Chemistry I-Honors
Intro Acid-Base Practice Test
Solution Set
1.
C
Which of the following salts produces a neutral solution when dissolved in water?
A) Na2CO3
B) NaF;
C) Mg(NO3)2;
D) NH4Cl;
E) Fe(HSO4)2
basic
basic
neutral
acidic
acidic
SB/WA
SB/WA
SB/SA
WB/SA
weak acidic anion
2.
A
If a weak base (MOH) ionizes only 0.62 %, what is the pH of a 0.400 M solution of this base?
A) 11.39;
B) 2.61;
C) 4.58;
D) 10.44;
E) 8.95;
F) need a KB value.
[OH-1] = (0.400 M)(0.0062) = 0.0025, then pOH = 2.61, then pH = 11.39
3.
B
-5
Benzoic acid has a KA = 6.6 x 10 . What is the pH of a 0.55 M solution of this acid?
A) 6.02; B) 2.22;
C) 4.74;
D) 10.98;
E) 1.30;
F) 3.91.
-5
6.6 x 10 = x2 / 0.55 , then x = [H+] = 0.0060, then pH = 2.22
4.
F
What is the percent ionization of a weak acid (HA) if a 0.0450 M solution has a pH of 5.31?
-7
-5
A) 0.450; B) 10.4; C) 4.87 x 10 ; D) 2.38; E) 4.68 x 10 ; F) 0.0109;
G) 0.055.
-6
pH = 5.31, [H+] = 4.9 x 10 ; % ionization =
5.
E
If 2.50 grams of perchloric acid (MW = 100.5 g/mole) are dissolved in enough water to make
12.0 liters of solution, what will be the pH of this solution?
A) 3.79;
B) 13.1;
C) 1.9;
D) 0.00207;
E) 2.68;
F) 3.35.
[H+] = (2.50 g / 100.5 g/mol) / 12.0 liters = 0.00207 M
6.
7.
D
F
-6
(4.9 x 10 / 0.0450 ) x 100 = 0.0109%
pH = 2.683
Which of the following compounds is not a strong acid?
A) hydrobromic acid;
B) hydrochloric acid; C) perchloric acid;
E) sulfuric acid;
F) both C and D.
D) carbonic acid;
What is the KA of a 0.500 M solution of a weak acid, HA, if its pH = 3.89?
-4
-4
-6
-8
-9
A) 5.1 x 10 ; B) 3.3 x 10 ;
C) 1.2 x 10 ; D) 5.9 x 10 ; E) 4.3 x 10 ;
-4
pH = 3.89, [H+] = 1.3 x 10
-4 2
KA = (1.3 x 10 ) / (0.500) =
3.3 x 10
-8
F) 3.3 x 10
-8
-28.
A
Which of the following compounds is not a strong base?
B) cesium hydroxide;
C) barium hydroxide;
D) potassium hydroxide;
E) sodium hydroxide;
F) both A and C.
A) ferric hydroxide;
9.
E
+1
What is the [H3O ] of an acidic solution if the pOH = 10.22?
-6
-11
A) 10.22;
B) 3.78;
C) 4.09 x 10 ;
D) 6.0 x 10 ;
+1
+1
-4
E) 1.7 x 10 ;
if pOH = 10.22, then pH = 3.78, then [H3O ] = [H ] = 1.7 x 10
-5
F) 4.9 x 10 .
-4
10. A
Given the following equation, what is the conjugate base to the reactant acid for the reaction
-1
22as written?
HSO4
+ HCO3 ---------> H2CO3 + SO4
acid
base
C.A.
C.B.
-1
-1
22A) SO4 ; B) HCO3 ; C) CO3 ; D) HSO4 ; E) H2O; F) H2CO3; G) H2SO4
11. A
When 0.020 moles of a substance are dissolved to make 2.00 liters of solution, the pH is
found to be 2.0. The substance is probably which one of the following:
A) strong acid; B) weak acid; C) weak base; D) strong base; E) neutral salt.
0.020 mol / 2.00 liter = 0.010, then -log(0.010) = 2.0
12. D
A beaker contains 100.0 ml of 0.200 M HNO3 solution. To this solution, 500.0 ml of distilled
water is added. What is the pOH of this diluted solution?
A) 12.602;
B) 11.65;
C) 13.0;
D) 12.523;
E) 12.48;
F) 12.35
+1
[H ] = [(0.200 M)(100.0 ml)] / 600.0 ml = 0.0333 M
pH = 1.477, pOH = 12.523
Acid
HC2H3O2
HCHO2
HClO
HF
13. C
14. D
Ka
-5
1.8 x 10
-4
1.8 x 10
-8
3.0 x 10 (weakest acid)
-4
6.8 x 10
Using the data above, which one of the following is the strongest base?
-1
-1
-1
-1
A) C2H3O2 ;
B) CHO2
C) ClO ;
D) F ;
E) H2O;
-1
-1
F) HC2H3O2
In the reaction BF3 + F ---------------> BF4 , the BF3 acts as a(n):
A) Arrhenius acid;
B) Arrhenius base; C) Bronsted-Lowry acid;
D) Lewis acid;
E) both C and D.
(e-1 pair acceptor)
-315. C
When ammonia (NH3) is dissolved in water, the hydroxide ion and ammonium ion are
produced to a small extent.
NH3 + H2O 
base
acid
+1
NH4
C.A.
-1
+ OH
C.B.
What is the role of the water molecule in this reaction?
A) Arrhenius acid;
B) Arrhenius base; C) Bronsted-Lowry acid;
E) both C and D.
16. A
Carbon dioxide would be classified as:
A) acidic anhydride;
B) basic anhydride;
D) Lewis acid;
C) neutral substance;
D) weak acid.
For the following statements, if it is incorrect, make it correct or indicate what is incorrect.
17. F
If an indicator is an organic acid (HIn) with its indicating range from 6.0-7.5, then putting
-
this indicator into a strongly acidic solution will cause [In ] to be much less than [HIn].
In acidic solutions, the equilibrium will cause the reaction to shift back towards the
molecule, making the concentration of HIn greater than the In-1.
18. T
When 1.0 mole of NaF is dissolved in 1.0 liter of water, red litmus paper changes to blue.
basic salt due to the presence of the fluoride ion which is a conjugate base
19. T
A characteristic that is common to both acids and bases is their ability to conduct electricity.
20. T
The NH3 molecule could be classified as amphoteric because it can act as either an acid or a
base.
NH3 + H+ --> NH4+
or NH3 ---> NH2-1 + H+
21. F
The pH of a 1.0 M solution of acetic acid is less than (closer to 1.0) than a 1.0 M solution of
HF.
(Use the KA values from problem #13.)
HF is the stronger acid, so it will have the lower pH.
....................................................................................................................
22. E
23. C
Lithium phosphate would be classified as:
A) weak acid; B) strong base; C) acidic salt; D) neutral salt;
F) weak base.
E) basic salt;
If the pH of a solution is 3.25, what is the pOH of the solution?
-4
-11
A) 3.25;
B) 5.6 x 10 ;
C) 10.75;
D) 11.25;
E) 1.8 x 10 ;
F) can't be determined unless the type of acid is known.
pH + pOH = 14
-424. What is the pH of a 0.050 M solution of sodium bicarbonate? KA for carbonic acid is 4.3 x 10
-7
-14
-7
-8
2
KH = KB = KW / KA = 1.0 x 10 / 4.3 x 10 = 2.3 x 10 = x / 0.050
-5
x = [OH-1] =3.4 x 10 M = > pOH = 4.47, then pH = 9.53
25. If a buffer is prepared by adding 0.350 moles of magnesium nitrite to 500 ml of a 1.00 M solution of
-4
nitrous acid, what is the pH of this solution? KA for nitrous acid is 4.5 x 10
pKA = 3.35
Mg(NO2)2
[NO2-1] = 0.700 moles / 0.500 liters = 1.40 M
pH = pKA + log (CB/A) = 3.35 + log ([1.40]/[1.00])
= 3.49
26. What is the normality of a solution of sulfuric acid if it takes 35.50 ml of a 3.45 M solution of KOH
to just neutralize 28.80 ml of the acid?
What is the molarity of this same acid solution?
NAVA = NBVB NA = [(3.45 N)(35.50 ml)] / (28.80 ml) = 4.25 N
(4.25 equiv / 1 liter) x (1 mol / 2 equiv) = 2.13 M
27. When 25.00 ml of 0.100 M acetic acid is just neutralized with a 50.00 ml of a solution of sodium
hydroxide, what is the pH of the neutralized solution? pKA for acetic acid is 4.74
@ equivalence point, a basic salt is produced - sodium acetate
[OAc-1] = (0.02500 liter)(0.100 M) / (0.0750 liter) = 0.0333 M
-14
-5
-10
2
KH = KB = KW / KA = 1.0 x 10 / 1.8 x 10 = 5.6 x 10 = x / 0.0333
-6
x = [OH-1] =4.3 x 10 M = > pOH = 5.36, then pH = 8.64
28. How many milliliters of distilled water will it take to dilute 30.0 ml of a solution of HCl with a pH of
2.25 to produce a solution with a pH of 3.77?
[H+] = 0.0056 M
[H+] = 1.7 x 10-4 M
M1V1 = M2V2 V2 = [(0.0056 M)(30.0 ml)] / (1.7 x 10-4 M) = 989 ml
Vwater = 989ml - 30ml = 959ml
29. What is the normality of a phosphoric acid solution if 12.00 grams of the acid is dissolved in enough
water to produce 400.0 ml of solution?
12.00 g / 98.0 g/mol / 0.400 liters = 0.306 M x 3 eq / 1 mol = 0.918 N
30. If 35.75 ml of a 0.2205 N solution of potassium hydroxide are used in a titration experiment to
neutralize 1.550 grams of an unknown solid acid, what is the gram equivalent mass of this acid?
GEW = 1.550 g / (0.03575 liters)(0.2205 N) = 197 g/eq
31. 25.00 ml of a solution of hydrochloric acid are needed to neutralize 31.00 ml of a 3.50 M solution of
aluminum hydroxide.
= 10.5 N
A) What is the normality of this acid solution?
NAVA = NBVB NA = 13.20 N
B) What is the salt produced in this reaction? AlCl3
C) Based on the type of salt produced, what will be the pH of the final, neutralized solution?
less than 7, since aluminum chloride is an acidic salt
-10
You have a 2.00 liters of a 2.00 M solution of HCN. HCN is known to have a KA of 4.0 x 10 .
Answer the following questions based on this information:
32. What is the pH of the solution?
KA = 4.0 x 10-10 = x2 / 2.00 x = [H+] = 2.8 x 10-5 M, then pH = 4.55
33. What is the percent ionization of this solution?
2.8 x 10-5 / 2.00 x 100 = 1.4 x 10-3 %
34. What is the KB for the cyanide ion?
KB = KW /KA = 1.0 x 10-14 / 4.0 x 10-10 = 2.5 x 10-5
-
35. Which is stronger - HCN as an acid or CN as a base?
The cyanide ion, since it has the larger K value.
36. What is the pOH of a 1.00 M solution of NaCN?
KB = 2.5 x 10-5 = x2 / 1.00
x = [OH-1] = 5.0 x 10-3 M, pOH = 2.30
If 1.0 liter of a 1.00 M solution of NaCN was added to the original acid solution,
37. How many moles of acid could be neutralized by this buffer?
1.00 mole (because you have 1.00 mole of the cyanide ion which is a conjugate base.)
38. How many moles of base could be neutralized by this buffer?
4.00 moles (based on the moles of acid in the original solution).
39. What is the final volume of this buffer?
3.00 liters
40. What is the pH of this buffer solution?
pH = pKA + log (CB/A) = 9.40 + log (1 / 4) = 8.80
41. If 0.500 mole of NaOH is added to this buffer solution, what will the final pH of the solution be?
pH = pKA + log (CB/A) = 9.40 + log (1.5 / 3.5) = 9.03
-
-4
42. 10.40 An aqueous solution has a [OH ] = 2.5 x 10 M. What is the pH of this solution?
pOH = 3.60, pH = 10.40
43. 10.06
If a weak monoprotic acid (HX) ionizes only 0.0262 %, what is the pOH of
a 0.440 M solution of this acid?
[H+] = (0.440 M)(2.62 x 10-4) = 1.15 x 10-4,
then pH = 3.94, then pOH = 10.06
44.
2.60
-5
Acetic acid has a KA = 1.8 x 10 . What is the pH of a 0.350 M solution of this acid?
KA = 1.8 x 10-5 = x2 / 0.350 x = [H+] = 2.5 x 10-3 M, then pH = 2.60
45.
2.06
If 7.00 grams of hydrobromic acid (MW = 80.9 g/mole) is dissolved in enough
water to make 10.0 liters of solution, what will be the pH of this solution?
(7.00 g / 80.9 g/mol) / 10.0 liters = [H+] = 0.00865,
pH = 2.06
46. 0.198 M
30.0 ml of a 0.125 M solution of NaOH is used to neutralize 18.9 ml of a solution of
nitric acid. What was the concentration (molarity) of the nitric acid?
MA = (0.125 M)(30.0 ml) / 18.9 ml = 0.198 M
47.
1.33 N
If 4.00 grams of barium hydroxide (MW = 171.3 g/mole) are just neutralized with
35.0 ml of a solution of HCl, what is the normality of the acid solution?
4.00 g / 171.3 g/mol = 0.02335 mol cpd = 0.0467 mol OH-1
when neutralized = 0.0467 mol H+ / 0.0350 liter = 1.33 M = 1.33 N
48.
11.89
When 440.0 ml of 0.168 M HCl is mixed with 560.0 ml of 0.146 M NaOH, what is the
pH of the mixed solution?
moles H+ = (0.4400 liters)(0.168 M) = 0.0739
moles OH-1 = (0.5600 liters)(0.146 M) = 0.0818
= 0.0079 moles xcs hydroxide in 1.00 liter,
then pOH = 2.11, then pH = 11.89
49.
8.56
-3
When 1.5 x 10 moles of nitrous acid are just neutralized with 25.00 ml of a 0.060 N
solution of potassium hydroxide, what will be the pH of the final solution?
(Remember to consider the products of this neutralization.)
-5
The KA for HNO2 is 4.5 x 10 . produces a basic salt- KNO2
[NO2-1] = ( 1.5 x 10-3 mole / 0.0250 liter) = 0.060 M
-14
-5
-10
2
KH = KB = KW / KA = 1.0 x 10 / 4.5 x 10 = 2.2 x 10 = x / 0.060
-6
x = [OH-1] =3.6 x 10 M = > pOH = 5.44, then pH =
50.
135.6 g/eq
8.56
A student takes 0.6554 grams of a weak monoprotic acid (HA) and dissolves it in
distilled water to make 100.0 ml of solution. The solution is then neutralized
with 33.56 ml of a 0.1440 M NaOH solution. What is the gram equivalent mass
of the acid?
GEW = 0.6554 g / (0.03356 liters)(0.1440 M) =
135.6 g/eq
51.
-
4.44
-5
A 1.00 liter solution of HOAc/OAc buffer is prepared. The Ka for HOAc is 1.8 x 10 .
-
If twice as many moles of acid (HOAc) as basic anion (OAc ) were used, what is the
pH of the solution?
pH = pKA + log (CB/A) = 4.74 + log (1/2) =
52.
25.7 g/eq
What is the gram equivalent mass of chromium (VI) hydroxide? Cr(OH)6
MW/6 = 154.0 g/mol / 6 = 25.7
53.
9.652
4.44
g/eq
73.00 mg of cupric hydroxide are dissolved in enough water to make 400.0 ml of
solution. This weak base is known to ionize only 1.20%. What is the pH of this
solution?
[OH-1] = [(73.00 x 10-3 g)/(97.5 g/mol)/ 0.400 liter x 2] x 1.20%
= 4.49 x 10-5 M then pOH = 4.348 then pH = 9.652
54.
2.15
If the pKA of an acid is 3.66, what is the pH of a 0.225 M solution of this acid?
if pKA = 3.66, then KA = 2.2 x 10-4 = x2 / 0.225
3.1% ionized
x = [H+] = 0.0070 M pH = 2.15
55.
10.18
What is the pH of a 1.0 M solution of potassium bicarbonate?
pKA for carbonic acid is 6.37.
-14
-7
-8
2
KH = KB = KW / KA = 1.0 x 10 / 4.3 x 10 = 2.3 x 10 = x / 1.00
-4
x = [OH-1] =1.5 x 10 M = > pOH = 3.82 ,
then pH =
10.18
56.
1.022
+
What is the [H3O ] of a 1.00 M solution of a weak monoprotic acid, HA, if the
pKA for this acid is known to be 2.000? I need to see 3 sig figs in the answer!
Cannot ignore x !!! use the quadratic or graph it.
KA = 0.0100 = x2 / 1.00 - x
x2 + 0.0100 x - 0.0100 = 0
x = 0.0951 M
57.
6.7 %
pH = 1.022
-3
If a weak monoprotic acid, HA, has a KA = 3.6 x 10 , what is the percent
ionization of a 0.75 M solution?
Again, cannot ignore x !!! use the quadratic or graph it.
KA = 3.6 x 10-3 = x2 / 0.75 - x
x2 + 0.0036 x - 0.0027 = 0
x = 0.0502 M
% ionization = 0.0502 / .75 x 100
= 6.7
58.
1.30
59.
2.8 x 10-12
What is the pH of the above solution?
pH = -log(0.0502) = 1.30
Using the above data, what is the KB for NaA?
-14
-3
KB = KW / KA = 1.0 x 10 / 3.6 x 10 =
60.
8.22
%
2.8 x 10-12
Using the above data, what would be the pH of a 1.00 M solution of NaA?
2.8 x 10
-12
= x2 / 1.00
-6
x = [OH-1] =1.7 x 10 M = > pOH = 5.78 ,
then pH =
8.22
61.
4.00
A saturated solution of hydrosulfuric acid (H2S) has a concentration of 0.10 M.
What is the pH of this solution?
(find pH from 1st ionization)
KA1 = 1.0 x 10
-7
= x2 / 0.10
x = [H+] = 1.0 x 10-4 M pH = 4.00
62.
1.3 x 10-13
What is the [S-2] in the solution in #61?
[S-2] = 2nd KA = 1.3 x 10-13
63.
1.363
What is the pH of a 0.200 M solution of KHSO4? (need to use the quadratic!)
Again, cannot ignore x !!! use the quadratic or graph it.
KA = 1.2 x 10-2 = x2 / 0.200 - x
x2 + 0.00120 x - 0.00240 = 0
x = 0.0434 M
pH = 1.363