Acids – Bases Unit Practice Test
AP ChemistryFree Response
For the following two questions: The set of lettered choices is a list of chemical compounds that are soluble in water and take part in acid base reactions. Select the one lettered choice that best fits each statement. A choice may be used once, more than once or not at all.
A. AlCl3 B. CH3OH C. HC2H3O2
D. K2C2O4 E. NaHCO3
1. Lewis acid but not a Brønsted-Lowry acid A
2. Reacts with an acid to form an ester B
The following two questions refer to aqueous solutions containing 1:1 mole rations of the following pairs of substances. Assume all concentrations are 1 M.
A. NH3 and NH4Cl B. H3PO4 and NaH2PO4 C. HCl and NaCl
D. NaOH and NH3
E. NH3 and HC2H3O2 (acetic acid)
3. The solution with the lowest pH C
4. A buffer at a pH > 8 A
5. As the number of oxygen atoms increases in any series of oxygen acids, such as HXO, HXO2, HXO3, ...., which of the following is generally true? E
A. The acid strength varies unpredictably.
B. The acid strength decreases only if X is a nonmetal. C. The acid strength decreases only if X is a metal.
D. The acid strength decreases whether X is a nonmetal or a metal. E. The acid strength increases.
6. What is the net ionic equation for the reaction that occurs when aqueous copper(II) sulfate is added to excess 6-molar ammonia? D
A. Cu2+ + SO
42¯ + 2 NH4+ + 2 OH¯ ---> (NH4)2SO4 + Cu(OH)2 B. Cu2+ + 4 NH
3 + 4 H2O --> Cu(OH)42¯ + 4 NH4+ C. Cu2+ + 2 NH
3 + 2 H2O --> Cu(OH)2 + 2 NH4+ D. Cu2+ + 4 NH
3 --> Cu(NH3)42+ E. Cu2+ + 2 NH
3 + H2O --> CuO + 2 NH4+
7. When solid potassium oxide is dissolved in water, which of the following is the strongest base in the resulting solution? (Think about net ionic equations) E
A. H+ B. K+ C. O
2-D. KH E. OH¯
9. If the acid dissociation constant, Ka, for an acid HA is 8 x 10¯4 at 25 °C, what percent of the acid is dissociated in a 0.50-molar solution of HA at 25 °C? E
A. 0.08% B. 0.2% C. 1%
8. Which is an acid anhydride? C
I. SO2 II. SO3 III. SO4 2-A. I only
B. II only C. I and II only
D. II and III only E. I, II, III
For the following two questions: A sample containing precisely 25.0 mL of a solution of H2SO4 of unknown concentration is analyzed using a solution of NaOH of known concentration. The NaOH is titrated slowly into the H2SO4 solution from a buret. When neutralization occurs as shown by change in color of the indicator, the final value of NaOH is recorded.
NaOH solution 0.125 M
Starting volume, NaOH 1.4 mL Final volume, NaOH 27.8 mL
10. Which identifies the spectator ions in this reaction? C
A. The only spectator ion is Na+ B. The only spectator ion is SO4 2-C. Both Na+ and SO
42- are spectator ions
D. Both Na+ and HSO
4- are spectator ions E. There are no spectator ions in this
reaction
11. If, unknown to the technician, some water had been added to the unknown sulfuric acid solution by mistake after its precise volume had been measured, which value in the table of observations, if any, would be changed? If there were a change, in which direction would the change occur? E
A. The final volume of NaOH would be reported larger. B. The final volume of NaOH would be reported smaller. C. The molarity of the NaOH solution would be reported larger. D. The molarity of the NaOH solution would be reported smaller. E. No change would be reported in any of the three values.
12. The safest and most effective emergency procedure to treat an acid splash in skin is to do which of the following immediately?
A. Dry the affected area with paper towels.
B. Sprinkle the affected area with powdered Na2SO4(s).
C. Flush the affected area with water and then with a dilute NaOH solution. D. Flush the affected area with water and then with a dilute NaHCO3 solution. E. Flush the affected area with water and then with a dilute vinegar solution.
13. When phenolphthalein is used as the indicator in a titration of an HCl solution with a solution of NaOH, the indicator undergoes a color change from clear to red at the end point of the titration. This color change occurs abruptly because:
A. phenolphthalein is a very strong acid that is capable of rapid dissociation
B. the solution being titrated undergoes a large pH change near the end point of the titration C. phenolphthalein undergoes an irreversible reaction in basic solution
D. OH¯ acts as a catalyst for the decomposition of phenolphthalein
14. When a sample of 0.0040 M HBr is diluted with an equal amount of water, the pH of the resulting mixture is closest to: B
A. 2.3 B. 2.7 C. 3.0
D. 3.3 E. 3.7
15. H2C2O4 + 2 H2O ↔ 2 H3O+ + C2O42¯ Oxalic acid, H2C2O4, is a diprotic acid with K1 = 5 x 10¯2 and K2 = 5 x 10¯5. Which of the following is equal to the equilibrium constant for the reaction represented above? C
A. 5 x 10¯2 B. 5 x 10¯5 C. 2.5 x 10¯6
Free Response (Calculator OK)
CH3NH2 + H2O CH3NH3+ + OH–
Methylamine, CH3NH2, is a weak base that reacts according to the equation above. The value of the ionization constant, Kb, is 5.25 X 10–4. Methylamine forms salts such as methylammonium nitrate, (CH3NH3+)(NO3–).
(a) Calculate the hydroxide ion concentration, [OH–], of a 0.225–molar aqueous solution of methylamine. (2pts)
Kb=[CH3NH3 +
][OH−] [CH3NH2]
Kb=5.25×10−4= [X][X] [0.225−X]≃
X2 0.225
Approximation rule, x is eliminated compared to 0.225 X = [OH–] = 1.09x10–2 M
solved using quadratic: X = [OH–] = 1.06x10–2 M
(b) Calculate the pH of a solution made by adding 0.0100 mole of solid methylammonium nitrate to 120.0 milliliters of a 0.225–molar solution of methylamine. Assume no volume change occurs. (2pts)
(c) How many moles of either NaOH or HCl (state clearly which you choose) should be added to the so-lution in (b) to produce a soso-lution that has a pH of 11.00? Assume that no volume change occurs. (3pts) HCl must be added.
(d) A volume of 100. milliliters of distilled water is added to the solution in (c). How is the pH of the solution affected? Explain. (2pts)
The ratio does not change in this buffer solution with dilution, therefore, no effect on pH. Method B
11.00=10.72+log[base] [acid]; log
[base] [acid]=0 .28 [base]
[acid]=1.905=
(0.225−X)
(0.0833+x); X=0.0227M 0
.0227 M x 0.120 L = 2.73x10–3 mol HCl
Method C
pOH=pKb+log[acid]
[base]; pKb=3.28
pOH=3. 28+log(0. 0833)
(0. 225) =2 .85; pH=11 .15
Method A
Kb=5.25×10−4=[0 .0833+X][0 . 0010] [0 .225−X]
X = 0.0228 M
0.0228 mol/L x 0.120 L = 2.74x10–3 mol HCl
Method A
[CH3NH3+] = 0.0100 mol / 0.1200 L = 0.0833 M or CH3NH2 = 0.120 L x 0.225 mol/L = 0.0270 mol
Kb=5.25×10−4=[0. 0833+X][X] [0.225−X] ≃
0.0833X 0 .225 X = [OH–] = 1.42x10–3 M; pOH = 2.85; pH = 11.15
Method B
pH=pKa+log[base] [acid]
Ka=1×10
−14
5.25×10−4=1.91×10
−11; pK
a=10.72
pH=10.72+log(0.225)
(0.0833)=11.15
CH3NH2 CH3NH3+ OH–
Initial 0.225 0 0
Change –X +X +X
Net Ionic Equation Writing:
A. Equimolar solutions of sodium hydroxide and acetic acid are mixed.
(ii) What are the spectator ions?
_______Na+ _(sodium ion)_ _______________________________________________________
B. Pure ethyl alcohol and pure acetic acid are mixed.
(ii) What type of organic molecule is created?
_______ester (via esterification)__________________________________________________
C. A solution of sulfuric acid is mixed with a solution of ammonia.
(ii) What is the conjugate acid in the reaction?
_______NH4+ ______________ _______________________________________________________ (i) Balanced equation:
OH¯(aq) + CH3COOH(aq) ↔ CH3COO¯(aq) + H2O(l)
(ii) Balanced equation:
H+(aq) + NH
3(aq) ↔ NH4+(aq)
(i) Balanced equation:
