CHEMISTRY 104 – Practice Problems #3
Chapters 16 and 17
http://www.chem.wisc.edu/areas/clc (Resource page)
Prepared by Dr. Tony Jacob
Suggestions on preparing for a chemistry exam:
1. Organize your materials (quizzes, notes, etc.) and find out exactly what material/chapters are on the exam.
2. Usually, a good method to prepare for a chem exam is by doing lots of problems. Re-reading sections of a chapter is fine,
but re-reading entire chapters takes up large amounts of time that is generally better spent doing problems.
3. Old exams and quizzes posted by your instructor should be completely worked though. Old exams give you a sense of how
long the exam will be, the difficulty of the problems, the variability of the problems, and the style of your instructor.
Good Luck!
CHAPTER 16 – ACID AND BASE EQUILIBRIA
1. Complete the short answer questions.
a. What is the conjugate base of HCO3-? __________
b. What is the conjugate acid of HCOO-? __________
c. What is the conjugate acid of NH3? __________
d. Identify the acid, base, conjugate acid, and conjugate base in the reaction.
HNO2(aq) + NaF(aq)
NaNO2(aq) + HF(aq)
2. Answer the short multiple choice questions.
I. Which chemical does not act as a Lewis acid?
a. Fe(ClO4)3
b. H+
c. Al+3
d. Zn+2
e. PH3
II. Which chemical listed below is not amphiprotic?
a. HCO3b. HPO4-2
c. HS-
d. H2PO4-
e. all are
III. Which chemical when dissolved in water will produce a basic solution?
a.
b.
c.
d.
e.
3. Which statement is incorrect? If all of the statements are correct select answer “e”.
a. Pure water can undergo autoionization and will produce equal amounts of H+ and OH-.
b. The value of Kw is 1.0 x 10-14 at 25˚C but changes with temperature.
c. In pure water with a Kw = 4.0 x 10-14, the pH = 6.70.
d. As the acidity of water changes, the concentrations of H+ and OH- will change but their product will always
equal Kw.
e. All statements are true.
4. What is the pH of a 0.0025M Ba(OH)2 solution?
a. 11.70
b. 2.30
c. 2.60
d. 11.40
e. 0.005
5. 1.25grams HNO3 is dissolved into 250ml water. What is the pH of the resulting solution?
a. 1.10
b. 0.70
c. 0.079
d. 1.70
e. 12.90
(nap time)
6. Which of the following reactions is the correct reaction for the Kb of (CH3)2NH?
a. (CH3)2NH(aq) + H2O(l)
b. (CH ) NH(aq) + O-2(aq)
(CH3)2N-(aq) + H3O+(aq)
(CH ) N-(aq) + OH-(aq)
c. (CH3)2NH(aq) + H2O(l)
(CH3)2NH2+(aq) + OH-(aq)
3 2
3 2
d. (CH3)2NH(aq) + OH-(aq)
e. (CH ) NH +(aq) + OH-(aq)
3 2
2
(CH3)2N-(aq) + H2O(l)
(CH3)2NH(aq) + H2O(l)
7. Which of the following reactions is the correct reaction for the Ka of HCOOH?
a. HCOOH(aq) + H O(l)
COOH-(aq) + H O+(aq)
2
b. HCOOH(aq) +
OH-(aq)
c. HCOOH(aq) + H2O(l)
d. HCOOH(aq) + H2O(l)
e. HCOO-(aq) + H O+(aq)
3
3
HCOO-(aq) + H2O(l)
HCOOH +(aq) + OH-(aq)
2
HCOO-(aq) + H3O+(aq)
HCOOH(aq) + H2O(l)
8. The Kb for 2,6-dimethlyaniline (C6H3(CH3)2NH2) is 7.8 x 10-11.
a. What is the pKb for 2,6-dimethlyaniline?
b. Draw the conjugate acid for 2,6-dimethylaniline showing all bonds and atoms.
c. What is the Ka for the conjugate acid of 2,6-dimethylaniline?
d. What is the pKa for the conjugate acid of 2,6-dimethylaniline?
9. Use the following Ka’s to answer the questions below assuming each acid is prepared as a 1.0M solution.
Note: As Ka increases the strength of the acid increases.
HOCN
CCl3COOH
3.5 x 10-4
2.2 x 10-1
CH2(OH)COOH
1.5 x 10-4
HNO2
4.5 x 10-4
I. Which solution is the strongest acid?
a. HOCN
b. CCl3COOH
c. CH2(OH)COOH
d. HNO2
II. Which solution would have the highest pH?
a. HOCN
b. CCl3COOH
c. CH2(OH)COOH
d. HNO2
III. Which solution will have the weakest conjugate base.
a. HOCN
b. CCl3COOH
c. CH2(OH)COOH
d. HNO2
10. Order the chemicals from lowest to highest pH: HClO4, NaCN, NH3, C6H5COOH, KNO3, LiOH, HF, NaF
(Ka HCN = 4.9 x 10-10; Kb NH3 = 1.8 x 10-5; Ka C6H5COOH = 6.3 x 10-5; Ka HF = 6.8 x 10-4)
_________
lowest pH
_________
_________
_________
_________
_________
_________
_________
highest pH
11. When 50.0ml 0.15M NaOH is combined with 100.ml 0.070M KOH what is the pH of the resulting solution?
a. 0.22
b. 13.83
c. 13.34
d. 12.99
e. 9.99
(coffee time)
12. a. Will Na2O(s) create a basic or acidic solution when dissolved into water?
b. Write the reaction for CaO(s) reacting with water.
c. Nonmetal oxides will form __________ (acidic or basic; choose one) solutions when dissolved in water.
d. Complete and balance the reaction: H2O(l) + P2O5(s)
e. Complete and balance the reaction: H2O(l) + SO3(g)
13. I. a. What is the pH of a 0.10M hydrofluoric acid, HF, solution? b. What is the %ionization for this
solution? (Ka HF = 6.8 x 10-4)
II. a. What is the pH of a 0.20M aniline, C6H5NH2, solution? b. What is the %ionization for this solution?
(Kb C6H5NH2 = 4.3 x 10-10)
14. I. A 0.10M acrylic acid, CH2CHCOOH, solution has a pH of 2.63. What is the Ka of acrylic acid?
a. 2.3 x 10-3
c. 2.4 x 10-3
b. 2.63
d. 5.6 x 10-5
e. 0.10
II. If a 0.015M solution of a weak acid, HA, has a percent ionization of 1.2%, what is the Ka for this acid?
a. 4.8 x 10-2
b. 0.015
c. 1.82
d. 1.8 x 10-4
e. 2.2 x 10-6
15. Which reaction corresponds to the Kb2 of arsenic acid, H3AsO4?
a. H2AsO4-(aq) + H2O(l)
b. HAsO4-2(aq) + OH-(aq)
c. AsO4-3(aq) + H2O(l)
d. H2AsO4-(aq) + H2O(l)
e. HAsO4
-2(aq)
+ H2O(l)
HAsO4-2(aq) + H3O+(aq)
AsO4-3(aq) + H2O(l)
HAsO4-2(aq) + OH-(aq)
H AsO (aq) + OH-(aq)
3
4
H2AsO4-(aq)
+ OH-(aq)
16. What is the pH of 0.025M sodium acetate, NaCH3COO(aq), solution? (Ka CH3COOH = 1.8 x 10-5)
17. When each chemical is dissolved in water will it yield an acidic, basic, or neutral solution?
a. KCl b. H2C2O4 c. NH4NO3 d. NaClO4 e. N(CH3)3 f. Li2S g. CH3COONa
18. Which chemical species will have the strongest conjugate base?
a. HBrO2
b. HClO2
c. HClO4
d. HIO
e. HBrO
19. Given the K values, determine for each reaction below whether it will be reactant- or product-favored?
Ka (HC3H5O3) = 1.4 x 10-4; Ka (HClO) = 3.0 x 10-8; Ka (HCN) = 4.9 x 10-10; Ka (HClO2) = 1.1 x 10-2;
Kb ((CH3)3N) = 6.4 x 10-5; Ka (HCOOH) = 1.8 x 10-4
I. LiOH(aq) + HC3H5O3(aq)
H2O(l) + LiC3H5O3(aq)
II. HCN(aq) + NaCl(aq)
HCl(aq) + NaCN(aq)
+
III. ClO2 (aq) + (CH3)3NH (aq)
HClO2(aq) + (CH3)3N(aq)
IV. HClO(aq) + HCOO-(aq)
(ice cream time!)
ClO-(aq) + HCOOH(aq)
20. Which of the following statements is incorrect? If all are correct, select choice “e”.
a. The greater the polarity in a binary acid, H–X, the stronger the acid.
b. The weaker the bond strength in a binary acid, H–X, the stronger the acid.
c. The greater the stability of the resulting anion in an oxyacid after H+ has dissociated, the stronger the acid.
d. The more oxygen atoms in an oxyacid, the stronger the acid.
e. All statements are true.
21. For each chemical when it is dissolved in water will it yield an acidic, basic, or neutral solution?
I. a. (CH3NH3)(ClO2) (Kb CH3NH2 = 4.4 x 10-4; Ka HClO2 = 1.1 x 10-2)
b. (NH4)(OCN)
II. a. NaHCO3(aq)
b. NaHC2O4(aq)
(Kb NH3 = 1.8 x 10-5; Ka HOCN = 3.5 x 10-4)
(Ka1 = 4.2 x 10-7; Ka2 = 4.8 x 10-11)
(Ka1 = 5.9 x 10-2; Ka2 = 6.4 x 10-5)
CHAPTER 17-BUFFERS
22. When KClO is added to a HClO solution the pH will
a. increase
b. decrease
c. stay the same
d. need to know [HClO] to determine this
23. Which of the following will not result in a buffer? (Assume 1L of each solution)
a. 0.2M HCl + 0.4M NH3
b. 0.2M NaF + 0.4 M HF
c. 0.2M CH3COOH + 0.1M NaOH
d. 0.2M HCl + 0.1M NaF
e. 0.2M CH3COOH + 0.3M CH3COONa
24.Which buffer would be the best choice for a solution to have a pH ≈ 9.5?
a. HOCN/NaOCN (Ka = 3.5 x 10-4)
b. C6H5NH3Cl/C6H5NH2 (Kb = 4.0 x 10-10)
c. HCOOH/HCOONa (Ka = 1.8 x 10-4)
d. HClO/NaClO (Ka = 3.5 x 10-8)
e. NH4Cl/NH3 (Kb = 1.8 x 10-5)
25. a. What is the pH of a 1.0L solution containing 0.50M acetic acid, HC2H3O2, and 0.40M sodium acetate,
NaC2H3O2? (Ka HC2H3O2 = 1.8 x 10-5)
b. To this solution, 7.3g of HCl is added. Write the net ionic reaction that occurs. What is the pH of the new
solution?
26. a. What is the pH of a 100.ml solution containing 0.20M KOCN and 0.10M HOCN, cyanic acid?
(Ka HOCN = 3.5 x 10-4)
b. To this solution, 0.20g of NaOH is added. Write the net ionic reaction that occurs. What is the pH of the
new solution?
27. A solution is prepared by dissolving 22.5 grams NaHCO3 (molar mass = 84g/mol) into 1.5L 0.25M H2CO3.
What is the pH of the buffer prepared? (H2CO3: Ka1 = 4.2 x 10-7; Ka2 = 4.8 x 10-11)
(eat some pizza)
28. A buffer with a pH of 4.25 was prepared from lactic acid, HC3H5O3, and sodium lactate, NaC3H5O3. If
[NaC3H5O3] = 0.25M, what concentration of HC3H5O3 is needed for the desired pH? (Ka HC3H5O3 = 1.4 x 10-4)
29. A 1.0L buffer is to be prepared from acetic acid (HC2H3O2) and sodium acetate (NaC2H3O2) with
pH = 5.10. If a 1.0L solution of 0.50M HC2H3O2 solution is prepared, how many grams of NaC2H3O2 should
be added to create the desired buffer? (Ka HC2H3O2 = 1.8 x 10-5)
CHAPTER 17-TITRATIONS
30. I. What is the pH at the equivalence point for the titration of benzoic acid, C6H5COOH, with NaOH?
a. less than 7
b. equal to 7
c. greater than 7
d. depends on Ka
II. What is the pH at the equivalence point for the titration of trimethylamine, (CH3)3N, with HCl?
a. less than 7
b. equal to 7
c. greater than 7
d. depends on Kb
31. For the titration of 0.10M CH3COOH with 2.0M NaOH it required 30.0ml of NaOH to reach the
equivalence point. What is the pH after 15.0ml of NaOH was added? (Ka CH3COOH = 1.8 x 10-5)
a. 9.0 x 10-6
b. 14.3
c. 5.05
d. 4.74
e. 2.87
32. During a titration, the following net ionic reactions occurred. For each reaction determine the Keq.
a. HF(aq) + OH-(aq) → H2O(l) + F-(aq)
Ka HF = 6.8 x 10-4
b. CH3NH2(aq) + H+(aq) → CH3NH3+(aq)
Kb CH3NH2 = 4.4 x 10-4
33. On the titration curve pictured identify the following
locations:
a. The place where only a weak acid is present. (label it A)
b. The buffer region of the titration curve. (label it B)
c. The equivalence point. (label it C)
d. The halfway to the equivalence point. (label it D)
e. Where the person doing the titration has overshot the
endpoint. (label it E)
f. The place where [weak acid] = [weak base]. (label it F)
g. The place where only a weak base is present. (label it G)
h. The region where the strong base dominates. (label it H)
34. How many milliliters of 0.750M KOH is required to reach the equivalence point during a titration of
100.0ml of a 0.100M HBr solution?
a. 13.3 ml
b. 75.0 ml
c. 86.7 ml
d. 100. ml
e. 113 ml
(take a walk)
35. I. Using the titration curve to the right, this is a titration curve for:
a. a weak base titrated with a strong acid
b. a weak acid titrated with a strong base
c. a strong base titrated with a strong acid
d. a strong acid titrated with a strong base
e. a polyprotic acid titrated with a strong base
II. The chemical being titrated is:
a. NH3 (Kb = 1.8 x 10-5)
b. HClO3
d. HC2H3O2 (Ka = 1.8 x 10-5)
e. HCN (Ka = 4.0 x 10-10)
c. H2CO3 (Ka1 = 4.5 x 10-7)
III. Which indicator would be a good choice for this titration?
a. methyl orange (Kind = 2.0 x 10-4)
b. bromophenol blue (Kind = 1.0 x 10-4)
c. methyl red (Kind = 7.9 x 10-6)
d. bromothymol blue (Kind = 1.0 x 10-7)
e. phenolphthalein (Kind = 5.0 x 10-10)
36. When does the pKa equal the pH of a weak acid solution, HA, titrated by a strong base?
a. pH = 1
b. Ka = pH
c. [H+] = [A-]
d. [HA] = [A-]
e. none of these
37. a. An indicator, HIn, has a Kind = 1.0 x 10-3, and is yellow in the HIn form and green in the In- form. What
color would a solution containing this indicator be at pH = 1.70?
b. During the titration of HCOONa with HCl, the pH at the equivalence point was 2.39. Would the indicator
from part “a” be a wise choice for this titration? Explain.
38. A 100. ml solution of 0.50M HNO2 is titrated with 2.0M NaOH. (Ka HNO2 = 4.5 x 10-4)
a. What volume of NaOH is needed to reach the equivalence point?
b. What is the pH of the solution initially before the titration is started?
c. What is the pH of the solution at the halfway to the equivalence point?
d. What is the pH of the solution at the equivalence point?
e. What is the pH of the solution if the volume of NaOH added was 1.0ml passed the equivalence point (i.e., the
volume of NaOH added is the amount to reach the equivalence point + 1.0ml more)? (Note: not all instructors have
students do calculations passed the equivalence point)
39. A solution of ethylamine, (CH2CH3)NH2, is titrated with 0.75M HCl. (Kb (CH2CH3)NH2 = 4.3 x 10-4)
a. If it required 30. ml of HCl to reach the equivalence point when 90.ml of (CH2CH3)NH2 was titrated, what
was the original concentration of the ethylamine?
b. What is the pH of the solution initially before the titration is started?
c. What is the pH of the solution after a total of 5.0ml of HCl is added? (Note: not all instructors have students do
calculations in the middle of a titration)
d. What is the pH of the solution at the halfway to the equivalence point?
e. What is the pH of the solution at the equivalence point?
f. What is the pH of the solution after a total of 30.50ml of HCl is added? (Note: not all instructors have student do
calculations passed the equivalence point)
(almost done)
CHAPTER 17-SOLUBILITY PRODUCT – Ksp (not all instructors do Ksp, Ch. 17.4-17.6)
40. a. What is the molar solubility of MgF2 in water? (Ksp MgF2(s) = 5.2 x10-11)
b. What is the solubility (g/L) of MgF2 in water?
41. I. The concentration of Sr2+(aq) in a saturated solution of SrF2 is 5.8 x 10-4 M. What is the Ksp for SrF2(s)?
II. 4.28 x 10-12 grams/L of PbS (MW = 239.3g/mol) will dissolve in water. What is the Ksp for PbS(s)?
42. I. In the set of 4 chemicals which is most soluble in water?
a. CuI (Ksp = 1.3 x 10-12) b. CuCl (Ksp = 1.7 x 10-7) c. CuCN (Ksp = 3.5 x 10-20) d. CuBr (Ksp = 6.8 x 10-9)
II. In the set of 3 chemicals which is most soluble in water?
a. AgI (Ksp = 8.5 x 10-17) b. AgCl (Ksp = 1.8 x 10-10) c. Ag2C2O4 (Ksp = 5.4 x 10-12)
43. In which solution will FeCO3(s) be most soluble?
a. 1M Na2CO3
b. 0.1M Na2CO3
c. 0.01M Fe(NO3)2
d. 0.001M Fe(NO3)2
44. Which of the following solids is more soluble in acid solutions than in pure water?
a. PbF2
b. PbCl2
c. PbBr2
d. AgCl
e. AgI
45. I. What is the molar solubility of PbI2(s) when dissolved in 0.020 M Pb(NO3)2? (Ksp PbI2 = 8.7 x 10-9)
II. What is the molar solubility of LaF3 if it is dissolved in a 0.0050 M NaF? (Ksp LaF3 = 2 x10-19)
46. I. Will a precipitate of Ag2CO3(s) form if a solution contains 1.0 x 10-5 M Na2CO3(aq) and 1.0 x 10-4 M
AgNO3(aq)? (Ksp Ag2CO3 = 8.1 x 10-12)
II. Will a precipitate of MgF2(s) form if 100. ml 1.0 x 10-3M Mg(NO3)2(aq) is mixed with 300. ml 5.0 x 10-4 M
NaF(aq)? (Ksp MgF2 = 3.7 x 10-8)
47. I. If a solution is 1.0 x 10-2 M Ag+ and Na2SO4 is added, what concentration of [SO4-2] is necessary to just
precipitate Ag2SO4(s)? (Ksp Ag2SO4 = 1.2 x 10-5)
II. What molar concentration of PO4-3 is needed to just precipitate Ag3PO4(s) from a 0.010M Ag+ solution?
(Ksp Ag3PO4 = 1.3 x 10-20)
48. If a solution contains 1 x 10-5M Cu+2, Pb+2, and Mn+2, which solid will precipitate first when Na2CO3 is
added to the solution? (Ksp CuCO3 = 2.3 x 10-10, Ksp PbCO3 = 7.4 x 10-14, Ksp MnCO3 = 2.2 x 10-11)
a. CuCO3
b. PbCO3
c. MnCO3
d. Na2CO3
e. all ppt at the same time
(yea done!)
ANSWERS
1. a. CO3-2 b. HCOOH c. NH4+
d. acid: HNO2, base: NaF (or F-); conjugate acid: HF, conjugate base: NaNO2 (or NO2-)
2. I. e {metal cations can accept electron pairs - “a”, “c”, and “d”; H+ is an acid - “b”; PH3 is isovalent with NH3 and has a lone
pair of electrons it can donate – Lewis base}
e {each chemical can release a H+ and each can accept a H+}
II.
III. b
3. e
4. a
5. a
6.
7.
{“b”–amine → basic; “a”–strong acid; “c”–nonmetal oxide → acidic soln; “d”–carboxylic acid; “e”–N with 4 groups → acidic}
{“b” – K(T); “c” – 4.0 x 10-14 = [H+][OH-] = x2; x = 2.0 x 10-7 = [H+]; pH = -log(2.0 x 10-7) = 6.70}
{Ba(OH)2 → Ba+2 + 2OH-; [OH-] = 2(0.0025) = 0.005M; pOH = -log(0.005) = 2.30; pH = 11.70}
{1.25g HNO3 x (1mol HNO3/63.0g HNO3) = 0.01984mol HNO3; [HNO3] = 0.01984mol/0.250L = 0.07937M;
+
[H ] = 0.07937M; pH = -log(0.07937) = 1.100}
c {Kb reactions are written as B + H2O
HB+ + OH- with the base is a reactant and accepts a H+ from water}
d {Ka reactions are always written as HA + H2O
A- + H3O+ where the acid of interest is a reactant and gives up a H+
to the base that is present; in reaction “a”, the wrong H is removed as the acidic H is the H of a carboxylic acid –COOH}
a. pKb = 10.11 {pKb = -log(7.8 x 10-11) = 10.11}
8.
b. see drawing on right
c. Ka = 1.3 x 10-4 {Ka = 1.0 x 10-14/7.8 x 10-11 = 1.28 x 10-4}
d. pKa = 3.89 {pKa = -log(1.28 x 10-4) = 3.892}
9. I. b {larger Ka → stronger acid}
II. c {highest pH means most basic or weakest acid; the weakest acid has the smallest Ka}
III. b {the strongest acid has the weakest conjugate base; the strongest acid has the largest Ka}
10. HClO4 < HF < C6H5COOH < KNO3 < NaF < NH3 < NaCN < LiOH
11.
12.
{start by grouping the chemicals into categories and in this order: SA < WA < neutral salts < WB < SB;
HClO4 and LiOH are SA and SB, respectively and go to the ends; KNO3 is a neutral salt (pH = 7) and goes in the middle; larger
Ka → stronger acid → lower pH: C6H5COOH Ka = 6.3 x 10-5 versus HF Ka = 6.8 x 10-4 so HF is a stronger acid; larger Kb →
stronger base → higher pH: NH3 Kb = 1.8 x 10-5 versus CN- Kb = Kw/Ka = 1 x 10-14/4.9 x 10-10 = 2.0 x 10-5 versus FKb = Kw/Ka = 1 x 10-14/4.8 x 10-4 = 2.1 x 10-11; so CN- is a stronger base than NH3 which is stronger than F-}
d {find [OH-] = mol OH-/Lsoln; Lsoln = (100+50)/1000 = 0.150L; mol NaOH = M x L = (0.15M)(0.050L) =
0.0075mol NaOH = 0.0075mol OH-; mol KOH = M x L = (0.070M)(0.100L) = 0.0070mol KOH = 0.0070mol OH-;
total mol OH- = 0.0075 + 0.0070 = 0.0145mol OH-; [OH-] = 0.0145/0.150 = 0.09667M; pOH = -log(0.09667) = 1.015;
pH = 12.985}
a. basic solution {metal oxides form basic solutions when dissolved in water}
b. CaO(s) + H2O(l)
Ca(OH)2
c. acidic
d. 3H2O(l) + P2O5(s)
2H3PO4(aq)
e. H2O(l) + SO3(g) → H2SO4(aq)
H+ + F-; do ICE Table: I row: 0.1, 0, 0; C row: -x, +x, +x; E row: 0.1-x, x, x;
Ka = [H+][F-]/[HF] → 6.8 x 10-4 = (x)(x)/(0.1-x); ok to make approximation? (6.8 x 10-4)100 < 0.1 – yes, make approximation:
6.8 x 10-4 = (x)(x)/(0.1) → x = 8.25 x 10-3 [H+]; pH = 2.08}
13. I. a. pH = 2.07
{HF
{%ionization = x/[HA]0 x 100% = (8.25 x 10-3/0.1) x 100% = 8.25%}
a. pH = 8.97 {C6H5NH2 + H2O
C6H5NH3+ + OH-; do ICE Table: I row: 0.2, ---, 0, 0; C row: -x, -x, +x, +x;
E row: 0.2-x, ---, x, x; Kb = [C6H5NH3+][OH-]/[C6H5NH2] → 4.3 x 10-10 = (x)(x)/(0.2-x); ok to make approximation?
(4.3 x 10-10)100 < 0.2 – yes, make approximation: 4.3 x 10-10 = x2/(0.2) → x = 9.27 x 10-6 = [OH-]; pOH = 5.03; pH = 8.97}
b. 8.3%
II.
b. 0.0046%
{%ionization = x/[B]0 x 100% = (9.27 x 10-6/0.2) x 100% = 4.6 x 10-3% = 0.0046%}
H+ + CH2CHCOO-; do ICE Table: I row: 0.1, 0, 0; C row: -x, +x, +x; E row: 0.1-x, x, x;
[H+] = 10-2.63 = 2.344 x 10-3 = x; Ka = [H+][CH2CHCOO-]/[CH2CHCOOH] = x2/(0.1-x);
Ka = (2.344 x 10-3)2/(0.10-2.344 x 10-3) = 5.626 x 10-5}
II. e {HA
H+ + A-; do ICE Table: I row: 0.015, 0, 0; C row: -x, +x, +x; E row: 0.015-x, x, x; Ka = [H+][A-]/[HA];
Ka = (x)(x)/(0.015-x); Ka = x2/(0.015-x); need “x”; %ion = [(amt dissociates)/[HA]o x 100%]; amt that dissociates = x in an ICE
Table; 1.2 = (x/0.015)100%; x = 0.00018; Ka = (0.00018)2/(0.015-0.00018) = 2.19 x 10-6}
15. e {Kb reactions are written as B + H2O
HB+ + OH- with the base as a reactant reacting with water and accepting a H+
and generating OH-; Kb1: AsO4-3 + H2O(l)
HAsO4-2(aq) + OH-(aq);
Kb2: HAsO4-2 + H2O(l)
H2AsO4-(aq) + OH-(aq); Kb3: H2AsO4- + H2O(l)
H3AsO4(aq) + OH-(aq); }
16. 8.57 {CH3COO- + H2O
CH3COOH + OH-; do ICE Table: I row: 0.025, ---, 0, 0; C row: -x, -x, +x, +x;
E row: 0.025-x, ---, x, x; Kb = [OH-][CH3COOH]/[CH3COO-]; Kb = 1.0 x 10-14/1.8 x 10-5 = 5.56 x 10-10;
Kb = 5.56 x 10-10 = (x)(x)/(0.025-x); ok to make approximation? (5.56 x 10-10)100 < 0.025 – yes, make approximation:
5.56 x 10-10 = x2/(0.025); solve for x: x = 3.727 x 10-6 = [OH-]; pOH = 5.429 → pH = 8.571}
14. I. d {CH2CHCOOH
17. a. neutral {both K+ and Cl- are spectators}
b. acidic {starts with a H → acid}
c. acidic {NO3- is a spectator; NH4+ is N with 4 groups → acid}
d. neutral {both Na+ and ClO4- are spectators}
e. basic {N with 3 groups attached → basic}
f. basic {Li+ is a spectator; S-2 + H2O
HS- + OH- → basic}
g. basic {basic; CB of CH3COOH}
18. d {strongest conjugate base means find the weakest acid; HClO4 – strong acid; between HBrO2 and HClO2, for oxyacids as
ENX ↑ ⇒ stronger acid and since ENBr < ENCl HBrO2 will be the weaker acid; between HBrO2 and HBrO, for oxyacids as
#O ↑ ⇒ stronger acid and since HBrO has fewer O atoms it will be the weaker acid; between HBrO and HIO, for oxyacids as
ENX ↑ ⇒ stronger acid and since ENI < ENBr HIO will be the weakest acid and have the strongest conjugate base}
19. I. product favored {LiOH is strong base and reaction goes to weaker acids/weaker bases}
II. reactant favored {HCl is strong acid and reaction goes to weaker acids/weaker bases, in this case the reactant side}
III. reactant favored {goes to weaker acid/weaker base side; Ka HClO2 = 1.1 x 10-2; find Ka (CH3)3NH+ =
Kw/Kb = 1.0 x 10-14/6.4 x 10-5 = 1.6 x 10-10; (CH3)3NH+ is weaker acid since it has a smaller Ka and the reaction goes to the
weaker side, in this case the reactant side; can also do this by comparing the bases: Kb (CH3)3N = 6.4 x 10-5; find
Kb ClO2- = Kw/Ka = 1.0 x 10-14/1.1 x 10-2 = 9.09 x 10-13; ClO2- is weaker base since it has a smaller Kb and the reaction goes
to the weaker side, in this case the reactant side}
IV. reactant favored {goes to weaker acid/weaker base side; Ka HClO = 3.5 x 10-8; Ka HCOOH = 1.8 x 10-4; HClO is the
weaker acid since it has a smaller Ka and the reaction goes to the weaker side, in this case the reactant side; can also do this by
comparing the bases: find Kb HCOO- = Kw/Ka = 1.0 x 10-14/1.8 x 10-4 = 5.56 x 10-11; find
Kb ClO- = Kw/Ka = 1.0 x 10-14/3.5 x 10-8 = 2.86 x 10-7; HCOO- is weaker base since it has a smaller Kb and the reaction goes
to the weaker side, in this case the reactant side}
20. e {“a” – greater polarity occurs as ENX ↑ ⇒ stronger acid; “b” – weaker bond strength occurs as the size of X ↑ ⇒ stronger
acid; “c” – the anion of an oxyacid is more stabilized if #O ↑ ⇒ stronger acid; “d” – as #O ↑ the anion that forms after the H+
dissociates is more stabilized ⇒ stronger acid}
{[CH3NH3][ClO2] → CH3NH3+ + ClO2-; CH3NH3+ - acid; ClO2- - base; find K values and compare –
whichever is bigger wins! Kb CH3NH2 = 4.4 x 10-4 → Ka CH3NH3+ = (1.0 x 10-14)/(4.4 x 10-4) = 2.27 x 10-11;
Ka HClO2 = 1.1 x 10-2 → Kb ClO2- = (1.0 x 10-14)/(1.1 x 10-2) = 9.09 x 10-13; since Ka CH3NH3+ > Kb ClO2- →
the solution is acidic}
b. acidic {[NH4][OCN](aq) → NH4+(aq) + OCN-(aq); NH4+ - acid; OCN- - base; find K values and compare – whichever is
bigger wins! Ka HOCN = 3.5 x 10-4 → Kb OCN- = 1.0 x 10-14/3.5 x 10-4 = 2.86 x 10-11;
Kb NH3 = 1.8 x 10-5 → Ka NH4+ = 1.0 x 10-14/1.8 x 10-5 = 5.56 x 10-10; since Ka NH4+ > Kb OCN- → solution is acidic}
II. a. basic {HCO3- can undergo two reactions: HCO3H+ + CO3-2 with Ka2 = 4.8 x 10-11 and
HCO3- + H2O
H2CO3 + OH- with Kb2 = Kw/Ka1 = 1.0 x 10-14/4.2 x 10-7 = 2.4 x 10-8; whichever K is larger wins; Kb2
is larger → basic solution}
b. acidic {HC2O4- can do two reactions: HC2O4H+ + C2O4-2 with Ka2 = 6.4 x 10-5 and
HC2O4- + H2O
H2C2O4 + OH- with Kb2 = Kw/Ka1 = 1.0 x 10-14/5.9 x 10-2 = 1.7 x 10-13; whichever K is larger wins!
Ka2 is larger → acidic solution}
22. a {Can think of this in 1 of 2 ways: Adding KClO which is a conjugate base and whenever a base is added the pH will
rise. Can also consider the dissociation reaction: HClO
H+ + ClO- and then add ClO- from KClO. From Le Chatelier’s
21. I. a. acidic
Principle, this will drive the reaction to the left and decrease the amount of H+; as [H+] decreases → pH will increase.}
23. d {HCl will react with NaF: HCl + NaF → HF + NaCl; moles of each chemical: 0.20mol HCl; 0.1mol NaF; do SCF Table:
S row: 0.2, 0.1, 0, 0; C row: -0.1, -0.1, +0.1, +0.1; F row: 0.1, 0, 0.1, 0.1; after reaction is complete: HCl and HF exist: a SA and
WA which is not a buffer}
24. e {the pH buffer range is: pH = pKa ± 1; calculate pKa ± 1 for each buffer: a) pKa = 3.46, range: 2.46 ↔ 4.46;
b) pKa = 4.60, range: 3.60 ↔ 5.60; c) pKa = 3.74, range: 2.74 ↔ 4.74; d) pKa = 7.46, range: 6.46 ↔ 8.46;
e) pKa = 9.26, range: 8.26 ↔ 10.26; the range of the buffer NH4Cl/NH3 contains a pH value of 9.5}
25. a. pH = 4.65 {pH = pKa + log([base]/[acid]); pH = -log(1.8 x 10-5) + log([0.4]/[0.5]) = 4.65}
b. pH = 4.20 {net ionic reaction: H+ + C2H3O2- → HC2H3O2; determine moles: mol HCl = 7.3gHCl x (1mol HCl/36.5g HCl) =
0.2mol HCl; mol C2H3O2- = (0.4M)(1L) = 0.4mol; mol HC2H3O2 = (0.5M)(1L) = 0.5mol; use SCF Table; S row: 0.2, 0.4, 0.5;
C row: -0.2, -0.2, +0.2; F row: 0, 0.2, 0.7; use H–H eqn: pH = pKa + log([base]/[acid]);
pH = -log(1.8 x 10-5) + log(0.2/0.]) = 4.20}
26. a. pH = 3.76 {pH = pKa + log([base]/[acid]); pH = -log(3.5 x 10-4) + log([0.2]/[0.1]) = 3.76}
b. pH = 4.15 {net ionic reaction: OH- + HOCN → H2O + OCN-; determine mols:
mol NaOH = 0.2gNaOH x (1mol NaOH/40g NaOH) = 0.005mol NaOH; mol HOCN = (0.1M)(0.1L) = 0.01mol HOCN;
mol OCN- = (0.2M)(0.1L) = 0.02mol OCN-; use SCF Table; S row: 0.005, 0.01, ---, 0.02; C row: -0.005, -0.005, +0.005, +0.005;
F row: 0, 0.005, ---, 0.025; use H–H eqn: pH = pKa + log([base]/[acid]); pH = -log(3.5 x 10-4) + log(0.025/0.005) = 4.15}
27. pH = 6.23 {use H-H equation: pH = pKa + log([base]/[acid]); find [base] = [NaHCO3];
mol NaHCO3 = 22.5g NaHCO3 x (1mol NaHCO3/84g NaHCO3) = 0.268mol NaHCO3; [NaHCO3] = 0.268mol/1.5L = 0.179M;
use Ka1 for the pKa in H-H eqn; pH = -log(4.2 x 10-7) + log(0.179/0.25) = 6.23}
28. 0.10M {pH = pKa + log([base]/[acid]); 4.25 = -log(1.4 x 10-4) + log(0.25/x); 4.25 = 3.85 + log(0.25/x); 0.396 = log(0.25/x);
2.49 = 0.25/x; x = 0.0995M = 0.10M}
29. 94g NaAc {pH = pKa + log([base]/[acid]); 5.10 = -log(1.8 x 10-5) + log(x mol/0.50 mol); 5.10 = 4.74 + log(x/0.50);
0.36 = log(x/0.50); 2.29 = x/0.50; x = 1.145mol NaC2H3O2 x (82g NaC2H3O2/1mol NaC2H3O2) = 93.9g NaC2H3O2}
30. I. c. greater than 7 {at equivalence point only the conjugate base of benzoic acid, C6H5COO-, exists; solution is basic}
II. a. less than 7 {at equivalence point only the conjugate acid of trimethylamine, (CH3)3NH+, exists; solution is acidic}
31. 4.74 {15.0ml is halfway to the equivalence pt; at ½ way to the equivalence pt: pH = pKa; pH = -log(1.8 x 10-5) = 4.74}
32. a. 6.8 x 1010
b.
H+ + F- (Ka for HF); H+ + OHH2O (reverse of Kw → 1/Kw); add 2 rxns and
-4
multiply K’s; Keq = (Ka for HF) x (1/Kw for H2O) = (6.8 x 10 ) x [1/1.0 x 10-14)] = 6.8 x 1010}
4.4 x 1010 {Rxns: CH3NH2 + H2O
CH3NH3+ + OH- (Kb for CH3NH2); H+ + OHH 2O
(reverse of Kw → 1/Kw); add 2 rxns and multiply K’s; Keq = (Kb for CH3NH2) x (1/Kw for H2O) =
(4.4 x 10-4) x [1/1.0 x 10-14)] = 4.4 x 1010}
{Rxns: HF
33.
34. a {KOH + HBr → H2O + KBr; find mol HBr: 0.1L x 0.1M = 0.01mol HBr; to neutralize HBr, moles of acid equals mol of
base in 1:1 for this reaction (it can be 2:1 depending on the acids/bases): 0.01mol HBr x (1mol KOH/1mol HBr) = 0.01mol KOH;
using M = mol/L where mol = 0.01mol HBr and M = 0.75M HBr → solve for L →
L = 0.01mol/0.75M = 0.01333L x (1000ml/1 L) = 13.33ml}
35. I. b {not multiple equivalence points so “d” and “e” are incorrect; pH starts low (acid) so “a” and “c” are incorrect}
II. d {at the halfway to the equivalence point, the pH = pKa; the equivalence point is at 0.1mol of base added so halfway point is at
0.05mol added; the pH at that point is ≈ 4.7 so the pKa = 4.7; the Ka = 104.7 = 2.0 x 10-5; therefore it is acetic acid; note: pKb for
NH3 = 4.74 which means its pKa = 9.26}
III. e {indicator pH range = pKind ± 1; the indicator pH range should encompass the pH at which the equivalence point lands;
phenolphthalein: pKind = -log(5.0 x 10-10) = 9.3; phenolphthalein pH range: 8.3 ↔ 10.3 includes the equivalence pt at pH = 9}
36. d {from the H-H equation}
37. a. yellow {The effective range of an indicator is pKind ± 1, and the pKind for this indicator = 3.00, this indicator is effective
for pH’s 2.00 < pH < 4.00. On the acidic side of the range, pH = 2.00, the indicator is in the HIn form which is yellow. It stays
yellow at pH’s below 2.00 so at pH = 1.70 it is still yellow.}
b. Yes. Since the effective range of an indicator is pKind ± 1, and the pKind for this indicator = 3.00, this
indicator is effective for pH’s 2.00 < pH < 4.00. Since the pH change at the equivalence point occurs at
2.39, this indicator is a wise choice for this titration.
38. a. 25
{mol HNO2 = (0.1L)(0.50M) = 0.05mol HNO2; at equivalence point: mol HNO2 = mol NaOH = 0.05;
vol NaOH = mol NaOH/M NaOH = 0.05/2.0 = 0.025L = 25ml}
b. pH = 1.82 {ICE table in M with HNO2
NO2- + H+; ICE Table: I row: 0.50, 0, 0; C row: -x, +x, +x; F row: 0.50-x, x, x;
plug into Ka expression: 4.5 x 10-4 = x2/(0.50-x); ok to make approximation? (4.5 x 10-4)100 < 0.5 – yes, make approximation
and solve for x; x = 0.015; pH = 1.82}
c. pH = 3.35 {pH = pKa = -log(4.5 x 10-4) = 3.35}
d. pH = 8.47 {find volume needed to get to equivalence point: (0.1)(0.5) = 0.05mol HNO2; at eq pt mol HNO2 = mol of NaOH
needed; 0.05mol/2M = 0.025L = 25ml NaOH needed to reach eq pt; at equivalence point; HNO2 + OH- → H2O + NO2-; do SCF
table in mol; S row: 0.050, 0.050, ---, 0; C row: -0.050, -0.050, ---, +0.050; F row: 0, 0, ---, 0.050;
find [NO2-] = 0.050/(0.025+0.100)L = 0.40M; do ICE table in M with NO2- + H2O
HNO2 + OH-; ICE Table:
I row: 0.40, ---, 0, 0; C row: -x, -x, +x, +x; F row: 0.40-x, ---, x, x; plug into Kb expression: Kb = Kw/Ka = 2.22 x 10-11;
2.22 x 10-11 = x2/(0.40-x); make approximation nd solve for x; x = 2.98 x 10-6; pOH = 5.53; pH = 8.47}
e. pH = 12.20 {HNO2 + OH- → H2O + NO2-; do SCF Table in mol; S row: 0.050, 0.052, ---, 0;
C row: -0.050, -0.050, ---, +0.052; F row: 0, 0.002, ---, 0.050; 0.050mol NO2- and 0.002mol NaOH; strong base dominates so use
NaOH; [NaOH] = 0.002mol/(0.100+0.026) = 0.01587M; [OH-] = 0.01587M; pOH = 1.799; pH = 12.20}
39. a. 0.25
{[(CH2CH3)NH2] = mol (CH2CH3)NH2/L (CH2CH3)NH2; L (CH2CH3)NH2 = 0.090L;
mol HCl = (0.75)(0.030) = 0.0225mol HCl; at equivalence point: mol HCl = mol (CH2CH3)NH2;
mol (CH2CH3)NH2 = 0.0225mol; [(CH2CH3)NH2] = 0.0225mol (CH2CH3)NH2/(0.090)L (CH2CH3)NH2= 0.25M}
b. 11.97 {(CH2CH3)NH2 + H2O
(CH2CH3)NH3+ + OH-; use Kb and ICE Table; 4.3 x 10-4 = x2/0.25-x = x2/0.25;
x = 0.01037; pOH = 1.98; pH = 12.02}
c. 11.33 {HCl + (CH2CH3)NH2 → (CH2CH3)NH3+ + Cl-; do SCF Table in mol; S row: 0.00375, 0.0225, 0;
C row: -0.00375, -0.00375, + 0.00375; F row: 0, 0.01875, 0.00375; a buffer → use H–H eqn;
pH = pKa + log([B]/[A]) = 10.63 + log (0.01875/0.00375) = 11.33}
d. 10.63
e.
{pH = pKa at the halfway to the equivalence point; pH = 10.63}
5.68 {HCl + (CH2CH3)NH2 → (CH2CH3)NH3+ + Cl-; do SCF Table in mol; S row: 0.0225, 0.0225, 0;
C row: -0.0225, -0.0225, + 0.0225; F row: 0, 0, 0.0225; a weak acid exists so use an ICE table, Ka, and units are M;
[(CH2CH3)NH3+] = 0.0225mol/(0.030 + 0.090) = 0.1875M; rxn: (CH2CH3)NH3+
(CH2CH3)NH2 + H+; ICE Table: I row:
0.1875, 0, 0; C row: -x, +x, +x; F row: 0.1875-x, x, x; Ka = 2.326 x 10-11= x2/(0.1875 – x); ok to make approximation?
(2.3 x 10-11)100 < 0.1875 – yes, make approximation: 2.326 x 10-11 = x2/0.1875; x = 2.088 x 10-6; pH = 5.68}
f. 2.51 {HCl + (CH2CH3)NH2 → (CH2CH3)NH3+ + Cl-; do SCF Table in mol; S row: 0.02288, 0.0225, 0;
C row: -0.0225, -0.0225, + 0.0225; F row: 0.000375, 0, 0.0225; 0.0225mol (CH2CH3)NH3+ and 0.000375mol HCl; strong acid
dominates so use HCl; [HCl] = 0.000375mol/(0.03050+0.09000) = 0.003112M; [H+] = 0.003112M; pH = 2.51}
40. a. 2.4 x 10-4 M {a. MgF2
Mg+2 + 2F-; Ksp = [Mg+2][F-]2 → do ICE table; 5.2 x 10-11 = (x)(2x)2 = 4x3 →
x = 2.35 x 10-4}
b. 0.015g/L {2.35 x 10-4 mol MgF2/L x (62.3g MgF2/1mol MgF2) = 0.0146g/L}
41. I. 7.8 x 10-10
II.
{SrF2
Sr+2 + 2F-; Ksp = [Sr+2][F-]2 → do the ICE table with M; Ksp = 4x3 and x = 5.8 x 10-4;
Ksp = 4(5.8 x 10-4)3 = 7.80 x 10-10}
3.20 x 10-28 {Ksp = [Pb+2][S-2]; find molar solubility (M) from grams/L: 4.28 x 10-12 g PbS/L x (1mol PbS/239.3g PbS) =
1.789 x 10-14 M PbS; from PbS(s)
Pb+2 + S-2 → [Pb+2] = 1.789 x 10-14M and [S-2] = 1.789 x 10-14M;
Ksp = (1.789 x 10-14)(1.789 x 10-14) = 3.198 x 10-28}
{since these all have the same Ksp expressions (i.e., x2) their Ksp’s can be compared directly; larger Ksp → more soluble}
II. c {since AgI and AgCl have the same Ksp expressions (i.e., x2) their Ksp’s can be compared directly; larger Ksp → more
soluble so AgCl is more soluble; to compare AgCl and Ag2C2O4 the “x” molar solubility has to be calculated for each and
compared; molar solubility, “x”, for AgCl = 1.34 x 10-5M; molar solubility, “x”, for Ag2C2O4 = 1.1 x 10-4M; since the molar
solubility of Ag2C2O4 is greater, it is more soluble}
43. d {from Le Chatelier’s principle the greater the concentration of Fe+2 or CO3-2, the lower solubility of the FeCO3}
44. a {PbF2 reacts with acids and produces a weak acid, HF; the other salts would produce strong acids. Strong acids
dissociate 100% and don’t form as products, which means the reaction doesn’t happen and the solid is unaffected by the H+.}
42. I. b
45. I. 3.3 x 10-4 M {PbI2(s)
Pb+2(aq) + 2I-(aq); Ksp = [Pb+2][I-]2 → do the ICE table in M with the starting
[Pb+2] = 0.02 → I row: ---, 0.02, 0; C row: -x, +x, +2x; E row: ---, 0.02+x, 2x; 8.7 x 10-9 = (0.02 + x)(2x)2; make approximation:
8.7 x 10-9 = (0.02)(2x)2 = 0.08x2 → x = 3.3 x 10-4}
II. 1.6 x 10-12 M {LaF3(s)
La+3(aq) + 3F-(aq); Ksp = [La+3][F-]3 → do the ICE table with the starting [F-] = 0.005 →
I row: ---, 0, 0.005; C row: -x, +x, +3x; E row: ---, 0.005+x, 3x; 2 x 10-19 = (x)(0.005 + 3x)3; make approximation;
2 x 10-19 = (x)(0.005)3; 2 x 10-19 = 1.25 x 10-7x → x = 1.6 x 10-12}
46. I. No precipitate. {Q = [Ag+]2[CO3-2] = (1 x 10-4)2(1 x 10-5) = 1 x 10-13; since Q < Ksp there will be no precipitate}
II. No precipitate. {Calculate Q; Q = [Mg+2][F-]2; [Mg+2]: M1V1 = M2V2; (1 x 10-3)(100) = (M2)(400);
[Mg+2] = 2.5 x 10-4M; [F-]: M1V1 = M2V2; (5 x 10-4)(300) = (M2)(400); [F-] = 3.75 x 10-4M;
Q = [Mg+2][F-]2 = (2.5 x 10-4)(3.75 x 10-4M)2 = 3.5 x 10-11; since Q < Ksp there will not be a precipitate}
47. I. [SO4-2] > 1.2 x 10-1 {Precipitation begins when Q > Ksp; the solution is saturated when Q = Ksp.
Ag2SO4
2Ag+ + SO4-2; let Q = Ksp = [Ag+]2[SO4-2] → 1.2 x 10-5 = (1 x 10-2)2[SO4-2] →
[SO4-2] = 1.2 x 10-1 M; when [SO4-2] just exceeds 1.2 x 10-1 M then Q > Ksp and a precipitation forms}
II. 1.3 x 10-14M {Ag3PO4
3Ag+ + PO4-3; Ksp = [Ag+]3[PO4-3]; Ksp/[Ag+]3 = [PO4-3] = 1.3 x 10-20/(0.01)3 =
1.3 x 10-14M}
{since each Ksp expression and each [M+2] are the same, the one with the smallest Ksp is least soluble and will ppt
first; in this case, PbCO3; Q expressions can be set up to find at what [CO3-2] each metal carbonate will ppt but it is not
necessary to do these calculations when all [M+2] and Ksp expressions are the same; here are the Q calculations:
Q = [Cu+][CO3-2]; 2.3 x 10-10 = (1 x 10-5)[CO3-2]; [CO3-2] = 2.3 x 10-5M;
Q = [Pb+2][CO3-2]; 7.4 x 10-14 = (1 x 10-5)[CO3-2]; [CO3-2] = 7.4 x 10-9M;
Q = [Mn+2][CO3-2]; 2.2 x 10-11 = (1 x 10-5)[CO3-2]; [CO3-2] = 2.2 x 10-6M;
when these concentrations of [CO3-2] are reached, the solids will ppt; since the solid requiring the smallest [CO3-2] to just
precipitate will precipitate first the PbCO3 precipitates first}
48. b