Recitation Activity 4 (Chem 121)
1.
Chapter 4
If you mixed together aqueous solutions of the following reactants, in which cases would
you expect to observe a precipitation reaction? If a reaction does take place write a
balanced chemical equation and the net ionic equation corresponding to the reaction.
a. Lead (II) nitrate + Copper (II) chloride
Pb(NO3 )2 (aq) + CuCl2 (aq) → PbCl2 (s) + Cu(NO3 )2 (aq)
[Full Equation]
2+
–
Pb (aq) + 2Cl (aq) → PbCl2 (s) [Net Ionic Equation]
b. Sodium hydroxide + Barium hydroxide
NaOH (aq) + Ba(OH)2 (aq) → No Reaction (both products are soluble)
2. Which combination will produce a precipitate?
A)
B)
C)
D)
E)
NaC2 H3 O2 (aq) and HCl (aq)
NaOH (aq) and HCl (aq)
AgNO3 (aq) and Ca(C2 H3 O2 )2 (aq)
KOH (aq) and Mg(NO3 )2 (aq)
NaOH (aq) and HCl (aq)
3. Oxidation cannot occur without _______________________. Explain?
(a) air
(e) water
(b) oxygen
(c) reduction
(d) acid
4. Please provide the oxidation state for each element in the following equation and indicate
which element is oxidized and which is reduced:
0
+1 +1 -2 +1 -2
+1 +5 -2 +1 -1
P4
+
HClO + H2 O
à
H3 PO4 + HCl
Ox: P goes from 0 to +5
Red: Cl goes from +1 to -1
5. How many moles of bromide ions are in 0.500 L of a 0.300 M solution of AlBr3 ? How
many bromide ions are there?
3+
AlBr → Al + 3Br
3
0.300 mol AlBr
3
3 mol Br
-
x 0.500 L = 0.150 mol AlBr x
= 0.450 mol Br
3 1 mol AlBr
1L
3
23
- 6.022 x 10 ions
23 0.450 mol Br x
= 2.71 x 10 Br ions
1 mol Br
-
6. You are given the following reagents and labware:
•
•
•
•
•
•
•
A bottle containing zinc powder
A bottle containing liquid mercury
A bottle containing copper powder
A graduated cylinder containing 306 mL of 0.500 M silver nitrate solution
A graduated cylinder containing 150 mL of 6.0 M hydrochloric acid solution
Two empty 500 mL Erlenmeyer flasks
A balloon
Given this inventory explain how you would complete the following tasks. In each case write
a balanced chemical equation to represent the chemical reaction that takes place.
a.
Mix one of the metals (Zn, Hg, or Cu) with one of the solutions in such a way that
you could inflate the balloon. What is the identity of the substance that inflates
the balloon? If you use an excess of the metal and all of the solution available how
many moles of gas would you produce?
Zn(s) + 2HCl(aq) → ZnCl (s) + H (g)
2
2
Hydrogen gas will inflate the balloon.
mol HCl
6.0
x 0.150 L = 0.900 mol HCl
L
1 mol H
0.900 mol HCl x
= 0.45 mol H
2 mol HCl
2
2
b. Mix one of the metals (Zn, Hg, or Cu) with one of the solutions in such a way that
you could obtain metallic silver. If you use an excess of the metal and all of the
solution available what mass of silver would you produce? What ions remain in the
solution at the end of this process?
Looking into the activity series, Zn or Cu could be used, giving the following equation:
Cu(s) + Ag(NO ) (aq) → Cu(NO ) (aq) + Ag(s)
3
2
3
2
This would produce Cu2+ and NO3 - ions.
If Zn was used, Zn2+ and NO3 - ions would be produced.
0.153 mol AgNO x
3
1 mol Ag
107.868 g Ag
x
= 16.5 g Ag(s)
1 mol AgNO
1 mol Ag
3
7. Consider the following titration experiment (note: there are parallels with the
procedures you will follow in experiment #7).
a.
A strontium hydroxide solution is prepared by dissolving 10.45 g of Sr(OH)2
in water to make 50.00 mL of solution. What is the molarity of this
solution?
10.45 g Sr(OH) x
2
1 mol Sr(OH)
= 0.0859 mol Sr(OH)
121.64 g Sr(OH)
2
2
2
0.0859 mol Sr(OH)
= 1.718 M
0.0500 L
2
b. Next the strontium hydroxide solution you prepared in part (a) is used to
titrate a nitric acid solution of unknown concentration. Write a balanced
chemical equation to represent the reaction between strontium hydroxide
and nitric acid solutions.
Sr(OH)2 (aq) + 2HNO3 (aq) ? 2H2 O(l) + Sr(NO3 )2 (aq)
c.
If 23.9 mL of the strontium hydroxide solution was needed to neutralize a
31.5 mL aliquot of the nitric acid solution what is the concentration
(molarity) of the acid?
In a neutralization rxn mol acid = mol base, or mol H+ = mol OH-
1.713 mol Sr(OH)
L
2
1L
x 23.9 mL Sr(OH) x
2
conc. HNO =
3
0.0315 L
= 0.08212 mol OH
x
1000 L
+
0.08212 mol OH = 0.08212 mol H
0.08212 mol H
2 mol OH
+
= 2.61 M
1 mol Sr(OH)
2
-