HCCS CHEM 1405
PRACTICE EXAM III (Ch. 7-9)
Multiple Choice: Choose the best answer. Key is highlighted in bold characters.
1. BaCl2 (aq) can be described as __________________ because BaCl2 (s) is ______ in water.
A) the solute in a heterogeneous mixture, inoluble
B) the solute in a heterogeneous mixture, soluble
C) the solute in a homogeneous mixture, soluble
D) the solute in a homogeneous mixture, inoluble
Hint: p. 175. Definition.
2. Which of the following compounds is insoluble in water?
A) BaCl2
B) AgBr
C) KCl
Hint: Solubility rules, Table 7.3, p. 177. Also see Q 13 & 24.
B) AlCl3
3. Which of the following compounds is soluble in water?
A) Mg(OH)2
B) AgNO3
C) CaSO4
Hint: Solubility rules, Table 7.3, in page 177. Also see Q 12 & 24.
D) PbCl2
4. What is the % w/w concentration for a solution that contains 100.0 g of water and 3.45 g of solute?
A) 3.33%
B) 3.45%
C) 45.3%
D) 54.3%
Hint: p.p. 178-179, example 7.1 or apply
formula %w/w = {mass of solute (g) / mass of solution (g)} x 100; %w/w = {mass of solute (g) / 100g
mass of solution (g)} x 100%
5. What is the volume of a 10% w/v aqueous solution that contains 37.8 g of HNO 3?
A) 3.78 mL
B) 37.8 mL
C) 378.0 mL
D) 3780.0 mL.
Hint: p.p. 181-182, example 7.5 or apply
formula %w/v = {mass of solute (g) / volume of solution (mL)} x 100;; %w/v = {mass of solute (g) /
100 ml of solution } x 100%
6. How many grams of hydrochloric acid, HCl, are there in 300.0 mL of a 4.000 M solution?
A) 20.3 g
B) 36.6 g
C) 43.74 g
D) 300 g.
Hint: p.p. 182-184. See Q 16. Remember to convert mL to L by apply 1000 mL = 1 L. Or apply
formula molarity = mole of solute/ liter of solution. Thus, grams of solute = molarity x liter of solution
x molar mass of solute.
7. What respectively is the molarity of chloride ion, Cl- , and barium ion, Ba2+, in 500.0 mL of a 0.100
M barium chloride, BaCl2 solution?
A) 0.200 M barium ion and 0.100 M chloride ion.
B) 0.200 M chloride ion and 0.100 M barium ion.
C) 0.500 M barium ion and 0.500 M chloride ion.
D) 0.500 M barium ion and 1.000 M chloride ion.
Hint: Solution stoichiometry. (i.e looking the coefficients in a balanced chemical equation or by simply
looking at the chemical formula of sodium sulfate and its component ions.) Note that the coefficients
in a balanced chemical equation represent the mole or molarity relationship among each substance.
8. How many moles, respectively, of sulfate ion and sodium ions in 0.4 L of a 3 M sodium sulfate,
Na2SO4, solution?
A) There are 1.2 moles of sulfate ion and 2.4 moles of sodium ion.
B) There are 2.4 moles of sulfate ion and 1.2 moles of sodium ion.
C) There are 1.2 moles of sulfate ion and 1.2 moles of sodium ion.
D) There are 2.4 moles of sulfate ion and 2.4 moles of sodium ion.
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Hint: p.p. 181-182 and stoichiometry (i.e looking the coefficients in a balanced chemical equation or
by simply looking at the chemical formula of sodium sulfate and its component ions.)
9. What is the molarity of a 0.1 L solution in which 7.29 grams of HCl is dissolved?
A) 5 M
B) 4 M
C) 3M
D) 2 M
Hint: p.183. Example 7.9. Definition or apply formula molarity = mole of solute/ liter of solution,
where mole = grams/ molar mass.
10. What is the final volume of NaCl (aq) after diluting a 0.10 L of 2.0 M NaCl (aq) into a 0.5 M
solution?
A) 0.05 L
B) 0.2 L
C) 0.4 L
D) 1L
Hint: p.p. 184-185. Examples 7.11 & 7.12. Definition: the mole number of solute does not change
before and after dilution, so applying the formula Mi x Vi = Mf x Vf where c means concentrated
solution and d means dilute solution; M means concentration in molarity and V means volume in liter.
11. Which of the following statement is wrong?
A) A saturated solution exhibits equilibrium between the dissolved and the undissolved solute.
B) An unsaturated, or say, homogeneous solution, is the solution that frequently seen in our daily life,
can dissolve more soluble solute if added.
C) The solubility of most of solid compounds increases with increasing temperature.
D) The solubility of gases decreases with increasing temperature.
E) A solution must be in a liquid form, that is, it can neither be in gaseous nor solid form.
Hint: p.p. 174-176.
12.Which of the following statement is false?
A) Magnesium hydroxide and calcium sulfate precipitate when magnesium sulfate is added to a
saturated calcium hydroxide solution.
B) Silver chloride precipitates when sodium chloride is added to a saturated silver nitrate solution.
C) Potassium chloride precipitates when potassium hydroxide is added to an ammonium
chloride solution.
D) HgBr is insoulble in water while HgBr2 is soluble in water.
Hint: Solubility rules, Table 7.3, in page 177. See Q 12 & 13.
13. What is the ionic (or say net-ionic) equation for AgNO3
A) Ag+ (aq) + Cl- (aq) AgCl (s)
B) Ag+ (aq) + Cl- (aq) AgCl (aq)
C) NO3- (aq) + H+ (aq) HNO3 (aq)
D) NO3- (aq) + H+ (aq) HNO3 (l)
E) AgNO3
O3 (aq)
O3 (aq)?
This is a very, very important question. Also see p. 188 in-text example and example 7.14.
Hint: p.p. 188-189 & must consult solubility rules, Table 7.3, in page 177. Remember that in writing
the net-ionic equation, one must separate the component ions in the soluble strong electrolytes,
which are labeled as (aq). Do not separate the component ions in the weak electrolytes as there are
less than 3% of ions are dissociated. Also do not alter the chemical formulas of solids (s) , liquids (l)
and gases (g), that is, leave them alone.
Since AgNO3 (aq), HCl (aq) and HNO3 (aq) are all soluble strong electrolytes and thus we must
separate them into their component ions in addition to considering the stoichiometry. As the equation
is already balanced, we can simply rewrite it as
Ag+ (aq) + NO3- (aq) + H+ (aq) + Cl- (aq) AgCl (s) + H+ (aq) + NO3- (aq)
Here, we see NO3- (aq) and H+ (aq) appear at both sides (reactant- and product-side) identically, and
thus we can cancel them to simplify the equation. Thus, we call NO 3- (aq) and H+ (aq) as spectator ions
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because they do not actually participate in the chemical reaction; they are present simply to maintain
the electrical neutrality.
Here, we call AgNO3
O3 (aq) as the molecular equation because
there are no charges shown in any of the chemical formula in the equation. We call Ag+ (aq) + NO3(aq) + H+ (aq) + Cl- (aq) AgCl (s) + H+ (aq) + NO3- (aq) as the complete ionic equation because we
see the spectator ions in the equation. We call Ag+ (aq) + Clnet-ionic equation
because it eliminate the spectator ions.
Another example: The molecular equation: NH3
H4Cl(aq)
The net-ionic equation: NH3 (aq) + H+ (aq) NH4 + (aq)
The spectator ion is Cl - .
Hint: p.p. 187-188 & must consult solubility rules, Table 7.3, in page 177. Be very carefully that even
though NH3 (aq) is labeled as (aq), it is a weak electrolyte. Thus, we must leave it alone as NH3.
14. The average osmotic pressure of blood is 13.4 atm at 30 oC. What concentration of glucose
(nonelectrolyte, i = 1) will be isotonic (p. 195) with blood?
A) 0.539 M
B) 0.431 M
C) 0.259M
D) 0.134 M
Hint: p.p. 192-194. Or apply the formula osmotic pressure = i x M x Rx T, where M is molarity, R is
the ideal gas constant as 0.082 atm.L/mol.K, and T is temperature in K.
Thus 13.4 = 1 x M x 0.082 x (30 + 273).
Note that i is the van Hoff’s factor or say the factor that corrects for the number of moles of particles
per mole of solute. For example,
i = 1 for nonelectrolytes like glucose because it does not form ions in solutions.
i = 2 for NaCl, KOH, CaCl2, KNO3, etc. because they are soluble strong electrolytes that do form ions
in aqueous solutions.
i = 3 for CaCl2, Ba(OH)2, Na2SO4, etc. because they are soluble strong electrolytes that do form ions in
aqueous solutions.
i = 4 for AlCl3, etc. because it is a soluble strong electrolyte that does form ions in aqueous solutions.
1 < i < x+y for soluble weak electrolytes, MxWy. For example, i is between 1 and 2 for CH3COOH
because it dissociates about 1-3% into CH3COO- and H+ ions in aqueous solutions.
15. In the reaction 3O2(g) 2O3(g) (ozone), the rate of appearance of ozone was 1.50 mol/ min. What
was the rate of disappearance of oxygen?
A) 1.50 mol/ min
B) 2.25 mol/ min
C) 3.50 mol/ min
D) 4.50 mol/ min
16. The activation energies for each of two reactions were found to be ( a) 24 kJ and ( b) 53 kJ. If the
temperatures of both reactions are identical, which has the greater rate of reaction?
(A) Reaction a
(B)Reaction b
(C) the two reactions have the same reaction rate
17. Write the equilibrium- constant expressions for each of the following reactions:
(A) [NH3][CO2]
(B) [NH3][CO2]/[NH2COONH4]
(B) (C) [NH3]2[CO2]/[NH2COONH4] (D) [NH3]2[CO2]
18. Write the equilibrium- constant expressions for each of the following reactions:
(A)1/[O2]0.5 (B) [Cu2O]/[O2]0.5[Cu]2
(C) [Cu2O]/[O2][Cu]
(D) 1/[O2]
19. In the reaction NH4Cl(s)NH3(g) +Cl2(g) the equilibrium concentrations were found to be
[NH3]=[Cl2]=3.71x 10-3M. Calculate the value of the equilibrium constant.
(A) 5.67x 10-3M
(B) 1.92x 10-4M (C) 1.38x 10-5M (D) 3.71x 10-6M
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19. For the following reaction,ΔH>0
(a) LeChatelier’s principle states that when the temperature is raised in a reaction at
equilibrium, the equilibrium shifts to lower the temperature. In this reaction, that means
the equilibrium shifts to the right (products).
(b) LeChatelier’s principle states that when a reactant or product is added to a reaction at
equilibrium, the equilibrium shifts to use up the compound that was added. Therefore,
this reaction will respond by using up more H2O and shift to the right (products).
(c) LeChatelier’s principle states that when a reactant or product is added to a reaction at
equilibrium, the equilibrium shifts to use up the compound that was added.Therefore, this
reaction will respond by using up more of the products and shift to the left (reactants).
(d) LeChatelier’s principle states that when the temperature is lowered in a reaction at
equilibrium, the equilibrium shifts to raise the temperature. In this reaction, that means
that the equilibrium shifts to the left (reactants).
(e) The addition of a catalyst lowers the activation energy of a reaction, which alters the
rate of both the forward and reverse reactions. Therefore, this change does not alter the
equilibrium.
Hint:
20. Calculate the hydronium ion concentration of a 0.020 M aqueous solution of
Ca( OH)2 .
(A) 2.5x 10-3 M (B) 1.5x 10-13 M (C) 2.5x 10-13 M (D) 2.5x 10-16 M
Hint: [ H3O+ ] =2.5x 10-13 M
21. Calculate the pH of 0.035M Ba(OH)2 solution.
(A)3.68
(B)5.89
(C)9.54
(D)12.84
Hint: the bases in this question is a strong base, which means that every base molecule
dissociates into OH– ions and a positive counter ion. Thus, [OH–] will equal the
concentration of the acid times the number of OH– ions per molecules.
22.
Identify the conjugate base of formic acid (HCOOH) in the reaction
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(Section 9.5)
HCOO-
+
(A) NH3
(B)
(C) NH4
An acid is a compound that donates a proton; its conjugate base is the compound that
is formed upon loss of a hydrogen. Both HCOOH and NH4
+ give up a proton to form HCOO– and NH3 respectively. Therefore HCOOH and HCOO–
are one conjugate acid–base pair, while NH4+ and NH3 are another.
23. Identify the conjugate acid-base pairs in the following equation:
24.The pKa=4.76 of acetic acid is , what’s the Kb for the hydrolysis of acetate ion in
water at 25C?
(A)5.75x1010 (B)5.75x10-10 (C)2.35x10-13 (D)5.75x10-16
Hint:KaKb=Kw
25. Write equilibrium- constant expressions for the complete dissociation of phosphoric
acid in water.
26. Predict whether aqueous solutions of the following salts are acidic, neutral, or basic: (
a) Ca(NO3 )2 ( calcium nitrate); ( b) NH4NO3 ( ammonium nitrate); ( c) KCN ( potassium
cyanide); ( d) CH3COONa ( sodium acetate); ( e) ( HCOO 2Mg ( magnesium formate).
Hint: ( a) Neutral; ( b) acidic; ( c) basic; ( d) basic; ( e) basic
27. Calculate the pH of an aqueous solution consisting of 0.050 M K 2 HPO 4 and 0.075
M KH2PO4 .
(A)2.34
(B)5.69
(C)7.02
(D)9.87
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28. What was the molar concentration of 20.0 mL of an aqueous H3PO4 solution that
required 42.3 mL of a 0.0850 M standard solution of KOH for complete neutralization?
(A)0.060
(B)0.080
(C)1.20
(D)10.00
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