Download Saturday Study Session 1 1st Class Reactions

Saturday Study Session 1
1st Class Reactions
Things to remember
• Solids, liquids, and gases can NOT be broken into
ions
• SOLUTION – if it says a solution, then it CAN be
broken into ions if it is soluble in water.
• Only ionic compounds can become separate ions
in a solution.
• The 6 strong acids (HCl, HBr, HI, HNO3, HClO4,
H2SO4) and the strong bases (group 1 + OH-and
Ba, Sr, Ca + OH-) are always written as separate
ions.
Electrolytes
Formula Equation (Molecular
Equation)
•
•
•
Gives the overall reaction stoichiometry but not
necessarily the actual forms of the reactants and
products in solution.
Reactants and products generally shown as
compounds.
Use solubility rules to determine which compounds
are aqueous and which compounds are solids.
AgNO3(aq) + NaCl(aq)
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AgCl(s) + NaNO3(aq)
4
Complete Ionic Equation
•
All substances that are strong electrolytes are
represented as ions.
Ag+(aq) + NO3-(aq) + Na+(aq) + Cl-(aq)
AgCl(s) + Na+(aq) + NO3-(aq)
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HCl --- Strong electrolyte
100% dissociation
Net Ionic Equation
•
Includes only those solution components undergoing a
change.
– Show only components that actually react.
Ag+(aq) + Cl-(aq)
•
AgCl(s)
Spectator ions are not included (ions that do not participate
directly in the reaction).
– Na+ and NO3- are spectator ions.
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7
Precipitant
White Precipitant
AgCl, Mg(OH)2,
BaSO4
Opening activity
– Solid calcium oxide is added to water
CaO + H2O  Ca(OH)2 molecular equation
CaO + H2O  Ca2+ + 2 (OH)- net ionic
– Potassium chlorate(s) is heated
2KClO3 2KCl + 3O2 molecular and ionic
– Calcium carbonate(s) is heated
CaCO3 CaO + CO2 molecular and ionic
– Sodium hydroxide(aq) is heated
2NaOH  Na2O + H2O molecular
2Na+ + 2(OH)-  Na2O + H2O
ionic
Aluminum metal is added to solution of copper II
chloride
2Al + 3CuCl2  2 AlCl3 + 3Cu molecular
2Al + 3Cu2+  2Al3+ + 3 Cu ionic
– Fluorine gas is bubbled into solution of sodium bromide
F2 + 2NaBr  2NaF + Br2 molecular
F2 + 2Br-  2F- + Br2
– Solutions of Hydrofluoric acid is added to ammonium hydroxide
HF + NH4OH  NH4F + H2O molecular
HF + NH4OH  NH4 + + F- + H2O ionic
– Butane is burned in Air
2C4H10 + 13O2  8CO2 + 10H2O molecular and ionic
Stoichiometry
• Exercise 1
A 500.0-g sample of potassium phosphate
is dissolved in enough water to make 1.50 L
of solution. What is the molarity of the
solution?
Stoichiometry
• Exercise 1
A 500.0-g sample of potassium phosphate
is dissolved in enough water to make 1.50 L
of solution. What is the molarity of the
solution?
M = Molarity =
moles of solute
liters of solution
1.57 M
Problem #1
• Step 1 - find molecular mass of K3PO4
• 212.274 g/mol
• Step 2 - convert from g of K3PO4 to mol
• 500.0 g / 212.274 g/mol = 2.355 mol
• Step 3 - divide mol by liters to get molarity
• 2.355 mol / 1.50 L = 1.57 M
• Exercise 2
What is the minimum volume of a 2.00 M
NaOH solution needed to make 150.0 mL
of a 0.800 M NaOH solution?
• Exercise 2
What is the minimum volume of a 2.00 M
NaOH solution needed to make 150.0 mL
of a 0.800 M NaOH solution?
60.0 mL
• M1V1 = M2V2
• (2 M)(X) = (0.800M)(150 mL)
• X = (.800M)(150mL)
(2.00M)
X = 60.0 mL
Exercise 3
(Part I)
10.0 mL of a 0.30 M sodium phosphate solution
reacts with 20.0 mL of a 0.20 M lead(II) nitrate
solution (assume no volume change).
– What precipitate will form?
lead(II) phosphate, Pb3(PO4)2
– What mass of precipitate will form?
1.1 g Pb3(PO4)2
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All group I metals are soluble
All nitrates are soluble
2Na3PO4(aq) + 3Pb(NO3)2(aq) → 6NaNO3(aq) + Pb2(PO4)2(s)
6Na+ + 2PO4-3 + 3Pb+2 + 6NO3- → 6Na+ + 6NO3- + Pb3(PO4)2
2PO4-3 + 3Pb+2 → Pb3(PO4)2
•
•
•
•
R
I
C
E
•
•
•
•
0.0040 -3x = 0 therefore x = 0.001333333
0 + 1x = 0.0013333 mol Pb3(PO4)2
Molecular mass of Pb3(PO4)2 equals 811.55 g/mol
0.0013 mol x 811.55 g/mol = 1.1 g of Pb3(PO4)2
2PO4-3 + 3Pb+2 → 1Pb3(PO4)2
.0030
.0040
0
-2x
-3x
+1x
.000033 excess
0
0.0013
Exercise 4
(Part II)
10.0 mL of a 0.30 M sodium phosphate solution
reacts with 20.0 mL of a 0.20 M lead(II) nitrate
solution (assume no volume change).
– What is the concentration of nitrate ions
left in solution after the reaction is
complete?
Nitrate 0.27 M
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• Step 1 – find moles of NO3- present
• 0.200 M Pb(NO3)2 = 2 NO3- x 0.200 M = 0.400 M
• 0.400 mol/L x .0020 L = .00080 mol of NO3-
• Step 2 – find new volume after mixing
solutions
• 20.0 ml + 10.0 ml = 30.0 ml or 0.003 L
• Step 3 - find molarity of NO3• 0.0008 mol / .003 L = 0.27 M
(Part III)
10.0 mL of a 0.30 M sodium phosphate solution
reacts with 20.0 mL of a 0.20 M lead(II) nitrate
solution (assume no volume change).
– What is the concentration of phosphate
ions left in solution after the reaction is
complete?
0.011 M
Excess was .000033 mol / .03 L = .011 M
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24
Which of the following ions form compounds with
Pb2+ that are generally soluble in water?
a)
b)
c)
d)
e)
S2–
Cl–
NO3–
SO42–
Na+
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25
Which of the following solutions contains
the greatest number of ions?
a)
b)
c)
d)
400.0 mL of 0.10 M NaCl.
300.0 mL of 0.10 M CaCl2.
200.0 mL of 0.10 M FeCl3.
800.0 mL of 0.10 M sucrose.
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26
A 0.50 M solution of sodium chloride in an open beaker
sits on a lab bench. Which of the following would decrease
the concentration of the salt solution?
a) Add water to the solution.
b) Pour some of the solution down the sink drain.
c) Add more sodium chloride to the solution.
d) Let the solution sit out in the open air for a couple
of days.
e) At least two of the above would decrease the
concentration of the salt solution.
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27
After completing an experiment to determine gravimetrically
the percentage of water in a hydrate, a student reported a value of
38 percent. The correct value for the percentage of water in the
hydrate is 51 percent. Which of the following is the most likely
explanation for this difference?
a.
Strong initial heating caused some of the hydrate
sample to spatter out of the crucible.
b.
The dehydrated sample absorbed moisture after
heating.
c.
The amount of the hydrate sample used was too small.
d.
The crucible was not heated to constant mass before
use.
e.
Excess heating caused the dehydrated sample to
decompose.
Acid–Base Reactions (Brønsted–Lowry)
“Bro-Pro”
Brønsted deals with Protons
•
•
•
•
•
Acid—proton donor
Base—proton acceptor
Conjugate Acid – formed when base gains a H+
Conjugate Base – formed when acid loses a H+
HF + NH3  NH4+ + FAcid
Base
Conjugate Conjugate
Acid
Base
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29
For the titration of sulfuric acid (H2SO4) with
sodium hydroxide (NaOH), how many moles of
sodium hydroxide would be required to react with
1.00 L of 0.500 M sulfuric acid to reach the
endpoint?
1.00 mol NaOH
2NaOH + 1H2SO4 → Na2SO4 + 2H2O
It takes 2 mol of OH- to react with 1 mol 2H+
(.500M)(1.00L) = .500 mol H+
so .500 mol x 2 =
1.00 mol NaOH
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Titration – Laboratory Procedure
To determine the concentration
of a solution
Redox reaction or acid base
Neutralization reaction
Redox Characteristics
•
•
•
•
Transfer of electrons
Transfer may occur to form ions
Oxidation – increase in oxidation state (loss of
electrons); reducing agent
Reduction – decrease in oxidation state (gain of
electrons); oxidizing agent
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32
Reaction of Sodium and Chlorine
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33
Find the oxidation states for each of the elements
in each of the following compounds:
•
•
•
•
•
K2Cr2O7
CO32MnO2
PCl5
SF4
Find the oxidation states for each of the elements
in each of the following compounds:
•
•
•
•
•
K2Cr2O7
CO32MnO2
PCl5
SF4
K = +1; Cr = +6; O = –2
C = +4; O = –2
Mn = +4; O = –2
P = +5; Cl = –1
S = +4; F = –1
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35
Which of the following are oxidation-reduction
reactions? Identify the substance oxidized and
reduced.
a)Zn(s) + 2HCl(aq)  ZnCl2(aq) + H2(g)
b)Cr2O72-(aq) + 2OH-(aq)  2CrO42-(aq) + H2O(l)
c)2CuCl(aq) CuCl2(aq) + Cu(s)
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Free Response
2003B
Answer the following questions that relate to chemical reactions.
(a) Iron(III) oxide can be reduced with carbon monoxide according to
the following equation.
Fe2O3(s) + 3 CO(g) → 2 Fe(s) + 3 CO2(g)
A 16.2 L sample of CO(g) at 1.50 atm and 200.°C is combined with
15.39 g of Fe2O3(s).
(i) How many moles of CO(g) are available for the reaction?
(ii) What is the limiting reactant for the reaction? Justify your answer
with calculations.
(iii) How many moles of Fe(s) are formed in the reaction?
. A 0.345 g sample of anhydrous BeC2O4, which contains an inert impurity, was dissolved
in sufficient Water to produce 100.0 mL of solution. A 20.0 mL portion of this solution
was titrated with KMnO4(aq). The balanced equation for the reaction that occurs is:
16H+(aq) + 2 MnO4-(aq) + 5 C2O42-(aq)  2 Mn2+(aq) + 10 CO2(g) + 8 H2O(l)
The volume of the 0.0150 M KMnO4(aq) required to reach the equivalence point of the
titration was 17.80 mL
i. Identify the the substance oxidized and the substance reduced in the titration reaction.
ii. For the titration at the equivalence point, calculate the number of moles of each of
the following that reacted:
a. MnO4b. C2O42iii. Calculate the total number of moles of C2O42- that were present in the 100.0 mL of
prepared solution.
iv. Calculate the mass percent of the BeC2O4 in the impure 0.345 g sample