What is the correct name and
bonding of BF3?
0
1. boron trifluoride,
covalent compound
2. boron trifluoride,
ionic compound
3. boron fluoride,
covalent compound
4. boron fluoride, ionic
compound
What is the correct name and bonding
of BF3?
Answer:
Covalent molecule since all atoms are
nonmetal.
Does not contain polyatomic ions.
1 boron atom and 3 fluorine atoms, use
prefixes, except mono of first element.
→Boron trifluoride
1
You have a solution of 0.10 M F– and 0.10 M CO32-.
You add 0.10 M magnesium nitrate dropwise into the
solution. Ksp for MgF2 is 7.4 x 10–9 and for MgCO3 is
3.5 x 10–8. Which of the following will precipitate first?
0
1. magnesium nitrate
2. magnesium
carbonate
3. magnesium fluoride
4. cannot be
determined from
the information
given
5. none of these
You have a solution of 0.10 M F– and 0.10 M CO32-. You
add 0.10 M magnesium nitrate dropwise into the
solution. Ksp for MgF2 is 7.4 x 10–9 and for MgCO3 is 3.5
x 10–8. Which of the following will precipitate first?
Answer:
Ksp (MgF2) = [Mg2+][F-]2
7.4 x 10-9 = (x)(2x)2 → x=1.2 x 10-3 M
Ksp (MgCO3) = [Mg2+][CO32-]
3.5 x 10-8 = (x)(x) → x=1.9 x 10-4 M
x= [Mg2+]=solubility of Mg salt
Smaller x is less soluble - will precipitate first
→ MgCO3
2
A comparision of 1.0 M HOCl and 0.010 M
HOCl would be expected to show ____________.
1.
2.
3.
4.
0
that the percent ionization
in both solutions is the
same
that the percent ionization
of 1.0 M HOCl is greater
than that of 0.010 M HOCl
that the percent ionization
of 0.010 M HOCl is greater
than that of 1.0 M HOCl
that a comparison of the
percent ionization is not
possible for the two
solutions
A comparision of 1.0 M HOCl and 0.010 M
HOCl would be expected to show ____________.
Answer:
Ka is independent of molarity/initial
concentration.
Less concentrated solution (more dilute) will
have higher number of molecules of products
to reach Ka.
Higher number of molecules of products means
higher percentage dissociation.
3
Which of the following Ka values belongs
to the weakest acid?
1.
2.
3.
4.
5.
0
1.4 x 10-4
1.8 x 10-5
4.0 x 10-4
6.2 x 10-10
6.4 x 10-5
Which of the following Ka values belongs
to the weakest acid?
Answer:
Smaller Ka means less dissociation →
weaker acid.
4
It is desired to prepare a system that functions as 0
a buffer at a pH of 5.6. Of the following systems,
the best choice would be ___________________.
1. HC2H3O2 (pKa = 4.7),
C2H3O2– (pKb = 9.3)
2. HOBr (pKa = 8.6),
OBr– (pKb = 5.4)
3. H2PO4– (pKa = 7.2),
HPO42– (pKb = 6.8)
4. NH2OH2+ (pKa = 5.9),
NH2OH (pKb = 8.1)
It is desired to prepare a system that functions as
a buffer at a pH of 5.6. Of the following systems,
the best choice would be ___________________.
Answer:
pH = pKa + log([A-]/[HA])
Best buffer has [A-] = [HA]
Choose acid/salt combination with pKa closest to
the desired pH
→ NH2OH2+ (pKa = 5.9), NH2OH (pKb = 8.1)
5
Addition of NaNO2 to an aqueous solution
of HNO2will cause _______________.
0
1. both [NO2–] and the
pH to increase
2. both [NO2–]and the
pH to decrease
3. both [HNO2] and the
pH to decrease
4. [HNO2] to decrease
and the pH to
increase
Addition of NaNO2 to an aqueous solution
of HNO2will cause _______________.
Answer:
At equilibrium
HNO2 (aq) ! H+ (aq)+ NO2- (aq)
Addition of NO2- from NaNO2 will
increase [NO2-] and cause the
equilbrium to shift to the left, increasing
[HNO2], and decreasing [H+], which
increases pH
6
The indicator phenol red (pKa =7.9) can be used
to detect the equivalence point for the titration of 0
a strong base with a strong acid, but not for the
titration of a weak base with a strong acid. This is
because ______________________.
1.
2.
3.
4.
the weak acid product is not
strong enough to cause a color
change in the indicator
the color change in the
indicator occurs before the
equivalence point is reached
the color change in the
indicator occurs after the
equivalence point is reached
the indicator changes color
twice during the course of the
titration
The indicator phenol red (pKa =7.9) can be used
to detect the equivalence point for the titration of
a strong base with a strong acid, but not for the
titration of a weak base with a strong acid. This is
because ______________________.
Answer:
The indicator will change color at its pKa.
At the equivalence point of a titration of a
weak base and strong acid, the pH is
determined by the pKa of the conjugate
acid of te weak base.
If this acid is weak, or the pKa is too small,
the pH will be too high to induce a
color change.
7
1.
2.
3.
4.
5.
Which of the following
statements is true?
0
When two opposing
processes are proceeding at
identical rates, the system is
at equilibrium.
Catalysts are an effective
means of changing the
position of an equilibrium.
The concentration of the
products equals that of the
reactants, and is constant at
equilibrium.
An endothermic reaction
shifts toward reactants
when heat is added to the
reaction.
None of the above
statements is true.
Which of the following
statements is true?
Answer:
At equilibrium the rates of the forward and
reverse reactions are equal, resulting in a
macroscopically static and microscopically
dynamic system.
Catalysts affect rate, but not the position of the
equilibrium.
Concentrations of products and reactants are
constant, but not necessarily equal.
An endothermic reaction requires heat to
proceed, so an increase in temperature would
shift the reaction towards the products.
8
According to the Bronsted -Lowry
definition, an acid is _________.
1.
2.
3.
4.
5.
0
a substance that increases
the hydroxide ion
concentration in a
solution.
a substance that increases
the hydrogen ion
concentration in a
solution.
a substance that can
accept a proton from
another species in
solution.
a substance that can
donate a proton to
another species.
an electron pair acceptor.
According to the Bronsted -Lowry
definition, an acid is _________.
Answer:
Bronsted-Lowry defined an acid as a
proton donor and a base as a proton
acceptor.
Effects on the [H+] or [OH-] (compared to
water on it’s own) may occur, but are
not the definition.
9
In the following reaction, which species is
the reducing agent?
0
3Cu (s) + 6H+ (aq) + 2HNO3 (aq) → 3Cu2+ (aq) + 2NO (g) + 4H2O (l)
1.
2.
3.
4.
5.
H+
Cu2+
N in NO
Cu
N in HNO3
In the following reaction, which species is
the reducing agent?
3Cu (s) + 6H+ (aq) + 2HNO3 (aq) → 3Cu2+ (aq) + 2NO (g) + 4H2O (l)
Answer:
The reducing agent is the reactant that is
oxidized.
Oxidation: Cu (s) → Cu2+ (aq) +2eReducing agent → Cu (s)
10
What is the percent H by mass
of NH4HCO3?
1.
2.
3.
4.
5.
0
5.1%
6.3%
9.5%
11.9%
none of these
What is the percent H by mass
of NH4HCO3?
Answer:
Atomic mass of NH4HCO3
= 14+(5 x 1)+12+(3 x 16) = 79 amu
Mass of H per molecule= 5 x 1 = 5 amu
% mass H = (5/79) x 100% = 6.3%
11
What is the sum of the coefficients in the
balanced reaction between lead(II) nitrate
and potassium iodide?
1.
2.
3.
4.
5.
0
2
3
4
5
6
What is the sum of the coefficients in the
balanced reaction between lead(II) nitrate
and potassium iodide?
Answer:
Pb(NO3)2 + 2KI → PbI2 + 2KNO3
Sum of coefficients = 1 + 2 + 1 + 2 = 6
12
The value of Ksp for AgI is 1.5 x 10–16. Calculate
the solubility, in moles per liter, of AgI in a 0.50
M NaI solution.
1.
2.
3.
4.
5.
0
1.7 x 10–8 mol/L
1.2 x 10–8 mol/L
2.8 x 10-9 mol/L
3.0 x 10–16 mol/L
7.5 x 10–17 mol/L
The value of Ksp for AgI is 1.5 x 10–16. Calculate
the solubility, in moles per liter, of AgI in a 0.50
M NaI solution.
Answer:
Initial
Change
Equilibrium
[Ag+]
0
+x
x
[I-]
0.50
+x
0.50 + x
Ksp=1.5 x 10-16 = [Ag+][I-]=(x)(0.50 + x)
≈ (x)(0.50) = 1.5 x 10-16
x = [Ag+] = 3 x 10-16 M = solubility of AgI
13
In a 0.20 M aqueous solution of HC3H3O2,
[H3O+] is found to be 3.3 x 10–3 M. Ka for
this acid is therefore ________________.
1.
2.
3.
4.
5.
0
3.3 x 10–3
1.7 x 10–2
8.3 x 10-2
2.2 x 10–6
5.5 x 10–5
In a 0.20 M aqueous solution of HC3H3O2,
[H3O+] is found to be 3.3 x 10–3 M. Ka for
this acid is therefore ________________.
Answer:
HC3H3O2 (aq) ! H+ (aq)+ C3H3O2- (aq)
Ka= [H+][C3H3O2-]/[HC3H3O2]
For every H+ formed, one C3H3O2- is also
formed
Ka = (3.3 x 10-3)2 / 0.20 = 5.5 x 10-5
14
Consider the titration of 100.0 mL of 0.250
M aniline (C6H5NH2; Kb = 3.8 × 10–10) with
0.500 M HCl. Calculate the pH of the
solution at the equivalence point.
1.
2.
3.
4.
5.
11.62
8.70
2.68
–0.85
none of these
Consider the titration of 100.0 mL of 0.250 M aniline
(C6H5NH2; Kb = 3.8 × 10–10) with 0.500 M HCl. Calculate
the pH of the solution at the equivalence point.
Answer:
At equivalence point, equal moles of acid and base are present and pH determined by
conjugate acid of the aniline
C6H 5 NH2
HCl
C6H5NH3 +
Initial (mmol)
25.0
25.0
0
Change (mmol)
-25.0
-25.0
+25.0
Final (mmol)
0
0
25.0
Final volume = 100.0 mL + (25mmol/0.500M) = 100.0 mL + 50.0 mL = 150.0 mL
C6H 5 NH3+ (aq)
Initial
Change
Equilibrium
!
C6H 5NH2 (aq) + H + (aq)
[C6H5NH3+]
0.167
-x
0.167 - x
[C6H5NH2]
0
+x
x
[H+ ]
0
+x
x
Ka = Kw / Kb = 1 x 10-14 / 3.8 x 10-10 = 2.6 x 10-5
Ka=2.6 x 10-5 = [H+ ][C6H 5 NH2]/ [C6H 5 NH3+] =x2/0.167
x = [C6H5NH2]=[H+]=2 x 10-3 M → pH = -log(2 x 10-3) = 2.67
15
Consider the reaction 2 BrCl (g) !Br2 (g) + Cl2 (g)
at T = 25ºC. This reaction has an equilibrium
constant K = 0.0172 at this temperature. If PBrCl=
0.400 atm, PBr2= 0.050 atm, and
PCl2= 0.050 atm, the reaction _________.
0
1. will proceed by
forming additional Br2
(g) and Cl2 (g).
2. will proceed by
forming additional
BrCl (g).
3. is at equilibrium.
4. none of the above
Consider the reaction 2 BrCl (g) ! Br2 (g) + Cl2 (g) at
T = 25ºC. This reaction has an equilibrium constant
K = 0.0172 at this temperature. If PBrCl= 0.400 atm,
PBr2= 0.050 atm, and
PCl2= 0.050 atm, the reaction _________.
Answer:
Kp= PBr2 PCl2 / (PBrCl)2
Qp = (0.050 atm)(0.050 atm)/(0.400 atm)2
Qp = 0.0156
Q p < Kp
so reaction will shift to the right (products)
16
Calculate the mass in grams of 3.65 x 1020
molecules of SO3.
1.
2.
3.
4.
5.
0
6.06 x 10-4 g
4.85 x 10-2 g
20.6 g
1650 g
none of the above
Calculate the mass in grams of 3.65 x 1020
molecules of SO3.
Answer:
3.65 x 1020 molecules x (1 mol / 6.02 x 1023 molecules
= 6.06 x 10-4 moles of SO3
6.06 x 10-4 moles SO3 x (80 g / 1 mol SO3)
= 4.85 x 10-2 g SO3
17