) / (0.171 - 2x)² 2400 = x² / (0.171 - 2x)² sqrt(2400) = x / (0.171 - 2x) sqrt(2400) (0.171 - 2x) = x (0.171)sqrt(2400) = [1 + 2(sqrt(2400))]x x = 0.084636 [NO] = 0.171 - 2(0.084636) = 1.73e-3 M [N2] = [O2] = x = 8.46e-2 M I .175 0 0 C -2x -x -x E .175-2x x x

A) What is Keq of S^2- + 2H^+ = H2S if: H2S=HS^- + H^+ K1=9.23*10^-8 HS^-= S^2- + H^+ K2=1.19*10^-19 B) What is Keq of PbCl2+ 2Ag^+ = 2AgCl+ Pb^2+ if: PbCl2= Pb^2+ + 2Cl^- K1=1.75*10^-10 AgCl2= Ag^+ + Cl^- K2 = 1.22 *10^-4

A) 9.10*10^25 (1/K1)*(1/K2) B) .01176 1/(K2)^2

Haber Process: Kp=Kc(RT)^DeltaN Kc=9.60, 573K, R=.08206 N2+3H2= 2NH3

0.00434 *DeltaN is 2-4=-2 from 1+3 on reactant side and 2 from products (Products - reactants) *Temp must be in Kelvin

What is the concentration of HCHO2 solution with pH=1.93? Ka= 1.8*10^-4 HCHO2= H+ + CHO2-

0.80M I x 0 0 C -.012 .012 .012 E x-.012=x .012 .012 HCHO2=x= (.012)^2/ (1.8*10^-4)

0.420 moles CO and hydrogen gas added to 1.0L. At eq, CO decreased to .290 moles. What is Keq? CO+2H2 = CH3OH

17.5 Keq= .13/(.29)(.16)^2

The reaction A2(g)+B(g)⇌A(g)+AB(g) has an equilibrium constant of Kp=2. There is a mixture containing 2 A atoms (red), 2 A2 molecules, and 4 AB molecules (red and blue). How many B atoms should be added to the diagram to illustrate an equilibrium mixture?

2 atoms

What is H^+ in .01 M Ca(OH)2 solution? Ca(OH)2= Ca^2+ + 2OH^-

5.0*10^-13 -log[.01*2] = 1.69 pOH 14-1.69 = 12.3 pH 10^-12.3 = 5*10^-13 *Same applies for MgOH2

0.18M solution of HF has pH = 1.96 What is Ka? HF= [H+] + [F-]

7.2*10^-4 HF H+ F- .18 0 0 -.011 +.011 +.011 .169 .011 .011 Ka = [.011][.011]/[.169] = 7.2*10^-4 -Ka is less than one, so it goes to left to reach equilibrium (Towards reactants)

A negative enthalpy value indicates an:

A negative enthalpy value indicates an exothermic reaction.

A mixture of 0.10 mol of NO, 0.050 mol of H2, and 0.10 mol of H2O is placed in a 1.0-L vessel at 300 K. The following equilibrium is established: 2NO(g)+2H2(g)←−→N2(g)+2H2O(g) At equilibrium [NO]=0.062M. A) Calculate the equilibrium concentration of H2. B) Calculate the equilibrium concentration of N2. C) Calculate the equilibrium concentration of H2O D) Calculate Kc.

A) 1.2×10−2 M B) 1.9×10−2 M C) 0.138 M D) Kc=650

Definition of Acids and Bases

Arrhenius Definition (1887) Acids must dissolve, generate H+ Bases must dissolve, generate OH- • Brønsted-Lowry Definition (1923) Acids donate H+ generate H+ Bases accept H+ generate OH- • Lewis Definition (1923) Lewis acids e- pair acceptors Lewis base e- pair donor

K=2.1×10−20. What can be said about this reaction? Where does the equilibrium of this reaction lie?

At equilibrium the concentration of reactants is much greater than that of products Equilibrium of reaction lies to the left

What is the conjugate base of A) HCO3− B) H3PO4 C) HS^-

CO3^2− H2PO4^- S^2-

Changes in volume affect the value of the reaction quotient, Q, whereas temperature changes affect the value of the equilibrium constant, K.

Changes in volume affect the value of the reaction quotient, Q, whereas temperature changes affect the value of the equilibrium constant, K.

Methanol (CH3OH) can be made by the reaction of CO with H2: CO(g)+2H2(g)⇌CH3OH(g) To maximize the equilibrium yield of methanol, would you use a high or low pressure?

High Pressure Q1= 1/1*1^2=1 Q2= 2/2*2^2 Q1>Q2 Before the stress, the system was at equilibrium, and so Q1=K. The new value of the reaction quotient, Q2, is less than K. Consider what happens when Q<K. Shift to right=decrease in container size=increase in pressure

An unknown salt is either KBr, NH4Cl, KCN, or K2CO3 If a 0.100M solution of the salt is neutral, what is the identity of the salt?

KBr

Which will form neutral solution when dissolved in water? K2CO3, Cu(NO)3, NH4Cl, KCl

KCl (Strong base and strong acid) K2CO3 (K2 is strong base) Cu(NO)3 (NO3 is weak acid) NH4Cl (Weak base and strong acid)

Ka and Kb

Ka= [H+][A-]/[HA (Weak Acid)] Kb= [OH-][SA](Strong acid)/[B-](weak base) *Strong acids and bases don't have Ka/Kb value

Lactic acid (CH3CH(OH)COOH) has one acidic hydrogen. A 0.10 M solution of lactic acid has a pH of 2.44. Calculate Ka

Ka=1.38*10^−4 HA (aq)+H2O(l)⇌H3O^+(aq) +A−(aq) I 0.10 0 0 C (−x) (+x) (+x) E. 0.10−x x 3.63*10^−3 [A−]=3.63*10^−3M [HA]=0.10−3.63*10^−3=0.09637 M Ka=[H3O+][A−]/[HA] Ka=[3.63*10^−3][3.63*10^-3]/0.09637 = 1.4*10^−4

Gaseous hydrogen iodide is placed in a closed container at 425 ∘C, where it partially decomposes to hydrogen and iodine: 2HI(g)←−→H2(g)+I2(g). At equilibrium it is found that [HI]= 3.59×10−3 M , [H2]= 4.87×10−4 M , and [I2]= 4.87×10−4 M . What is the value of Kc at this temperature?

Kc = 1.84×10−2

A mixture initially contains A, B, and C in the following concentrations: [A] = 0.450 M , [B] = 1.25 M , and [C] = 0.300 M . The following reaction occurs and equilibrium is established: A+2B⇌C At equilibrium, [A] = 0.260 M and [C] = 0.490 M . Calculate the value of the equilibrium constant, Kc

Kc=2.49 [.49]/(.26)(.87)^2

Keq=.008 At eq, NOCl=.25M and NO=.1M What is eq concentration of Cl2? 2NOCl=2NO+Cl2

Keq= .008= (.1^2)x/(.25^2)= 0.05M

A 1L reaction vessel is filled with 0.50 moles SO3 at 1000K. At eq, SO3=0.2M. What is the value of Keq 2SO3 = 2SO2 + O2

Keq= 0.338 Initial: .50M 0. 0. Change: -.3M. +.3M. +.15M Equilibrium: .2M. .3M. .15M

System initially contains 1*10^-3 M H2 and 2*10^-3 I2. At eq, concentration of HI= 1.87*10^-3. Calculate Keq H2 (g)+ I2 (g) = 2HI (g)

Keq= 51 (Reaction going to right/products) I 1*10^-3 2*10^-3 0 C -9.35*10^-4 -9.35*10^-4 1.87*10^-3 E. 6.5 *10^-5 1.065*10^-3 1.87*10^-3

0.015M N2 and .035M H2 combined. At eq, NH3 is .02M. Determine Keq N2 (g)+ 3 H2 (g) = 2NH3

Keq= [.020]^2/[.005][.005]^3 = 6*10^5 Make ICE table Goes towards right

Keq equation for 6CO2 (g) + 6H2O (l) = C6H12O6 (s) + 6O2 (g)

Keq= [O2]^6/ [CO2]^6

Graph showing reactants (Beginning) above products (End of graph)

Kf>Kr Kf/Kr= equilibrium constant (Will be greater than 1)

Given the two reactions PbCl2⇌Pb2++2Cl−, K3 = 1.75×10−10, and AgCl⇌Ag++Cl−, K4 = 1.22×10−4, what is the equilibrium constant Kfinal for the following reaction? PbCl2+2Ag+⇌2AgCl+Pb2+

Kfinal = 1.18×10^−2 K3=[E]e[F]f/[A]a[B]b=[C]c[D]d/[A]a[B]b⋅[E]e[F]f/[C]c[D]d or K3=K1⋅K2

Given the two reactions H2S⇌HS^−+H^+, K1 = 9.23×10−8, and HS−⇌S2^−+H^+, K2 = 1.19×10−19, what is the equilibrium constant Kfinal for the following reaction? S2−+2H+⇌H2S

Kfinal = 9.10×10^25

At a certain temperature, the pH of a neutral solution is 7.29. What is the value of Kw at that temperature?

Kw = 2.6×10−15 Kw= [H+][OH−]

Methanol (CH3OH) can be made by the reaction of CO with H2: CO(g)+2H2(g)⇌CH3OH(g) The enthalphy change for the reaction is -90.7 kJ. To maximize the equilibrium yield of methanol, would you use a high or low temperature?

Low Temperature *The reaction is exothermic, so if you added heat it would cause a left shift, and you want to create a right shift

Lowering pH

Metal cations except group IA metals Ca^2+, Sr^2+, and Ba^2+ lower pH of aqueous solution -More positively charged = more hydrolysis = lower pH -Smaller ions= more hydrolysis = lower pH

What is the acid/ base and conjugate base/acid: A) NH3 (aq) + H2O (l)= NH4^+ (aq) + OH^- (aq) B) HCl (aq) + H2O (l)= H3O^+ (aq) + Cl^- (aq)

NH3= Base H2O= Acid NH4^+= Conjugate acid OH^-= Conjugate base HCl= Acid H2O= Base H3O^+ = Conjugate acid Cl^- = Conjugate base

Which is a stronger base? NO3^- or NO2^-? HCO3^- or CO3^2-?

NO2^- More oxygens = more acidic Also, HNO3 is a strong acid, so we know that NO3^- is a weak base CO3^2- Fewer hydrogens= more basic

What effect will NO3− ions have on the pH of a solution? What effect will CO32− ions have?

No effect Ions will increase the pH.

Consider the reaction A+2B⇌C for which in the initial mixture Qc=[C]/[A][B]^2=387 In which direction will reaction proceed to reach equilibrium? What will happen if you decrease Qc?

The reaction will proceed in reverse to form reactants. A decrease in the value of Qc will increase [A] and [B]

What is the H+ concentration for an aqueous solution with pOH = 3.24 at 25 ∘C?

[H+] = 1.7×10−11 M

If acetic acid is the only acid that vinegar contains (Ka=1.8×10−5), calculate the concentration of acetic acid in the vinegar. A particular sample of vinegar has a pH of 2.95.

[HC2H3O2] = 7.1×10−2 M [10^-2.95]^2/1.8x10^-5 = [Acid] Ka = 1.8x10^-5 = [H+][A-]/[HA] [H+] = 10^-2.95 = .001122 1.8x10^-5 = (.001122)^2 / ([HA] - .001122)

The pH Table and pH Scale

[OH^-][H^+] = 1.0*10^-14 pH + pOH = 14 (ka)(kb)= kw Magnitude difference of 100 (Going from 7 to 6 = 10^1) -Going from 2 to 6 = 100,000 (10^4)

7 Strong Acids

hydrochloric acid (HCl) nitric acid (HNO3) sulfuric acid (H2 SO4) hydrobromic acid (HBr) hydroiodic acid (HI) perchloric acid (HClO4) chloric acid (HClO3)

Calculate pH for strong base solution: 13.0 mL of 1.60×10−2 M Ca(OH)2 diluted to 490.0 mL .

pH = 10.929 It's OH2, so multiply 4.245*10^-4 X 2

Calculate pH for strong base solution: 1.135 g of KOH in 450.0 mL of solution.

pH = 12.6528

Calculate pH for strong base solution: 8.9×10−2 M KOH.

pH = 12.95

What is the pH of a .10M solution of HCHO2 if Ka=1.8*10^-4? HCHO2= [H+] + CHO2^-

pH = 2.37 1.8*10^-4 = x^2/.10 x=.0042 -log(.0042)= 2.37

Ka of HClO=3*10^-8. What is pH of .34 M NaClO (weak base)?

pH= 10.5 1.0*10^-14/3*10^-8=3.33*10^-7 = Kb 3.33*10^-7=x^2/.34 = 3.37 -log(3.37)=3.47 pOH

NH4^+ has Ka=5.5*10^-10. Calculate pH of .629M solution of NH3 NH3 + H2O (l)= NH4^+ + OH^-

pH= 11.5 Kb= 1.8*10^-5=x^2/.629 x=.00336M OH^- -log(.00336) = 2.47 pOH

Determine the pH of .15M solution of KF

pH= 8.17 Ka of HF= 6.8 *10^-4 KF is a weak base

A mixture formed by adding 59.0 mL of 2.5×10−2 M HCl to 120 mL of 1.5×10−2 M HI. Calculate pH:

pH=1.74 .059*(2.5*10^-2)=0.001475 .120*(1.5*10^-2)=0.0018 0.001475+0.0018=0.003275 .120+.059=179 mL or .179L 0.003275/.179=.0183 -log[.0183]= 1.7376

A new potential heart medicine, code-named X-281, is being tested by a pharmaceutical company, Pharma-pill. As a research technician at Pharma-pill, you are told that X-281 is a monoprotic weak acid, but because of security concerns, the actual chemical formula must remain top secret. The company is interested in the drug's Ka value because only the dissociated form of the chemical is active in preventing cholesterol buildup in arteries. To find the pKa of X-281, you prepare a 0.090 M test solution of X-281 at 25.0 ∘C. The pH of the solution is determined to be 3.00. What is the pKa of X-281? What is pKb of the conjugate base of X-281? (Assume 25 ∘C.)

pKa= 4.95 pKb= 9.05 Concentrations (M): HA(aq)⇌H+(aq)+A−(aq) initial: .09 0 0 change: -x +x +x final: .09-x x x 10^-3 = [H+] = 1.0×10−3 M [HA] = 8.9×10^−2 M pKa= -log(1.124*10^-5) -log[1.1236]=4.95

Lewis Concept

-Least restrictive -Acid: Electron Acceptor (Must have empty orbitals) -Base: Electron donor (getting more positive) (Must have non-bonding electron pair to donate) Lewis acid: must have empty orbitals to accept electron pair Lewis base: must have a non-bonding electron pair to donate -A Lewis acid is an electron-pair acceptor, and a Lewis base is an electron-pair donor. A substance that accepts an electron pair is an acid, such as Fe3+ in this example where a pair of electrons on CN− is transferred to the metal: Fe3+(aq) + 6CN−(aq) → Fe(CN)63−(aq) A substance that donates an electron-pair is a base, such as water in this example where a pair of electrons on the oxygen of water are transferred to the metal cation core: [Fe(H2O)5]2+(aq) + H2O(l) ⇌ [Fe(H2O)6]3+(aq) Ni(NO3)3(s)+6H2O(l)→Ni(H2O)63+(aq)+3NO3−(aq) Lewis Acid: Ni(NO3)3 Lewis Base: H2O (CH3)2NH(g)+HF(g)→(CH3)2NH2F(s) Lewis Acid: HF Lewis Base: (CH3)2NH Ag+ (aq) + Cl- = AgCl (s) Ag+ is Lewis acid, Cl- is Lewis base

Arrhenius Concept

-Most restrictive - Acid: H+ donor in H2O -Base: OH- donor in H2O For example, a substance with an ionizable protons is an acid: HNO2(aq) ⇌ NO2−(aq) + H+(aq) A substance that can either directly or indirectly increase hydroxide ions is a base: NH3(aq) + H2O(l)⇌ NH4+(aq) + OH−(aq) However, the Arrhenius concept only applies to aqueous solution. KOH(aq)+HNO3(aq)→KNO3(aq)+H2O(l) KOH=Arrhenius Base HNO3=Arrhenius Acid

Equilibrium

-Rate of formation of product=the rate of formation of reactants (Once equilibrium is achieved, the amount of each reactant and product remains constant) -As system approaches equilibrium, both forward and reverse reactions are occurring (Proceeding at same rate) A System in Dynamic Equilibrium 1. Products are being made, but reactants are being made at the same rate 2. Total amount of reactant and product doesn't change 3. Neither reactant nor product can escape the system 4. There is a ratio of concentration terms that is constant

PCl5 (g) ⇌ PCl3 (g)+ Cl2 (g) A) If Keq=0.497, will there be more PCL5 or PCL3 at equilibrium? B) If a 2L flask contains 1.72 moles of PCl5 and .7 moles of PCl3 at eq, what is concentration of Cl2 at equilibrium C) What will be the result of adding PCl3 to the reaction? D) What is K^1eq for the reverse reaction?

-Reactant favored -Equilibrium lies to left because Keq<1 Concentration of Cl2= 1.22 M .497= (x*.350)/(.860) C) It will shift to the left. Keq will not change because equilibrium will be re-established (graph will go straight up, then level off at level higher than initial) D) 1/.497= 2.01

Endothermic Reaction

-Seperation of solvent/solute molecules -DeltaH is positive

Catalyst

-Something at the beginning and at the end -Speeds up the reaction without being consumed in the process -Works by lowering the overall Ea for the reaction/Decreases activation energy -Both forward and backward rate constants are increased -Keq stays the same -Rate at which equilibrium is reached is increased -Does not effect enthalpy change -May provide new reaction pathway -Brings reactants physically closer together -Weakening existing chemical bonds -Twisting reactants into position from which reaction is more likely *Shortest Ea = fastest reaction

Calculating strongest conjugate base

-Strongest base=Smallest Ka value -Acidity increase going down a column -Electronegativity increase going up and right

Keq: Equilibrium Constant

-[products]/[reactants] -If Keq is small (less than 1), the equilibrium lies to the left (Reactants bigger than products) -If Keq is big (Keq>>1) equilibrium lies to the right (Reaction favors products/products predominate at equilibrium) -MUST BE IN GASEOUS OR AQUEOUS FORM -When multiplying a reaction by a factor X, K will be raised to the power of X (If you multiply reaction of K1 by 2, K2=K1^2) -At a given temperature, Keq depends only on the stoichiometry

Is the solution acidic, basic or neutral? A) NH4Br B) Ca(NO2)2 C) Ca(HCO3)2 D) CaBr E) NH4NO2 F) KCl G) LiBr H) NaOCl I) CaSO3 J) NH4Cl

A) Acidic (we know Br is an acid, and we know NH4 is weak base) B) Basic (We know Ca is strong base because its a metal on the far left, and we know the NO2 is not a strong acid, making it weak) C) Basic (Strong base with weak acid) D) Neutral E) Neutral (Weak base and weak acid) F) Neutral (Strong acid and base) G) Neutral (Strong acid and base) H) Basic (Na=strong base, OCl=weak acid) I) Basic J) Acid )NH4=weak base, Cl=strong acid) *STRONGER ACID = WEAKER CONJUGATE BASE *MORE HYDROGENS = MORE ACIDIC

Rank the following compounds in order of decreasing acid strength using periodic trends: HCl, HBr, BH3, H2S H2SO3, H2SO4, H2S What is the strongest acid? C) H2SO4, H3NO3, H2NO3 D) HBrO2, HIO2, HBrO3, HIO3

A) HBr, HCl, H2S, BH3 Acid strength increases as you move down a group. Acid strength increases from left to right across a period. B) H2SO4, H2SO3, H2S C) H2SO4 (More Oxygens often means more acidity) D) HBrO3

How will the following equilibria respond to an increased pressure caused by decrease in volume? A)3NO = N2O + NO2 B)N2+O2 = 2NO C) 2Cl2 + 2H2O = 4Cl + O2

A) Reaction will shift to the right (Towards products) -A decrease in volume = increase pressure, increase in concentration of molecules -System will respond by shifting to side with fewer molecules (To the right because 4>2) B) No influence (2=2) C) Reaction will shift left (towards reactants) *Adding to products shifts reaction to left *What would happen to these equilibria if the pressure was increased by the addition of an inert gas? nothing b/c "concentrations" don't change

The equilibrium constant for the reaction: 2NO(g)+Br2(g)⇌2NOBr(g) is Kc=1.3×10−2 at 1000 K At this temperature does the equilibrium favor NO and Br2, or does it favor NOBr? Calculate Kc for 2NOBr(g)⇌2NO(g)+Br2(g) Calculate Kc for NOBr(g)⇌NO(g)+1/2Br2(g)

A) The equilibrium favors NO and Br2 B) 77 (1/(1.3*10^-2) C) 8.8 (Sqr root of 77)

Calculate [OH−] for each of the following solutions, and indicate whether the solution is acidic, basic, or neutral. A) [H+]= 4.4×10−3 M B) [H+]= 1.1×10−9 M C) A solution in which [H+] is 1000 times greater than [OH−].

A) [OH−] = 2.3×10^−12 M Acidic B) [OH−] = 9.1×10−6 M Basic C) [OH−] = 3.2×10−9 M Acidic 1x10^-14 = [H+] [OH-] 1x10^-14 = 1000[H+] [OH-] 1x10^-14 = 1000x^2 1x10^-17 = x^2 3.16^-9 = x

Ephedrine, a central nervous system stimulant, is used in nasal sprays as a decongestant. This compound is a weak organic base: C10H15ON(aq)+H2O(l)←−→C10H15ONH+(aq)+OH−(aq) A 0.035 M solution of ephedrine has a pH of 11.33. A. What is the equilibrium concentration of OH− and C10H15ONH+? B. What is the equilibrium concentration of C10H15ON? C. Calculate Kb for ephedrine.

A. 2.14*10^-3 B. [C10H15ON] = 3.3×10−2 M C. Kb = 1.3×10−4 [OH^-]=10^-2.67= 2.14*10^-3 C10H15ON(aq)←−→C10H15ONH+(aq)+OH−(aq) I .035M 0M 0M C -2.14*10^-3 2.14*10^-3 2.14*10^-3 E 0.033M 2.14*10^-3 2.14*10^-3

Calculate the pH of each of the following strong acid solutions. A. 2.69×10−2 M HNO3 B. 0.590 g of HClO3 in 2.30 L of solution. C. 10.00 mL of 2.00 M HCl diluted to 0.530 L . D. A mixture formed by adding 59.0 mL of 2.5×10−2 M HCl to 120 mL of 1.5×10−2 M HI.

A. pH = 1.570 B. pH = 2.518 C. pH= 1.423 D. pH= 1.74 M1V1=M2V2

Which is the Lewis acid/Base CO2 + OH- = HCO3- 2NH3 + Ag+ = Ag(NH3)2+

CO2 = Lewis acid, OH- Lewis Base 2NH3 = Lewis base, Ag+ Lewis acid

Exothermic reaction: Effect of Increasing Temp and decreasing temp Endothermic reaction: Effect of Increasing Temp and decreasing temp

Exo: Increasing T shifts reaction left (More reactants, fewer products, lower Keq), decreasing shifts right (More products, fewer reactants, higher Keq) Endo: Increasing T shifts reaction right (More products, fewer reactants, higher Keq), decreasing shifts reaction left (More reactants, fewer products, lower Keq)

The following equation shows the equilibrium in an aqueous solution of ammonia: NH3(aq)+H2O(l)⇌NH4+(aq)+OH−(aq) Which of the following represents a conjugate acid-base pair?

H2O and OH− The simplest way to identify a conjugate acid-base pair is to find the species whose formulas differ only by one proton. -Acid donates a proton to form its conjugate base

What is the conjugate acid of HPO32− ?

H2PO3−

Arrange the compound in series of increasing acid strength: H2SO4, H2SeO3, H2SeO4

H2SeO3 < H2SeO4 < H2SO4 First, look at which has more oxygen: Oxyacids (Y-O-H): Increasing number of Oxygen atoms on Y increase acid strength The, look at electronegativity (If two are in same column, the one higher up has larger electronegativity) -Acid strength increase with increasing electronegativity of Y -In this example, Y is Se and S

Heterogeneous VS Homogeneous Reaction

Homogeneous reactions are chemical reactions in which the reactants and products are in the same phase, while heterogeneous reactions have reactants in two or more phases. Reactions that take place on the surface of a catalyst of a different phase are also heterogeneous. If the concentration of a substance remains constant, it is not included in the equilibrium expression. Pure solids (s), pure liquids (l), and solvents are not included in the equilibrium expression 3NO(g)⇌N2O(g)+NO2(g) (Homogeneous) CH4(g)+2H2S(g)⇌CS2(g)+4H2(g) (Homogeneous) Ni(CO)4(g)⇌Ni(s)+4CO(g) (Heterogeneous) HF(aq)⇌H+(aq)+F−(aq) (Homogeneous) 2Ag(s)+Zn2+(aq)⇌2Ag+(aq)+Zn(s) (Heterogeneous)

Does the acidity of each molecule increase or decrease as the electronegativity of the atom X increases?

Increase

How does the magnitude of Kp for the reaction 2HI(g)⇋H2(g)+I2(g) change if the equilibrium is written 6HI(g)⇋3H2(g)+3I2(g)?

It is cubed *The stoichiometry is tripled; the constant is the cubed!

If the volume is decreased, how will the reaction respond to relieve the stress? If the volume is increased, how will the reaction respond to relieve the stress?

It will consume reactants to make products. Rightward shift It will consume products to make reactants.

If you mix 1.2M N2 and 2.5 H2 to form .05 NH3, will the reaction form product or reactants as it proceeds toward equilibrium? Keq= 6*10^-5

Qeq= 1.33*10^-4 K>Q Reaction going towards products (to right)

What will be the result of an increase in pressure on the following reaction: 2CO2 (g)= 2CO (g) + O2 DeltaH = -514 KJ What will be the effect of temperature?

Reaction is exothermic, so heat is added to the right -Increase in pressure, moves to left -Decrease in pressure, moves to right -Temp: If you add temp, reaction becomes more endothermic, less exothermic, and shifts to the left -If you decrease heat, reaction becomes more negative, more exo, moves right -In endothermic reaction (absorbs heat), heat added to reactants, DeltaH>0, and solubility increases with T *Shift towards direction of excess heat

Le Chatelier's Principle: Temperature

Solid and Liquid Solutes: Endothermic (absorbs heat) dissolution processes NH4NO3 (s) + HEAT = NH4+ (aq) + NO3- (aq) DeltaH > 0 => solubility increases with T Exothermic (gives off heat) dissolution process: MgSO4 (s) = Mg2+ (aq) + SO42- (aq) + HEAT DeltaH < 0 => solubility decreases with T

Relating pKa and pKb

The degree to which a weak acid dissociates in solution is given by its acid-ionization constant, Ka. For the generic weak acid, HA, HA(aq)⇌A−(aq)+H+(aq) and the acid-ionization constant is given by Ka=[A−][H+]/[HA] Similarly, the degree to which a weak base reacts with H2O in solution is given by its base-ionization constant, Kb. For the generic weak base, B, B(aq)+H2O(l)⇌BH+(aq)+OH−(aq) and the base-ionization constant is given by Kb=[BH+][OH−]/[B] Another way to express acid strength is by using pKa=−logKa Another way to express base strength is by using pKb=−logKb

Hydrogen is used as a rocket fuel because it is very light and reacts explosively and completely with oxygen. For the combustion reaction 2H2(g)+O2(g)⇌2H2O(g) what is the likely magnitude of the equilibrium constant K?

The equilibrium constant is the ratio of product to reactants at equilibrium. Think about the relative amounts of products versus reactants in a reaction that goes essentially to completion and how this affects the size of K. K>10^3

For the reaction 2H2S(g)⇌2H2(g)+S2(g), K= 1.00×10−6 at 1000 K what can be said about this reaction at this temperature?

The equilibrium lies far to the left.

For the reaction C(s)+H2O(g)⇌CO(g)+H2(g), K= 2.44 at 1000 K what can be said about this reaction at this temperature?

The reaction contains significant amounts of products and reactants at equilibrium.

Ammonia can be produced via the chemical reaction N2(g)+3H2(g)⇌2NH3(g) During the production process, the production engineer determines the reaction quotient to be Q = 3.56×10^−4. If K = 6.02×10^−2, what can be said about the reaction?

The reaction is not at equilibrium and will proceed to the right. If Q is less than K, the ratio of products to reactants is smaller than at equilibrium. In this case, the reaction will proceed to the right to create more product and reach equilibrium. Conversely, if Q is greater than K, the ratio of products to reactants is larger than at equilibrium. In this case, the reaction will proceed to the left and revert some of the product back to reactants to reach equilibrium.

The reaction 2CH4(g)⇌C2H2(g)+3H2(g) has an equilibrium constant of K = 0.154. If 6.60 mol of CH4, 4.05 mol of C2H2, and 10.30 mol of H2 are added to a reaction vessel with a volume of 5.00 L , what net reaction will occur?

The reaction will proceed to the left to establish equilibrium. Qeq= 4.06

Understanding the magnitude of K

The value of the equilibrium constant, K, can give you valuable information about how far a reaction as written will go towards completion. In other words, the magnitude of K can tell you whether the reactants in a chemical equation will hardly react at all to form product, react somewhat, or react relatively completely to form product. Value of K Reaction favors Reaction lies to K<<<1 reactants left K ~ 1 neither reactants nor products center K>>>1 products right