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MHS  AP Chemistry
Name ________ KEY_______________
Period ___ Date __________
4  Reactions in Aqueous Solution
Station 1 – SOLUBILITY RULES
Circle the ionic compounds that are Insoluble (i.e. circle the precipitates): bold red are insoluble
MgI2
CuSO4
NH4Cl
Fe(OH)3
CsI
AgCl
CdS
CuCl2
PbSO4
Ba(OH)2
Na2SO4
NH4CrO4
Sr(NO3)2
Hg2I2
Na2CrO4
BaCO3
PbBr2
Ca(ClO3)2
AgCH3COO
MgS
Station 2 – WRITE THE IONIC EQUATION
For the following molecular equations, write the ionic equation (include phases, s, g, l or aq), underline the
spectators, and write the net ionic equation.
HI (aq) + NaOH (aq)  NaI (aq) + H2O (l)
H+ + I- + Na+ + OH-  Na+ + I- + H2O (l)
H+ + OH-  H2O (l)
Strong acid & strong base
CuSO4 (aq)
+ Sr(NO3)2 (aq)
 Cu(NO3)2 (aq)
+ SrSO4(s)
Cu2+ + SO42- + Sr2++ 2 NO3-  Cu2+ + 2 NO3- + SrSO4(s)
SO42- + Sr2  SrSO4(s)
2HF (aq)
+ Ba(OH)2 (aq)  BaF2 (aq) + 2 H2O (l)
2 HF (aq) + Ba2+ + 2 OH-  Ba2+ + 2 F- + 2 H2O (l)
HF (aq) + OH-  F- + H2O (l)
(HF is a weak acid so it won’t completely dissociate – so you don’t break it into ions in the ionic equation)
Station 3 – PREDICT THE PRODUCTS
Predict the products in these molecular equations. Balance the whole reaction and indicate (s), (l), (g), or (aq):
CuCO3 (aq) + HF (aq)  CuF2 (aq) + H2CO3  CuF2 (aq) + H2O (l) + CO2(g)
Ca(NO3)2 (aq) + (NH4)2SO4 (aq)  CaSO4 (s) + 2 NH4NO3 (aq)
2 CH3COOH (aq) + Zn(OH)2 (aq)  2 H2O (l) + Zn(C2H3O2)2 (aq)
Mg(OH)2 (s) + 2 (NH4)Cl (aq)  MgCl2 (aq) + 2 NH4OH**
MgCl2 (aq) + 2 NH3 (g) + 2 H2O (l)
**NH4OH is one of the 3 gas forming products
6 HBr (aq) + Al2S3 (aq)  3 H2S (g) + 2 AlBr3 (aq)
Zn(s) + PbSO4 (aq)  ZnSO4 (aq) + Pb (s)
2 Al(s) + 3 Ni(NO3)2 (aq)  2 Al(NO3)3 (aq) + 3 Ni (s)
Station 4 – MOLARITY PROBLEMS
Solve the following problems:
1. A 2.00 mole sample of NaOH is dissolved in enough water to make 500. mL of solution. What is the
concentration of the solution?
2.00 mol NaOH
= 4.00 M
0.500 L
2.
60.0 grams of NaOH (MM = 40.00 g·mol-1) is dissolved in enough water to make 0.750 L of solution.
What is the concentration of the solution?
60.0 g NaOH
1 mol NaOH
1
= 2.00 M
40.0 g NaOH
0.750 L
3. A 250. mL sample of a 0.125 M solution of NaOH contains ________ grams of NaOH.
0.125 mol NaOH
0.250 L solution
40.0 g NaOH
= 1.25 g NaOH
1 L solution
1 mol NaOH
4.
5.
What volume of 0.65 M H2SO4 contains 49.1 g of H2SO4?
49.1 g
1 mol
1L
= 0.77 L
98.1 g
0.65 mol
For each solution, identify the ions that exist in aqueous solution, and specify the concentration of each:
a) 0.5 M Al2S3
b) 0.3 M Mg3N2
1.0 M Al3+
1.5 M S2-
0.9 M Mg2+
0.6 M N3-
Station 8 – More Molarity Problems
1. a. If 5.00 grams of sodium hydroxide is dissolved to make 600 mL of solution, what is its molarity?
5.00 g NaOH
mol
40.0 g NaOH
1
0.600 L
= 0.2 M NaOH
b. What mass of potassium chloride has to be dissolved in water to produce 2.00 liters of a 2.45 M solution?
2.45 mol KCl
1L
2.0 L
74.6 g KCl
mol KCl
= 365 g KCl
c. If 24.63 grams of magnesium chloride is dissolved to make exactly 3.00 liters of solution calculate the
concentrations of the ions in the solution (in moles/liter).
24.63 g MgCl2
mol
95.3 g MgCl2
1
3.00 L
= 0.0861 M MgCl2
MgCl2  Mg2+ + 2Cl
[Mg2+] = 0.0861 M
[Cl] = 0.172 M
2. a. What is the molarity of the solution that results from adding 25 mL of a 0.15 M solution of sodium
hydroxide to sufficient water to make 500 mL of solution?
(0.15 mol/L) x (0.025 L) = M2 x (0.500 L)
M2 = 0.0075 M
b. What volume of a 2.50 M solution of hydrochloric acid is required to prepare 2.0 liters of a 0.30 M
solution?
(2.50 mol/K) x V1 = (0.30 M) x (2.0 L)
V1 = 0.24 L
3. When excess silver nitrate was added to a 25.0 mL sample of a solution of calcium chloride, 0.9256 gram of
silver chloride precipitated. What is the concentration of the calcium chloride solution?
2 AgNO3 (aq) + CaCl2 (aq)  2 AgCl (s) + Ca(NO3)2 (aq)
Need to calculate moles of CaCl2 to determine the molarity (moles CaCl2 per L
solution).
Use stoichiometry mass to mole calculation.
0.9256 g
AgCl
1 mol
AgCl
143.4 g
AgCl
1 mol
CaCl2
2 mole
AgCl
= 0.003225
mol CaCl2
0.003225 mol
0.025 L
= 0.129 M
Station 5 – OXIDATION NUMBERS
Determine the oxidation number of each element in the compound:
Oxidation numbers of all elements are shown. The underlined one matches the underlined element.
MgF2 +2, -1
CuSO4 +2, +6, -2
NH4+ -3, +1
CO2 +4, -2
AgCl +1, -1
Cr2O72- +6, -2
ClO3- +5, -2
CaH2 +2, -1
**H with a metal is -1
For each statement, classify the change of the underlined element as [O]xidation, [R]eduction, or [N]either:
__O___
CH4 (g) + O2 (g)  CO2 (g) + H2O (g) -4 to +4
__N__
NaOH + HCl  NaCl + H2O
__R__
gaining electrons
__O__
losing electrons
__R__
Zn(s) + 2HCl (g)  ZnCl2 (aq) + H2 (g) +1 to 0
__R__
2 Mg (s) + O2 (g)  2 MgO (s)
-2 to -2
0 to -2
Station 6 – Oxidation Reduction Reactions
Use the reactivity chart to the right and determine if each reaction will
occur or not. If it WILL occur: (a) identify the products, and (b) identify
which metal is reduced and which is oxidized.
1.
Mg(s) + Zn(NO3)2 
Mg(NO3)2 (aq) + Zn(s)
Mg: 0 to +2; loses electrons; oxidation
Zn: +2 to 0; gains electrons; reduction
 No Reaction
2.
Fe(s) + Mg(NO3)2
3.
Zn(s) + Cu(NO3)2
Zn(NO3)2 (aq) + Cu(s)
Zn: 0 to +2; loses electrons; oxidation
Cu: +2 to 0; gains electrons; reduction
4.
Cu(s) + Ca(NO3)2
No Reaction
Station 7 – Aqueous Reactions – Particle Views
1. Make a sketch of the following in beaker and show what species are present in the solution after the reaction
has reached equilibrium.
a) Hydrogen fluoride is diluted in water.
b. Hydrogen chloride is diluted in water.
Weak acid so some molecule is present
Strong acid so no molecule present, only ions
H+
H+
F–
Cl–
HF(aq)
2. Make a sketch of the following:
a) A 1.0 M solution of AgNO3 in a 1 L beaker.
b) A 1.0 M solution of CaBr2 in a 1 L beaker.
b) A 1.0 M solution of CaBr2 in a 1 L beaker.
Ca2+
+
Ag
Br–
NO3–
Br–
c) Write the balanced molecular equation for a mixture of AgNO3 (aq) and CaBr2 (aq).
2 AgNO3 (aq) + CaBr2 (aq)  2AgBr (s) + Ca(NO3)2 (aq)
d) Make a sketch of a 1 L beaker after the reaction is complete.
Products are AgBr (s); 1 Ca2+ and 2 NO3-
Ca2+
NO3NO32AgBr(s)
Station 9 – Titrations
1. What volume of a 0.291 M solution of NaOH is required to reach the equivalence point in a titration against
25.0 mL of a 0.350 M HCl?
At the endpoint: moles added H+ = moles added OH
+
+
moles H+ = (0.350 mol H /L) x (0.025 L) = 0.00875 mol H

moles OH = (0.291 mol OH /L) x (Vbase)
0.00875 mol H+= (0.291 M) x Vbase
Vbase = 0.0301 L
2.
What volume of 0.812 M HCl, in milliliters, is required to titrate 1.33 g of NaOH to the equivalence point?
NaOH(aq) + HCl(aq)  NaCl(aq) + H2O(l)
At the endpoint: moles added H+ = moles added OH
+
moles H+ = (0.812 mol H /L) x (Vacid)
moles OH =
1.33 g NaOH
1 mol NaOH
40.0 g NaOH
1 mol OH= 0.03325 mol OH1 mole NaOH
(0.812 mol H+/L) x (Vacid) = 0.03325 mol OH-
Vacid = 0.0409 L = 40.9 mL
3.
What volume of 0.955 M HCl, in milliliters, is needed to titrate 2.152 g of Ca(OH)2 to the equivalence
point?
Ca(OH)2(aq) + 2 HCl(aq)  CaCl2 (aq) + 2 H2O(l)
+
moles H+ = (0.955 mol H /L) x (Vacid)
moles OH =
2.152 g
Ca(OH)2
1 mol Ca(OH)2
2 mol OH-
74.1 g Ca(OH)2
1 mole Ca(OH)2
(0.955 mol H+/L) x (Vacid) = 0.05808 mol OH-
Vacid = 0.0608 L = 60.8 mL
= 0.05808 mol OH-
Station 10 – SPECTROSCOPY – Part 1
In the spectroscopy lab, we used the calorimeter and measured transmittance to determine the concentration of
blue dye in a solution. Our calorimeters have three options for wavelengths: Red - 620 nm, Green - 550 nm,
Blue - 470 nm. Based on the absorption spectrum for the three dyes in the figure below, which settings is
the best setting for a lab measuring the concentration of each? Absorbance measurements are most accurate
in the range 0.2–1.0. Explain your answer.
You want the wavelength that gives the largest range of
absorbance for that molecule.
Red 40:
Use Blue (470 nm) – absorbance
ranges from 0 to ~0.45
Crystal Violet:
Use Green (550 nm) – absorbance
ranges from 0 to ~0.5
Blue 1:
Use Red (620 nm) – absorbance
ranges from 0 to ~0.45
Station 11 – SPECTROSCOPY – Part 2
Using the standard calibration curve below, determine the concentrations of the following solutions.
Calibration Curve, A v. Concentration
The relationship between absorbance (A) and %
transmittance (%T) is:
A = - log ( )
a) Measured %T is 56%:
A = 0.25
concentration = 1.0 M
b) Measured %T is 35%:
A = 0.46
concentration = 1.9 M
c) Measured %T is 63%
A = 0.20
concentration = 0.84 M
Molar Concentration (M)
Slope of line =
Absorbance (y ) = 0.24 Concentration (x)
Use slope equation to determine concentration.
* or you can just read it off the graph.
Conc = Absorbance/0.24