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Transcript
www.apchemsolutions.com
Lecture 10
Aqueous Solutions and Chemical Reactions I
Tutorial
1) Which solution is the best conductor of electricity (electrolyte)? Justify your
answer.
a. 1.0 M Li2(CO3), 1.0 M NH3, or 1.0 M LiCl
All solutions are 1.0M, meaning that they all contain 1 mole of the formula unit in
question per litre of water. NH3 is a covalent compound that does not dissociate into
positive and negative ions; and thus, it does not conduct electricity. Li2(CO3) and LiCl
are both soluble ionic compounds that dissociate in water; however, Li2(CO3) dissociates
into three ions (Li2(CO3)(s) Æ 2 Li+(aq) + CO32-(aq)) whereas LiCl only dissociates into two
ions (LiCl(s) Æ Li+(aq) + Cl-(aq)). As a solutions ability to conduct electricity increases as
the concentration of ions in the solution increases, 1.0M Li2(CO3) is the best conductor.
Write balanced net ionic equations for the reactions in questions 2 through 4. Remember,
compounds containing group 1A ions, nitrate, and ammonium are always soluble.
2) A potassium hydroxide solution is mixed with a solution of zinc nitrate.
The potassium ion is always soluble as it is a Group 1A element. Nitrate is also soluble
with everything. Thus, K+ and NO3- must be spectator ions.
Zn2+(aq) + 2 OH-(aq) Æ Zn(OH)2(s)
3) A solution of lead (II) nitrate is poured into a solution of sodium iodide.
The sodium ion is always soluble as it is a Group 1A element. Nitrate is also soluble
with everything. Thus, Na+ and NO3- must be spectator ions.
Pb2+(aq) + I-(aq) Æ PbI2(s)
4) Solutions of ammonium chloride and silver nitrate are mixed.
Ammonium and nitrate are soluble with everything. Thus, NH4+ and NO3- must be
spectator ions.
Ag+(aq) + Cl-(aq) Æ AgCl(s)
© 2009, 2008 AP Chem Solutions. All rights reserved.
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5) Determine the oxidation numbers on each atom in the following compounds.
a. CF4
This is a covalent compound that does not contain hydrogen or oxygen atoms. Thus we
set the oxidation number of the most electronegative element equal to its charge as an
ion. Fluorine is the most electronegative element. As the fluoride ion has a charge of -1.
Thus, the oxidation number assigned to fluorine in this molecule is also -1. Because
there are four fluorine atoms in the compound and the overall charge on the species is
zero, the oxidation number assigned to carbon is +4.
b.H2SO3
Oxygen has an oxidation number of -2 and hydrogen has an oxidation number of +1.
(3 oxygen atoms) x (-2) = -6
(2 hydrogen atoms) x (+1) = +2
-6 + (+2) = -4
Because there is one Sulfur atom in the compound and the overall charge on the species
is zero, the oxidation number assigned to Sulfur is +4.
-4 + (+4) = 0
6) Identify the substance that is oxidized and the substance that is reduced in each
reaction.
a. Zn(s) + H2SO4(aq) Æ ZnSO4(aq) + H2(g)
Zn(s) is oxidized. Its oxidation number increases form 0 in Zn(s) to +2 in ZnSO4(aq).
ZnSO4(aq) is the same as Zn2+(aq) + SO42-(aq). The oxidation number of a monoatomic ion
(Zn2+) is equal to it charge as an ion.
H2SO4(aq) is reduced. The oxidation number on hydrogen is reduced from +1 in
H2SO4(aq) to 0 in H2.
b.Zn(s) + Cu2+(aq) Æ Zn2+(aq) + Cu(s)
Zn(s) is oxidized. The oxidation number increases form 0 to +2.
Cu2+ is reduced. Theoxidation number is reduced from +2 to 0.
© 2009, 2008 AP Chem Solutions. All rights reserved.
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