NAME _________________________________________
PRE-QUIZ FOR QUEST 2 ACID/BASE
DIRECTIONS: I took a number of these questions from around the web. Not all are like AP questions – but I
feel that these will certainly give you a run for your review money. I am attacking your review in this fashion –
because I want you to hear another voice … and this may do the trick. The answers are on the last pages.
1. Which of the following does NOT represent an acid followed by its conjugate base?
1) H3O+ / H2O
2) HCN / CN3) HCl / Cl4) HC2H3O2 / OH5) All are acids followed by their conjugate base
2. According to the Bronsted-Lowry definition, a base is
1) a substance that increases the hydroxide ion concentration of a solution
2) a substance that can accept a proton from an acid
3) a substance that can donate an electron pair to the formation of a covalent bond
4) a substance that increases the anion formed by the autoionization of the solvent
5) a substance that donates a proton
3. The equilibrium constant for the reaction A- + H+  HA is called:
1) Ka
2) Kb
3) 1/Ka
4) Kw/Kb
5) KwKa
This is the formation of the acid HA from its ions … thus the Kc of the forward reaction
is essentially equal to 1/the dissociation constant or rather 1/Ka
4. At 0°C, the Kw for water is 1.2 x 10-15. The pH of pure water at this temperature is
1) 7.00
2) 6.88
3) 7.56
4) 7.46
5) none of these
Kw = [H+][OH-] …so
1.2 x 10-15 = x2 or…. x = 3.46 x 10-8 ….thus
pH = - log [3.46 x 10-8] = 7.46
5. Given HCN(aq) + HCO3-(aq)  CN-(aq) + H2CO3(aq)
1) HCN
2) HCO33) CN4) H2CO3
5) H2O
If K< 1, what is the strongest base in this system?
Since K<1 then the reverse reaction of the equilibrium is favored…
Hence, CN- must be a superior base to HCO3-
1
6. The following acids are listed in order of decreasing acid strength in water.
HI > HNO2 > CH3COOH > HClO > HCN
According to Bronsted-Lowry theory, which of the following ions is the weakest base?
1) I2) NO23) CH3COO4) ClO5) CN-
thus: what is the strongest acid?
7. Which of the following is an acid?
1) HCOOH
2) CH3OH
3) KOH
4) NH3
5) CH3NH2
8. What is the strongest acid?
1) HClO2
2) HClO3
3) HClO4
4) HF
5) HOCl
9. Which of the following reactions does not proceed significantly to the right in aqueous solution?
1) HCl + H2O → H3O+ + Cl2) H3O+ + OH- → 2H2O
3) H2O + HSO4- → H2SO4 + OH4) HCN + OH- → H2O + CN5) H2SO4 + H2O → H3O+ + HSO4-
3 is the answer …. sulfuric acid (H2SO4) is a significantly
stronger acid than water … thus the reverse reaction in this
case, will be favored …thus proceed to the left and NOT to the
right.
10. What is the [H+] in a solution which shows a pH of 2.30?
1) 2.3M
2) 11.7M
3) 5.0 x 10-3M
4) 2.0 x 10-12M
5) none of these
pH = -log [H+] thus 2.30 = - log [H+] or
the antilog of 2.30…. (choice 3 is correct)
[H+] = 10-2.30 …that is take
2
11. What is the pH of a solution at 25 degrees Celsius in which [ OH-] = 3.4 x 10-5 M.
1) 4.5
2) 10.5
3) 9.5
4) 6.3
5) none of these
pH + pOH = 14 …. find the pOH by taking the -log[OH-] …thus pOH = -log[3.4 x 10-5]
pOH = 4.47 … thus 14 – 4.47 = 9.53
12. What is the pH of a 10.0 M solution of HNO3?
1) 10
2) 1.0
3) 0
4) -1.0
5) none of these
nitric acid is a strong acid … thus 1 x 101 …. take the -log[H+]
13. Nitrous acid, HNO2, has an ionization constant Ka = 4.0 x 10-4. The pH of a 0.25 M HNO2 solution is:
Ka = x2
thus:
4.0 x 10-4 = x2
and x2 = 1.0 x 10-4
1) 2.00
[HA]
0.25
2) 2.30
3) 2.70
so…x = 1 x 10-2 or 0.01 (which is the [H+]) thus: pH = -log[H+] or pH = 2
4) 3.70
5) none of these
14. In a solution prepared by dissolving 0.10 mole of an acid HX in enough water to make 1.00 L of solution,
the pH is observed to 1.35. What is the Ka for this acid?
1) 2.0 x 10-2
2) 3.6 x 10-2
3) 4.5 x 10-2
4) 5.0 x 10-12
5) None of these
Given that value of pH …this could be a strong-ish acid ….hence ignoring “x” may be dangerous
… Use the pH to determine the [H+] and then plug this value into the Ka expression.
1.35 = -log[H+] or rather … 0.045
Ka = x2
so Ka = (0.045)2
[HA]-x
0.055
this one is a bit off ….
15. A 0.05 M aqueous solution of a weak monoprotic acid is 1.2% ionized at equilibrium at 25 °C.
Ka for this acid is:
We can ignore “x” due to a very low percent ionization …
1) 0.034
0.12 = x
x = 6 x 10-3
thus: Ka = x2
= (6 x 10-3)2
2) 6.4 x 10-8
0.05
0.05
0.05
3) 7.3 x 10-33
4) 29
5) none of these
16. What is the pOH of a 0.10 M solution of Ba(OH)2?
1) 13.30
2) 0.70
3) 1.00
4) 13.00
5) none of these
pOH = -log [OH-] …but the strong base: barium hydroxide is
di-hydroxyl … so there is twice the concentration of OH- as the
solution’s molar concentration
[OH-] = 0.20 M ….- log(0.20)
3
17. The [OH-] in a 0.50 M pyridine solution (C5H5N; Kb = 1.7 x 10-9) is:
Kb = x2
[B]
1) 0.50 M
2) 2.9 x 10-5 M
3) 1.8 x 10-9 M
4) 3.3 x 10-10 M
5) none of these
so:
18. The equilibrium constant for the reaction
1.7 x 10-9 = x2 = x2 = 8.5 x 10-10 so x = 2.9 x 10-5
0.50
NH4+ + OH- ↔ NH3 + H2O
is:
1) 1/Kb for NH3
2) 1/Ka for NH4+
3) Kw/Ka for NH4
+
4) Kw/Kb for NH3
1 is the only accurate statement … since we are discussing the production
of the weak base, ammonia …. then the Kc = 1/Kb of the base’s
dissociation.
5) Kb for NH3/Kw
19. What is the pH of a 0.05 M solution of ascorbic acid, Vitamin C (Ka1 = 7.9 x 10-5; Ka2=1.6 x 10-12)
The pH is dependent upon the Ka1 only … Ka = x2
=
We can ignore the change in x due to
[HA]
a very small Ka
1) 1.3
2) 2.7
3) 3.1
4) 5.4
5) 6.5
20. Given:
HAc
Ka = 1.8 x 10-5
H2CO3 Ka1= 4.3 x 10-7
Ka2 = 5.6 x 10-11
Which of the following 0.01M solutions will have the highest pH?
1) HAc
2) NaAc
3) Na2CO3
4) H2CO3
5) NaHCO3
The highest pH will be given by the strongest base. The strongest base is the
conjugate of the weakest acid ….thus the strongest base = CO32- (which evolves
from the very weak acid HCO3-2 …that represented by Ka2)
21. If Ka for HCN is 6.2 x 10-10, what is Kb for CN-?
1) 6.2 x 10-24
2) 6.2 x 104
3) 1.6 x 10-5
4) 1.6 x 10-23
5) none of these
Kw = Ka Kb … thus assuming standard conditions of 25C, then
Kb = 1. x 10-14/ 6.2 x 10-10 or… choice 3
4
22. Which of the following substances can be dissolved in water to give a basic solution?
1) NH4Cl
2) NaBr
3) KF
4) NaHSO4
5) KNO3
Based upon hydrolysis, K+ has no affinity for the OH- of water, (the cation is of low
positive charge and large radius…. but the F- ion does bond with H+ of water, forming
the molecular form of a weak acid (HF), removing H+ from solution …driving the [OH-]
to a greater level than the [H+] … producing an alkaline (basic) solution. 1) Produces
an acid solution, as NH4+ is the conjugate acid of a weak base…2, 4 and 5) are neutral
salts, as both species are conjugates of strong acids or strong bases.
23. What is the pH of a 1.0 M aqueous solution of NaCl?
1) 7.0
2) greater than 7.0
3) less than 7.0
4) not enough information
5) too tired to answer
NaCl is a neutral salt … both ionic species Na+ and Cl- are conjugates of
either a strong acid or strong base (NaOH and HCl) …hence neither ion as
a strong affinity for bonding with the H+ or OH- of water …and no change
in pH will be manifested.
24. What is the pH of a 1.0 M aqueous solution of KNO2?
1) 7.0
2) greater than 7.0
3) less than 7.0
4) not enough information
Here we have the conjugate acid of a strong base (K+ relative to KOH)
and the conjugate base of a weak acid (NO2- relative to HNO2) …so K+ has
virtually no affinity to re-bond with the OH- of water ….but NO2- can bond
with H+, reconstituting the weak (molecular) acid HNO2. This creates an
increase in the OH- given the ratio between H+ and OH- thus a pH > 7
25. Which of the following is the STRONGEST Lewis acid?
1) Na+
2) Al3+
3) CH3COO4) Mg2+
5) Cl-
Large positive charge, small radius … empty valence orbitals …. This ion
maximizes the changes of weakening an O-H bond in a hydrating water
molecule, and allowing that lost H+ to be transferred to a solvating water
molecule ….
26. All of the following species can function as Bronsted-Lowry bases in solution EXCEPT:
1) H2O
2) NH3
3) S24) NH4+
5) HCO3-
Bases accept H+ … but in choice 4 … there is no longer any ability to bond with
(accept) H+ since all available orbitals are occupied.
27. What is the pH of a 0.36M solution of sodium acetate? (Ka for acetic acid = 1.8 x 10-5)
1) 9.15
2) 4.85
3) 2.59
4) 11.41
5) 7.00
Kw = KaKb …. Since we are dealing with the conjugate base of a weak acid
(acetate vs. acetic acid) …find the Kb: Kb = Kw/Ka or… 5.55 x 10-10
Use the Kb expression to solve for the [OH-]… Kb = x2/[salt] (ignoring Δx)
x (or the [OH-]) = 1.414 x 10-5 … find the pOH = -log[OH-] and then subtract
from 14 to find the pH
5
28. Which of the following will NOT give an acid solution when dissolved in water?
1) K2O
2) P2O5
3) CO2
4) NO2
5) SO3
Metal oxides have O2- and it has a huge affinity for protons … and in a reaction
with water, protons are extracted, leaving hydroxide ions behind….causing the
pH to rise.
29. HCN is a weak acid (Ka = 6.2 x 10-10) NH3 is a weak base (Kb = 1.8 x 10-5). A 1.0 M solution of NH4CN
would be
1) strongly acidic
2) weakly acidic
3) neutral
4) weakly basic
Ammonium ion in water can drop pH but cyanide ion can raise pH, via
hydrolysis. HCN is a much weaker acid than ammonia is a base. thus CN- is a
stronger conjugate base than ammonium ion is a conjugate acid … the
cyanide ion then can more efficiently engage in hydrolysis, which means
that H+ bond with it, thus leaving OH- or increasing [OH-]
30. The equilibrium constant for this reaction is approximately 0.001.
HPO42-(aq) + HCO3-(aq)  H2PO4-(aq) + CO32-(aq)
Which is the strongest conjugate base in this reaction?
The Kc suggests that the forward reaction is NOT favored (as it is rather
1) HPO42-(aq)
small ….Kc<1 … Thus the reverse reaction is favored …meaning that the
2) HCO3-(aq)
carbonate ion must be a superior base….and has a greater tendency to
3) H2PO4-(aq)
bond with H+
4) CO32-(aq)
5) there is no base in this reaction
31. Calculate the pH of a 0.00175 M solution of KOH.
1)
2)
3)
4)
5)
7.89
11.2
12.7
-2.76
2.76
pOH = - log[OH-] or rather…. pOH = -log[1.75 x 10-3] = 2.757
pH + pOH = 14
or rather …. 14 – 2.757 = 11.2
32. Calculate the [H+] in a solution that has a pH of 8.38.
1)
2)
3)
4)
5)
4.17 x 10-9
2.40 x 108
3.8 x 10-8
1.21 x 10-2
1.21 x 102
Take the antilog of 8.38 or: [H+] = 10-8.38
or type into your (CASIO) calculator: -8.38 …shift log
6
33. If an acid, HA, is 15% dissociated in a 2.00 M solution, what is Ka for the acid?
1)
2)
3)
4)
6.4 x 10-3
2.2 x 10-2
5.3 x 10-2
7.5 x 10-2
This is a significant ionization (greater than 5%) ….we must account for the
change in “x” …so run a % ionization and then plug into the Ka = x2
[HA]-x
2
0.15 = x
x = 0.30
Ka = (0.30)
2.00
1.7
34. What is the pH of a 0.00350 M HNO3 solution?
1)
2)
3)
4)
5)
2.456
1.018
6.780
4.903
-2.466
This is a strong acid …hence the [H+] = the molar concentration of the solution.
pH = - log(0.00350) ….
35. Calculate the [H3O+] of a 0.10 M solution of NH4Cl in H2O at 25°C (Kb for NH3 = 1.8 x 10-5)
1)
2)
3)
4)
5)
5.6 x 10-10
2.4 x 10-5
1.8 x 10-6
1.8 x 10-5
7.5 x 10-6
Kw/Kb = Ka ….. thus Ka = 5.56 x 10-10
Ka = x2
[ ]
thus 5.56 x 10-10 = x2
0.1
or 5.56 x 10-11 = x2
the square root of 5.56 x 10-11 = 7.456 x 10-6
36. What concentration of HOCl (Ka = 3.5 x 10-8) has the same pH as that of 2.50 x 10-4 M HNO3?
1)
2)
3)
4)
5)
6.8 10-8 M
3.5 x 10-2 M
1.79 M
0.560 M
3.93 M
The [H3O+] of the nitric acid solution (a strong acid) = 2.50 x 10-4 M … thus
the question becomes: What concentration of HOCl (Ka = 3.5 x 10-8 gives a
[H+] of 2.50 x 10-4?
Use the Ka expression: Ka = x2
[HA]
= 3.5 x 10-8 = (2.50 x 10-4)2
x
7
37. Based upon the following values of Ka, place the acids in order from weakest to strongest:




HCN
(Ka = 6.2 x 10-10)
HCOOH (Ka = 1.78 x 10-4)
HC2H3O2 (Ka = 1.8 x 10-5)
HNO2
(Ka = 4.6 x 10-4)
Weak = small Ka
Strong = large Ka
1) HCN < HC2H3O2 < HCOOH < HNO2
2) HCN < HNO2 < HC2H3O2 < HCOOH
3) HNO2 < HCOOH < HC2H3O2 < HCN
4) HC2H3O2 < HCN < HNO2 < HCOOH
5) HCOOH < HNO2 < HCN < HC2H3O2
38. Which of the following is not true for the dissociation of a strong acid?
1)
2)
3)
4)
5)
The conjugate base will be weak
Ka is large
The equilibrium lies far to the left
The equilibrium lies far to the right
[H+] >> [HA]
The definition of a strong acid indicates that virtually of
chance of establishing equilibrium exists, since there is
virtually 100% ionization….hence the reaction goes far to
the right …. making choice 3 untrue.
39. What are the conjugate bases in the following reaction?
1)
2)
3)
4)
5)
H2SO4 + H2O → HSO4- + H3O+
HSO4- and H3O+
H2SO4, H2O and HSO4H2O and HSO4H2SO4 and H2O
H2SO4 and H3O+
40. The pH of a 0.150 M solution of an aqueous weak monoprotic acid is 4.10. Calculate Ka for the acid.
1)
2)
3)
4)
5)
4.21 x 10-8
5.09 x 10-2
2.26 x 10-4
7.94 x 10-5
6.31 x 10-9
Take the antilog of the pH to determine the [H3O+]
[H3O+] = 7.9 thus: Ka = x2 =
[HA]
Ka = (7.9)2 = 4.21 x 10-8
0.150
8
41. At 25°C a solution has a [OH-] = 2.90 x 10-2. Calculate the pH of the solution.
1)
2)
3)
4)
5)
1.54
3.17
12.5
-4.54
9.03
Calculate the pOH using the –log[OH-] ….then use
14 = pH + pOH and subtract: 14 – pOH = pH
42. The pH of a 0.500 M solution of KC2H3O2 is:
1) greater than 7
2) less than 7
3) 7
K+ is the conjugate acid of a strong base …it has a rather large
radius and low positive charge, hence has no affinity to rebond with OH- … Acetate ion is the conjugate base of a weak
acid. This is not a terribly stable conjugate base and will
hydrolyze water, reconstituting the weak molecular acid of
acetic (ethanoic) acid, leaving OH- or increasing [OH-]
43. Which of the following is a conjugate acid/base pair?
1)
2)
3)
4)
5)
1. (4)
2. (2)
3. (3)
4. (4)
5. (3)
6. (1)
7. (1)
8. (3)
9. (3)
10. (3)
11. (3)
12. (4)
13. (1)
14. (2)
15. (5)
16. (2)
17. (2)
18. (1)
19. (2)
20. (3)
21. (3)
22. (3)
23. (1)
24. (2)
25. (2)
26. (4)
27. (1)
28. (1)
29. (4)
30. (4)
HCl/OClH2SO4/SO42NH4+/NH3
H3O+/OHnone of these
A conjugate acid base pair, must be separated from each other by 1 H+
(not 2!!!) … These species are often in equilibrium with each other.
31. (2)
32. (1)
34. (3)
33. (1)
34. (5)
35. (3)
36. (1)
37. (3)
38. (3)
39. (3)
41. (5)
42. (1)
43. (3)
9
FROM THE 2014 AP CHEMISTRY EXAM:
CH3CH2COOH(aq) + H2O(l)  CH3CH2COO- + H3O+
Propanoic acid, CH3CH2COOH, is a carboxylic acid that reacts with water according to the equation above. At
25 °C the pH OF a 50.0 mL sample of 0.20 M CH3CH2COOH is 2.79.
a) Identify a Bronsted-Lowry conjugate acid-base pair in the reaction. Clearly label which is the acid
and which is the base.
b) Determine the value of Ka for propanoic acid at 25°C
c) For each of the following statements, determine whether the statement is true of false. In each case,
explain the reasoning that supports your answer.
i) The pH of a solution prepared by mixing the 50.0 mL sample of 0.20 M CH3CH2COOH with
a 50.0 mL sample of 0.20 M NaOH is 7.00
ii) If the pH of a hydrochloric acid solution is the same as the pH of a propanoic acid solution,
then the molar concentration of the hydrochloric acid solution must be less than the molar
concentration of the propanoic acid solution.
A student is given the task of determining the concentration of a propanoic acid solution of unknown
concentration. A 0.0173 M NaOH solution is available to use as the titrant. The student uses a 25.00 mL
volumetric pipet to deliver the propanoic acid solution to a clean, dry flask. After adding an
appropriate indicator to the flask, the student titrates the solution with the 0.173 M NaOH, reaching the
end point after 20.52 mL of the base solution has been added.
d) Calculate the molarity of the propanoic acid solution.
10
11
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