M6 Exam

pdf

School

Portage Learning *

*We aren’t endorsed by this school

Course

121

Subject

Chemistry

Date

Jan 9, 2024

Type

pdf

Pages

9

Uploaded by ChefOwl3912

Report
Question 1 (6.1): (1): Identify each of the compounds below as ACID, BASE or SALT on the basis of their formula and explain your answer (a) HC6H502 (b) Be{(OH)2 (c) Mn(Cl02)2 Answer: (a) HC6H502 Acid Contains H + Polyatomic group (C6H502) (b) Be(OH)2 Base Contains Metal (Be) + OH Polyatomic group (c) Mn(Cl02)2 Salt Contains Metal (Mn) + Polyatomic group (ClO2) (2): Identify each of the compounds below as ACID, BASE or SALT on the basis of their formula and explain your answer (a) HMnOA4 (b) CaCr207 (c) LiOH Answer: (a) HMnO4 Acid Contains H + Polyatomic group (Mn0O4) (b) CaCr207 Base Contains Metal + Cr207 Polyatomic group (c) LiOH Base Contains Metal + OH Polyatomic group Question 2 (6.3): (1): For the Br@gnsted-Lowry acid base reactions shown below, list the stronger acid, stronger base, weaker acid and weaker base in the answer blanks provided: H2CO3 + HS- S HCO3- + H2S 1 Stronger acid, 2. Stronger base
3. Weaker acid 4. Weaker base Answer: 1. Stronger acid = H2CO3 2. Stronger base = HS 3. Weaker acid = HCO3 4. Weaker base = H2S (2): In the following reaction label each material as to whether it is a stronger acid, stronger base, weaker acid or weaker base and explain your answer. CO3A-2 + H20 = HCO3- + OH- Answer: CO37-2: weaker base - proton acceptor, causes lesser reaction H20: weaker acid - proton donor, causes lesser reaction HCO3-: stronger acid - proton donor, cause greater reaction OH-: stronger base - proton acceptor, causes greater reaction Question 3 (6.3): (1): For the compounds shown below, choose the strongest acid, explain why it is the strongest acid and write the formula of its conjugate base in the answer blank: 1. CH4 or NH3 or H20 strongest acid is: 2. Explanation: 3. Formula of conjugate base of the strongest acid: Answer: 1. H20 2. O has the higher electronegativity (of C, N or O) which makes the H- O bond most polar and most likely to form H+ 3. OH-
(2): For the compounds shown below, choose the strongest acid, explain why it is the strongest acid and write the formula of its conjugate base in the answer blank: HCIO4 or HBrO4 or HIO4 1. strongest acid is: 2. Explanation: 3. Formula of conjugate base of the strongest acid: Answer: 1. strongest acid is: HCIO4 2. Explanation: The electronegativity, of the halogen atoms, will be directly proportional with the power of the radicals/ ionic groups, (ClO4-, BrO4-, 104-), to attract electrons towards itself. 3. Formula of conjugate base of the strongest acid: HCIO4- (1): Show the calculation of the [H+] and pH of a .00198 solution of the strong acid H2S04- Answer: [H+] = 2 X [H20S4] = .00396 M pH =log (.00396) = 2.402 (2): Show the calculation of the pH of a 0.00275 M solution of the strong acid H2S04. Answer: [H+] = 2 x [H2S04] = 0.00550 M pH = - log (0.00550) = 2.260 Question 5 (6.6): (1):
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
In the titration of 15.0 mL of H2504 of unknown concentration, the phenolphthalein indicator present in the colorless solution turns pink when 26.4 mL of .130 M NaOH is added. Show the calculation of the molarity of the H2S04 Answer: (H2S04 is the acid and NaOH is the base) Ma X mLa / 1000 X Sa / Sb =Mb X mLb / 1000 Ma X 15.0 /1000 X 2/1 =.130 X 26.4 / 1000 M_H2054 = .114M (2): In the titration of 0.127 grams of AI{OH)3 in the presence of bromothymol blue indicator, the blue solution turns yellow after 22.3 mL of H2S04 of unknown concentration is added. Show the calculation of the molarity of the H2S04 solution? Answer: (H2S0O4 is the acid and Al{OH)3 is the base) Ma x mLa/1000xSa/Sb=gb/MWb Max22.3/1000x2/3=0.127 / 78.004 MH2504 =0.110 M Question 6 (6.7): (1): Determine the [OH-], pH and % ionization for 0.298 M ammonia (NH3), if Kb = 1.8 x 10-5. Answer: NH3 + H20 (lig) = NH4+ + OH- NH3 H20 NH4+ OH- original conc 298 M very large 0 0 change —-X 0 +X +X equilibrium conc | .298-x ignore X X Kb = 1.8 X 107-5 = [NH4+][OH-] / NH3 = (x) (x) / (.298 ) 1.8 X 107-5 = xA2 / (.298) XA2 = (1.8 X 107-5)(.298)
X=2.3 X 107-3 [OH-] = 2.3 X 10A-3 pOH = -log[OH-] = -log(2.3 X 10-3) = 2.64 pH =14 —pOH = 14 2.64 = 11.36 % ionization = [NH4+] / [NH3] X 100 = ((2.3 X 107-3) / (.298)) X 100 = .77% (2): Show the calculation of the [OH-], pH and % ionization for 0.118 M ammonia (NH3), if Kb = 1.8 x 10/-5. NH3 + H20 (lig) = NH4+ + OH- Answer: NH3 H20 NH4+ OH- original conc 118 M very large 0 0 change —X 0 +X +X equilibrium conc | .118-x ignore X X Kb = 1.8 X 107-5 = [NH4+][OH-] / NH3 = (x) (x) / (.118 - X) 1.8 X107-5=(x) (x) / (.118 —x) XA2= (1.8 X 107-5)(.118) X =1.46 X 10"-3 = [OH-] or [NH4+] [OH-] =1.46 X 107-3 pOH = -log[OH-] = -log(1.46 X 107-3) = 2.84 pH=14-pOH=14-2.84=11.16 % ionization = [NH4+] / [NH3] X 100 = ((1.46 X 10-3) / (.118)) X 100 = 1.24% Question 7 (6.8): (1): Predict and explain whether a solution of AIBr3 is acidic, basic or neutral. Answer: AIBr3 is acidic since Al3+ hydrolyzes to form a weak base giving us AI(OH)3 and H30+
(2): Predict whether solutions of the following salts are acidic, basic or neutral and explain your answer: ZnSO4 NaCN Answer: ZnS04: Acidic since Zn+2 hydrolyzes to form a weak base [Zn(OH)2] and H30+ NaCN: Basic since CN- hydrolyzes to form a weak acid (HCN) and OH- Question 8 (6.9): (1): What is the pH of a buffer prepared by mixing 250 mL of .115 M NH4Cl and 250 mL of .200 M of NH3? Answer: 250 mL of .115 M NHA4CI contains: .115 mole / L X .250 L + .02875 mole of NH4+ ion 250 mL of .200 M NH3 contains: .200 mole / L X .250 L =.0500 mole of NH3 When the two .250 mL solutions are combined the new volume becomes 500 mL so we have: .02875 mole of NH4+/.500 L = .0575 M NH4+ .0500 mole of NH3/.500 L =.100 M NH3 NH3 + H20 (lig) = NH4+ + OH- NH3 H20 NH4+ OH- Original Conc A0M Very large 0575 M 0 Change -X 0 +X +X Equilibrium Conc | .100 -x lgnore .0575 +x X Kb = 1.8 X 10A-5 = ([NH4+][OH-])/[NH3] = ((.0575 + x)(x))/(.100- x) 1.8 X 107-5 = ((.0575 + x)(x))/(.100- x) 1.8 X 107-5 = (.0575)(x)/(.100) x=3.13 X 107-5 = [OH-] OH-=3.13 X 107-5 POH = -log[OH-] = -log(3.13 X 10A-5)=4.50 pH 14 pOH = 14— 4.50 = 9.50
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
(2): What is the pH of 1.00 L of a buffer containing 0.110 M NH4Cl and 0.110 M NH3? NH3 + H20 (lig) & NH4++ OH- Kb=1.8X10/-5 Answer: NH3 H20 NH4+ OH- Original Conc 110 M Very large 110 M 0 Change -X +X +X +X Equilibrium Conc | .110 -x lgnore 110 +x X Kb = 1.8 X 10A-5 = ([NH4+][OH-])/[NH3] = ((.110 + x)(x))/(.110- x) 1.8 X 107-5 = ((.110 + x)(x))/(.110- x) 1.8 X 107-5 = (.110)(x)/(.110) OH- = 1.8 X 10A-5 POH = -log[OH-] = -log(1.8 X 10/-5)=4.74 pH—14—-pOH=14-4.74=9.26 Question 9 (6.9): (1): (1) What is the pH of 1.00 L of a buffer containing 0.110 M NH4Cl and 0.110 M NH3? (2) What will the pH be upon addition of 0.0200 mole HCl to 1.00 L of the buffer? (3) What will the pH be upon addition of 0.0200 mole NaOH to 1.00 L of the buffer? Show your work. NH3 + H20 (lig) = NH4++OH- Kb =1.8X 107-5 Answer: (1) NH3 H20 NH4+ OH- Original Conc 110 M Very large 110 M 0 Change -X +X +X +X Equilibrium Conc | .110 -x lgnore 110 +x X Kb = 1.8 X 102-5 = ([NH4+][OH-])/[NH3] = ((.110 + x)(x))/(.110- x) 1.8 X 107-5 = ((.110 + x)(x))/(.110- x)
1.8 X 107-5 = (.110)(x)/(.110) OH-=1.8 X107-5 pOH = -log[OH-] = -log(1.8 X 107-5)=4.74 pH—14 pOH = 14— 4.74=9.26 (2) NH3 H20 NH4+ OH- Original Conc 110 M Very large 1100 0 +.02 mole H+ .090 M 0 130 Change -X 0 +X +X Equilibrium Conc | .090 - x lgnore 130 + x X Adding OH- converts NH4+ to NH3 Kb = 1.8 X 10/7-5 = ([NH4+][OH-])/[NH3] = ((.130 + x)(x))/(0.090 - x) 1.8 X 107-5 = (.130)(x)/(.090 - x) 1.8 X 107-5 = (.130)(x)/(.090) OH-=1.25 X 107-5 pOH = -log[OH-] = -log(1.25 X 107-5)=4.90 pH—-14-pOH=14-4.90=9.10 (pH went down slightly when H+ was added) (3) NH3 H20 NH4+ OH- Original Conc 110 M Very large 1100 0 +.02 mole H+ 130 M 0 .090 Change -X 0 +X +X Equilibrium Conc | .130-x lgnore .090 + x X Adding OH- converts NH3 to NH4+ Kb = 1.8 X 10A-5 = ([NH4+][OH-])/[NH3] = ((.090 +x)(x))/(0.130 - x) 1.8 X 10A-5 = (.090)(x)/(.130 - x) 1.8 X 10A-5 = (.090)(x)/(.130) X = 1.8 X 10A-5 = (.090)(.130)/(.090) X=2.60 X 10A-5 OH-=2.60 X 107-5
pOH = -log[OH-] = -log(2.60 X 107-5) = 4.59 pH—14 —pOH = 14— 4.59 =9.41 (pH went up slightly when OH- was added) Question 10 (6.10): List and explain what type of acid and base are involved in the titration curve below and select an appropriate indicator to be used in this titration from the list below. Indicator Acid Color pH range of color Base color change Methyl violet Yellow 0.0-1.6 Blue Litmus Red 5.0-8.0 Blue Phenolphthalein Colorless 8.0-9.8 Pink Alizarin yellow R Yellow 10.1-12.0 Red " " 19 19 12 12 " 1" 10 10 9 ® u & s y. o [ [ st s 4 3 3 2 2 1 1 0 o 4] 10 20 » w0 50 0 Answer: The location of the equivalence point at pH < 7, indicates that this is a strong acid / weak base titration. The rapid drop occurs approximately between pH 8 and pH 3. The indicator chosen to signal this change should change color within that range indicating that we are limited in choice of indicator to Litmus but Methyl violet, Phenolphthalein and Alizarin Yellow R are unable to be used for this titration.
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help