Sample Question Answers - Unit 4
Upon successful completion of this unit, the students should be able to:
4.1 Define the terms solute, solvent, and solution.
1. The substance which gets dissolved.
4.2 Define the terms strong, weak, and non-electrolyte and solve related problems.
1. A substance which, when dissolved, completely separates into ions.
2. Slightly soluble means that only a small amount of solute is able to be dissolved in a
given amount of solvent. Weak electrolyte means that a substance which, when
dissolved (regardless of how much dissolves) only partially separates into ions (only
a few ions form, mostly molecules remain). If something is a weak electrolyte it can
still be very soluble (meaning that a large amount of solute is able to be dissolved in a
given amount of solvent) but only a few of those dissolved particles would separate
into ions. An example of something that is very soluble and also a weak electrolyte is
acetic acid.
3. b
4.3 Define Arrhenius acid and base.
1. A substance which produces hydrogen ions (H+) when dissolved in water.
Example: HNO3 Æ H+(aq) + NO3-(aq)
4.4 Define and list common examples of:
a. strong acids (HCl, HBr, HI, H2SO4, HNO3, HClO4)
b. weak acids (H3PO4, HF, HC2H3O2 [CH3COOH])
c. strong bases (Group 1 hydroxides [LiOH, NaOH, KOH, RbOH, CsOH], and
larger Group 2 hydroxides [Ca(OH)2, Sr(OH)2, Ba(OH)2])
d. weak bases (NH3).
1. acetic acid (HC2H3O2 or CH3COOH), phosphoric acid (H3PO4)
2. A substance which a) completely dissociates into ions when dissolved and b)
increases the concentration of hydroxide ions (OH-) in solution.
4.5 List and/or recognize all ionic compounds and strong acids as examples of strong
electrolytes.
1. NaCl, like all ionic compounds, is a strong electrolyte. Therefore, in solution, the
sodium and chloride ions should be completely dissociated from one another. In the
picture, the blue spheres (Na+) should be separated from the green spheres (Cl-).
4.13 Solve stoichiometry problems involving reactants and/or products in solution.
1.
? mL Ca(OH)2 soln = 3.4 g H2O x
mol H2O 1 mol Ca(OH)2 L Ca(OH)2 soln 1000 mL Ca(OH)2 soln
x
x
x
2 mol H2O
0.62 mol Ca(OH)2
L Ca(OH)2 soln
18.02 g H2O
= 1.5 x 10 2 mL Ca(OH)2 soln
2.
? mL NaOH soln. = 10 mL H2SO4 soln ×
10 -3 L H2SO4 soln 0.202 mol H2SO4 2 mol NaOH
L NaOH soln
x
×
x
mL H2SO4 soln
L H2SO4 soln
1 mol H2SO4 0.136 mol NaOH
x
mL NaOH soln
= 30 mL NaOH soln
10 -3 L NaOH soln
3.
? g Fe 2+ = 39.32 mL KMnO4 soln ×
5 mol Fe 2+
10 -3 L KMnO4 soln 0.03190 mol KMnO4 1 mol MnO4 x
x
×
L KMnO4 soln
1 mol KMnO4 1 mol MnO4 mL KMnO4 soln
= 0.006271 mol Fe 2+
Since all the Fe was converted to Fe 2+ we now know we had 0.006271 mol Fe
0.006271 mol Fe x
55.85 g Fe
= 0.3500 g Fe
mol Fe
0.3500 g Fe
x 100 = 38.78 % Fe
1.1081 g ore
4.
? g C2H5OH = 8.76 mL K2Cr2O7 soln ×
1 mol C2H5OH
10 -3 L K2Cr2O7 soln 0.04988 mol K2Cr2O7 1 mol Cr2O7 2x
x
×
L K2Cr2O7 soln
1 mol K2Cr2O7
2 mol Cr2O7 2mL K2Cr2O7 soln
x
46.07 g C2H5OH
mol C2H5OH
= 0.0101 g C2H5OH
0.0101 g C2H5OH
x 100 = 0.101 % alcohol in blood
10.002 g blood
4.14 Recognize and describe the nature of a precipitation reaction, acid-base reaction,
and redox reaction.
1. redox
2. redox
3. a. acid-base
b. precipitation
c. redox
d. precipitation
4. ___R___
S8(s) + 8O2(g) Æ 8SO2(g)
___P___
2NaCl(aq) + Pb(NO3)2(aq) Æ 2NaNO3(aq) + PbCl2(s)
___R___
CH4(g) + 2O2(g) Æ CO2(g) + 2H2O(l)
___AB__
HCN(aq) + LiOH(aq) Æ LiCN(aq) + H2O(l)
5. Many acid-base reactions (including all those we have covered so far) involve an acid
reacting with a base to produce a salt plus water. Precipitation on the other hand is
when a precipitate (i.e. a solid) forms upon the mixing of two aqueous solutions.
Yes, a reaction can be both. The reaction of sulfuric acid with barium hydroxide
produces the insoluble barium sulfate:
H2SO4(aq) + Ba(OH)2(aq) Æ BaSO4(s) + 2H2O(l)
4.15 Learn and be able to apply the following solubility rules for ionic compounds:
Soluble Ionic Compounds:
1. All common compounds of Group 1 ions (Li+, Na+, K+ etc…) and the
ammonium ion (NH4 +) are soluble.
2. All common nitrates (NO3 -), acetates (CH3COO -), and most perchlorates
(ClO4 -) are soluble.
3. All common chlorides (Cl -), bromides (Br -), and iodides (I -) are soluble, except
those of Ag +, Pb 2+, Cu +, Hg2 2+.
4. All common sulfates (SO4 2-) are soluble, except those of Ca 2+, Sr 2+, Ba 2+, and
Pb 2+.
Insoluble Ionic Compounds:
1. All common metal hydroxides are insoluble, except those of Group 1 and the
larger members of Group 2 (starting with Ca 2+).
2. All common carbonates (CO3 2-) and phosphates (PO4 3-) are insoluble, except
those of Group 1 and NH4 +.
3. All common sulfides are insoluble except those of Group 1, Group 2 and NH4 +.
1. a. insoluble
b. soluble
c. soluble
d. insoluble
2. To a sample of the municipal water you could add an aqueous solution that contains
an anion that would precipitate the Pb2+. For example, you could add a simple salt
solution such as NaCl(aq). The Cl- would combine with the Pb2+ to form PbCl2(s),
providing evidence of the Pb2+ in the water supply.
3. Yes, Fe(OH)2 will form.
4. The key here is that the two compounds you mix together must be soluble in water to
begin with and one of them must contain Zn2+ and the other CrO42-. In each case, for
the counter ion we can use any ion that we know is always soluble when part of an
ionic compound. For example we can pair Zn2+ with nitrate, NO3-, and we can pair
Na+ with the CrO42-. The reaction would be:
Na2CrO4(aq) + Zn(NO3)2(aq) Æ ZnCrO4(s) + NaNO3(aq)
5. Cs+, NH4+
6. This solubility rule, as well as all of those listed above as part of Objective 4.15,
relate to IONIC compounds. Carbon tetrachloride is a molecular compound.
4.16 Predict products and write a balanced molecular equation and net ionic equation
for precipitation reactions and acid-base reactions which are double displacement
reactions.
1. a. Ba2+(aq) + SO42-(aq) Æ BaSO4(s)
b. No precipitate and no reaction.
2. HCN(aq) + OH-(aq) Æ CN-(aq) + H2O(l)
3. Molecular Equation: 2HBr(aq) + Ca(OH)2(aq) Æ CaBr2(aq) + 2H2O(l)
Net Ionic Equation: H+(aq) + OH-(aq) Æ H2O(l)
4. a. products: KCl(aq) and Zn3(PO4)2(s)
b. products: CaCl2(aq) and H2O(l)
c. No reaction
d. products: SrSO4(s) and Al(NO3)3(aq)
5. HC2H3O2(aq) + OH-(aq) Æ C2H3O2-(aq) + H2O(l)
4.17 Identify spectator ions in a chemical reaction.
1.
1. a. Cr3+ and NO3b. All spectator ions (or no spectator ions depending how you look at it). The
important thing to know is that there was no reaction.
+
2. Na
3. Ca2+ and Br5. Ca2+
2. d
4.18 Define titration and perform calculations related to titrations.
1.
? mol HCl = 0.2844 g Na2CO3 x
mol Na2CO3
2 mol HCl
x
= 0.005366 mol HCl
106.0 g Na2CO3 1 mol Na2CO3
10 -3 L HCl soln
42.43 mL HCl soln x
= 0.04243 L HCl soln
mL HCl soln
0.005366 mol HCl
= 0.1265 M HCl
0.04243 L HCl soln
2. HNO2 + KOH Æ KNO2 + H2O
? mol KOH = 25 mL HNO2 soln x
10 -3 L HNO2 soln 0.100 mol HNO2 1 mol KOH
x
x
= 0.0025 mol KOH
mL HNO2 soln
L HNO2
1 mol HNO2
3. H2SO4(aq) + 2NaOH(aq) Æ Na2SO4(aq) + 2H2O(l)
? mol NaOH = 25.43 mL H2SO4 soln x
15.10 mL NaOH soln x
10 -3 L H2SO4 soln 0.567 mol H2SO4 2 mol NaOH
= 0.0288 mol NaOH
x
x
mL H2SO4 soln
L H2SO4 soln
1 mol H2SO4
10 -3 L NaOH soln
= 0.0150 L NaOH soln.
mL NaOH solution
0.0288 mol NaOH
= 1.92 M NaOH
0.0150 L NaOH soln
4. Titration – a technique in which one solution is used to analyze another (often to find
an unknown concentration).
4.19 State and apply the following rules for assigning oxidation numbers:
1. For free elements, the oxidation number is zero. Examples: Na (0), O2 (0), S8
(0).
2. For monatomic ions, the oxidation number is the charge on the ion. Examples:
Na+ (+1), Cl- (-1), Al3+ (+3).
3. For Group 1A elements, the oxidation number is always +1 when they are in
their ionic form. Examples: Na+ (+1), Rb+ (+1).
4. For Group 2A elements, the oxidation number is always +2 when they are in
their ionic form. Examples: Ca2+ (+2), Ba2+ (+2).
5. For oxygen, the oxidation number is usually a -2. Exceptions: as a free
element it is zero; for peroxides it’s a -1 (examples of peroxides: Na2O2, H2O2);
for superoxides it’s -1/2 (examples of superoxides: RbO2, CsO2); with fluorine
it’s positive (examples of oxyfluorine compounds: OF2, O2F2).
6. For hydrogen, the oxidation number is +1 when combined with other
nonmetals and -1 when combined with metals.
7. For fluorine, the oxidation number is always a -1 (except when it is a free
element).
8. For binary compounds (those with two different elements) [and in general when
two elements are covalently bonded], the element with the greater
electronegativity is assigned a negative oxidation number equal to its charge in
simple ionic compounds of that element. Example: In PCl3, Cl is more
electronegative than P and since Cl forms a -1 charge in simple ionic
compounds (such as NaCl, MgCl2), Cl is assigned an oxidation number of -1 in
PCl3.
9. The sum of the oxidation numbers equals the charge on the molecule or ion.
1. a. +2
b. +6
2. a.
b.
c.
d.
+3
-1
+6
-1
4.20 Define oxidation, reduction, reducing agent, and oxidizing agent and solve related
problems.
1. Oxidizing agents get reduced therefore it has gained electrons.
2. Upper right hand corner (not including the noble gases); where the most reactive
nonmetals are like fluorine and oxygen.
3. Oxidation – loss of electrons.
4.21 Identify the following in an oxidation-reduction reaction: element getting
oxidized, element getting reduced, oxidizing agent, reducing agent.
1. a. Cr (because it changes from a +3 oxidation state in Cr(OH)3 to a +6 oxidation
state in CrO42-)
b. ClO32. S2O32- (because sulfur gets oxidized; it changes from a +2 oxidation state in S2O32to a +2.5 oxidation state in S4O62-)
3. oxidized: C (because it changes from a -2 oxidation state in CH3OH to a zero
oxidation state in CH2O)
reduced: Cr (because it changes from a +6 oxidation state in Cr2O72- to a +3
oxidation state in Cr3+)
oxidizing agent: Cr2O72reducing agent: CH3OH
Additional Unit 4 Sample Questions:
1. 2HCl(aq) + Ba(OH)2(aq) Æ BaCl2(aq) + 2H2O(l)
1.25 L HCl soln x
1.35 mol HCl
= 1.69 mol HCl
L HCl soln
0.75 L Ba(OH)2 soln x
1.69 mol HCl x
1.25 mol Ba(OH)2
= 0.938 mol Ba(OH)2
L Ba(OH)2 soln
1 mol Ba(OH)2
= 0.845 mol Ba(OH)2
2 mol HCl
0.938 mol Ba(OH)2 x
have
2 mol HCl
= 1.88 mol HCl
1 mol Ba(OH)2
need
For HCl have < need therefore HCl is limiting reactant.
1.69 mol HCl x
2 mol H2O
= 1.69 mol H2O
2 mol HCl
2. 2AgNO3(aq) + CaCl2(aq) Æ 2AgCl(s) + Ca(NO3)2(aq)
? g AgCl = 30.0 mL AgNO3 soln x
10 -3 L AgNO3 soln 0.657 mol AgNO3 2 mol AgCl
x
x
mL AgNO3 soln
L AgNO3 soln
2 mol AgNO3
x
143.4 g AgCl
= 2.82 g AgCl
mol AgCl
3. 0.10 M KBr because it would have the most ions in solution. KBr is an ionic
compound and is therefore a strong electrolyte. HC2H3O2, acetic acid, is a weak
electrolyte so would have fewer ions in solution. CH3CH2OH, ethyl alcohol, and
C6H12O6, glucose, would have no ions in solution (except for negligible amounts of
H+ and OH- from the ionization of water) because they are nonelectrolytes.
4. Na2SO4(aq) + Pb(NO3)2(aq) Æ 2NaNO3(aq) + PbSO4(s)
23.7 mL Na 2SO 4 soln x
10 -3 L Na 2SO 4 soln 0.179 mol Na 2SO 4
= 0.00424 mol Na 2SO 4
x
mL Na 2SO 4 soln
L Na 2SO 4 soln
30.6 mL Pb(NO 3)2 soln x
0.00424 mol Na 2SO 4 x
10 -3 L Pb(NO 3)2 soln 0.130 mol Pb(NO 3)2
x
= 0.00398 mol Pb(NO 3)2
mL Pb(NO 3)2 soln
L Pb(NO 3)2 soln
1 mol Pb(NO 3)2
= 0.00424 mol Pb(NO 3)2
1 mol Na 2SO 4
0.00398 mol Pb(NO 3)2 x
1 mol Na 2SO 4
= 0.00398 mol Na 2SO 4
1 mol Pb(NO 3)2
For Pb(NO3)2 have < need therefore Pb(NO3)2 is limiting reactant.
0.00398 mol Pb(NO3)2 x
303.3 g PbSO4
1 mol PbSO4
x
= 1.21 g PbSO4
mol PbSO4
1 mol Pb(NO3)2
need
have
Three ions will be left in the solution after the reaction is complete: Na+, NO3-, and SO42-.
The Na+ and NO3- are spectator ions which have undergone no net change. Therefore
whatever their number of moles was at the start of the reaction will be the same as after
the reaction. Their concentrations will be different however because we now have a
different volume because we mixed two solutions together. To get the total volume we
will have to assume additive volumes (therefore the total volume will be 23.7 mL + 30.6
mL = 54.3 mL = 0.0543 L).
0.00424 mol Na2SO4 x
2 mol Na +
= 0.00848 mol Na +
mol Na2SO4
0.00848 mol Na +
= 0.156 M Na +
0.0543 L soln
0.00398 mol Pb(NO3)2 x
2 mol NO3= 0.00796 mol NO3 mol Pb(NO3)2
0.00796 mol NO3 = 0.146 M NO3 0.0543 L soln
Some of the SO42- has precipitated as part of PbSO4 but some remains in solution because
Na2SO4 was the excess reactant. So the number of mol of SO42- left over after the
reaction will be the difference between how much SO42- we started with and how much
precipitated as part of PbSO4.
0.00424 mol Na2SO4 x
1 mol SO4 2= 0.00424 mol SO4 2- before reaction
1 mol Na2SO4
0.00398 mol Pb(NO3)2 x
1 mol PbSO4
1 mol SO4 2x
= 0.00398 mol SO4 2- precipitated
1 mol Pb(NO3)2 1 mol PbSO4
0.00424 mol before - 0.00398 mol precipitated = 0.00026 mol SO4 2- remains in solution
0.00026 mol SO4 22= 0.0048 M SO 4
0.0543 L soln
5. a. Bright because sulfuric acid, H2SO4, is a strong electrolyte.
b. Molecular Equation: H2SO4(aq) + Ba(OH)2(aq) Æ BaSO4(s) + 2H2O(l)
Total Ionic Equation:
2H+(aq) + SO42-(aq) + Ba2+(aq) + 2OH-(aq) Æ BaSO4(s) + 2H2O(l)
Net Ionic Equation: same as Total Ionic Equation
As we add Ba(OH)2, the sulfuric acid gets neutralized and produces solid BaSO4
and H2O, neither of which have mobile ions to conduct the electric current.
Therefore the light begins to grow dim as the concentrations of ions decreases in
solution. When exactly one mole of Ba(OH)2 has been added, the light bulb will
be completely off because only the BaSO4 precipitate and the additional water
will be in solution.
c. Upon adding excess Ba(OH)2(aq) there is no longer H2SO4 available to
neutralize it. Therefore, Ba(OH)2, being a strong electrolyte, will now add Ba2+
and OH- ions into solution so the solution is able to conduct again.