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Study Guide: Chem 118 Lab Essentials
1. Solubility Rules
General Rules :
o Alkali metals and ammonium salts are usually soluble.
o Nitrates, acetates, and most perchlorates are soluble.
o Halides (chlorides, bromides, iodides) are soluble, except with
silver, lead, and mercury.
o Sulfates are generally soluble, except with calcium, strontium,
barium, and lead.
o Carbonates, phosphates, sulfides, and hydroxides are mostly
insoluble except with alkali metals and ammonium.
Note : Understanding which ions form precipitates is essential for
predicting reaction outcomes.
2. Converting Units (Area and Volume)
Basic Conversion Tips :
o
1 mL = 1 cm
3
(important for volume-related calculations).
o For area conversions, remember to square linear conversions
(e.g.,
1 ft
2
2
o For volume, cube the conversion factors (e.g.,
1 ft
2
2
3. Molecular and Net Ionic Equations
Molecular Equations : Show all reactants and products in their
complete, neutral forms.
Complete Ionic Equations : Break down aqueous compounds into
ions.
Net Ionic Equations : Eliminate spectator ions (those that do not
change during the reaction), leaving only the species that participate
directly.
4. Endothermic vs. Exothermic Reactions
Endothermic Reactions :
o Absorb heat from surroundings; temperature of surroundings
drops.
o Example: Dissolving ammonium salts in water.
Exothermic Reactions :
o Release heat to surroundings; temperature of surroundings rises.
o Example: Dissolving calcium chloride in water.
5. Using a Best-Fit Line from a Calibration Curve
Best-Fit Line : The line that best approximates data points on a graph,
minimizing the distance of each point from the line.
Equation Interpretation : From the best-fit equation y = mx + b , where y
is the dependent variable (e.g., absorbance), x is the concentration,
and m (slope) indicates sensitivity.
Calibration Curves : Used to determine unknown concentrations by
finding the corresponding y -value for a given x -value.
6. Molarity, Moles, and Volume
Key Relationship :
o M =
n
V
, where
M
is molarity,
n
is moles, and
V
is volume (L).
Conversions :
o To find moles from molarity and volume: n = M × V.
o To find volume from moles and molarity:
V =
n
M
Practical Use : Essential in preparing solutions and performing
titrations.
Example Problems
1. Solubility and Precipitation Reactions
Problem : Given the following reactants, predict whether a precipitate
will form and, if so, identify it:
o a)
Ag NO
3
( aq )+ NaCl ( aq ) →
o b) K
2
SO
4
( aq )+ Ba ( NO
3
2
( aq ) →
Solution :
o For (a), AgCl is insoluble, so a white precipitate of silver chloride
will form.
o For (b),
Ba SO
4
is insoluble, leading to a precipitate.
2. Molarity, Volume, and Moles
Problem : How many moles of NaOH are present in 250 mL of a 0.5 M
NaOH solution?
Solution : Use Molarity
( M )=
moles
L
o Moles of NaOH =0.5 M × 0.250 L =0.125 moles.
3. Net Ionic Equations
Problem : Write the molecular, complete ionic, and net ionic equations
for the reaction between
Pb
NO
3
2
( aq ) and KI ( aq )
Solution :
o Molecular Equation : Pb ( NO
3
2
( aq )+ 2 KI ( aq ) → PbI
2
( s )+ 2 K NO
3
( aq )
o Complete Ionic Equation :
Pb
2 +¿( aq )+ 2 NO
3
−¿( aq )+ 2 K
+¿ ( aq )+ 2 I
−¿( aq ) → PbI 2
( s ) + 2 K
+¿ ( aq )+ 2 NO 3
−¿ ( aq ) ¿
¿
¿
¿
¿
¿
o Net Ionic Equation : Pb
2 +¿( aq )+ 2 I
−¿( aq ) → Pb I 2
( s ) ¿
¿
4. Endothermic vs. Exothermic Reactions
Problem : Identify whether dissolving
NH
4
NO
3
in water is endothermic
or exothermic. Describe what happens to the temperature of the
solution.
Solution : Dissolving
NH
4
NO
3
in water is endothermic (absorbs heat),
so the solution temperature decreases.
5. Using a Calibration Curve
Problem : A calibration curve for a phosphate solution shows a linear
relationship with the equation
y =0.02 x +0. where
y is absorbance and
x is concentration in ppm. What is the concentration of phosphate if
the absorbance is 0.5?
Solution : Rearrange to solve for
x : x =
( y −0.01)
o x =
=24.5 ppm
6. Color Observations in Reactions
Problem : Describe the color change observed when copper(II) sulfate
is mixed with sodium hydroxide.
Solution : A blue precipitate of
Cu ( OH )
2
forms.
7. Density Conversions
Problem : If a sugar solution has a density of 1.2 g/mL, calculate the
mass of 250 mL of this solution.
Solution : Use mass = density × volume.
o mass =1.2 g / mL × 250 mL = 300 g
