Chapter 7 – SOLUBILITY & REACTIONS
Saturated Solution – a solution in which no more solute will
dissolve
 the solubility of a substance is the concentration of a saturated
solution of that substance
 solubility is usually measured in grams of solute per 100mL and is
very dependant on temperature e.g. the solubility of sodium
sulphate in water at 0ºC is 4.76g/100mL
 this means that 4.76g of sodium sulphate can be dissolved in
100mL of water at 0ºC…..and if you add more it won’t dissolve!
Solubility Curves
 Graphs of Solubility(Maximum Concentration) vs. Temperature
allow a fast and easy reference (See Figure 2 on page 316)
Answer questions
1 & 2 on Pages
316/317.
Solubility of Gases
 Gases can & do dissolve in liquid
Examples; chlorine in swimming pools
oxygen in rivers & streams
carbon dioxide in cans of pop
 Does temperature effect the solubility of a liquid as it does in
solids?
DO LAB EXERCISE 7.1.1 on Page 318.
SOLUBILITY IN WATER – GENERALIZATIONS & EXAMPLES
SOLUTE
LOWER HIGHER DOESN’T
IN
TEMP.
TEMP. DISSOLVE WATER
E.G. (as
xg/100mL)
180g@0ºC
487g@100ºC
SOLIDS
GASES
LIQUIDS
polar
nonpolar*
ELEMENTS
* = immiscible with water (forms a layer)
** = miscible (dissolves completely)
small
polar **
Low
solubility
in water
incl.
halogens
&
oxygen
Answer questions 6 & 7
on page 319.
Crystallization
 When the solute comes out of solution as hard particles this is
called crystallization
 as solvent evaporates there is not enough solvent to keep the
solute in solution
 this process can be speeded up by heating a solution to evaporate
off the solvent-- used industrially to produce table salt & table
sugar
Solubility Categories
 solubilities can range from high to extremely low
 we use the term insoluble to mean negligible solubility
 we can categorize the solubilities as;
HIGH SOLUBILITY
LOW SOLUBILITY
INSOLUBLE
Maximum concentration at SATP
of greater than or equal to
0.01mol/L
Maximum concentration at SATP
of less than 0.1mol/L
Substance that has a negligible
solubility at SATP
SOLUBILITY OF COMMON CATIONS & ANIONS AT SATP
COMMON ANIONS
Cl-1,Br-1,I-1
C
A
T
I
O
N
S
High
Solubility
> or =
0.1mol/L
(at SATP)
Low
Solubility
< or =
0.1mol/L
(at SATP)
Most
Ag+,Pb2+,
Tl+,Hg2+,
Hg+,Cu+
S2Group
I & II,
NH4+1
Most
OH-1
Group
I,
NH4+,
Sr2+,
Ba2+,
Tl+
Most
SO4-2
Most
Ag+,
Pb2+,
Ca2+,
Ba2+,
Sr2+,
Ra2+
CO32-,PO4-4
C2H3O2-
NO3-1
Group I
&
NH4+1
Most
All
Most
Ag+1
Most
Answer questions 11-13 on page 325.
Hard Water Treatment
 hard water is water containing a higher concentration than normal
of calcium and magnesium ions
 usually results in floating scum, when using soap in a bath, soap
not working up a good lather, etc.
 to counteract this water softeners can be used
 water can be softened by removing the excess calcium &
magnesium ions
 our municipalities use the soda-lime process but home watersoftening involves an ion exchange process where sodium ions
replace the calcium & magnesium
Answer questions 1-4 on Page 329.
7.3 Reactions in Solution
The solubility table on Page 324 can be useful in predicting whether
the products of a reaction will be soluble or insoluble.
Net Ionic Equations
 if we mix lead (II) nitrate with potassium iodide, we will produce a
yellow precipitate.
 If we look at our solubility table using the products we expect to
be formed: lead iodide + potassium nitrate we can hypothesize;
1. lead iodide will have a low solubility
2. potassium nitrate will have a high solubility
(1) Pb(NO3)2(aq) + 2KI(aq) -- PbI2(s) + 2KNO3(aq)
(2) Pb(NO3)2(aq) + NaI(aq) -- PbI2(s) + 2NaNO3(aq)
(3) Pb(C2H3O2)2(aq) + MgI2(aq) -- PbI2(s) + Mg(C2H3O2)2(aq)
 using Arrhenius’ Theory of Dissociation we can rewrite (1)
Pb(NO3)2(aq) + 2KI(aq) -- PbI2(s) + 2KNO3(aq)
as:
Pb2+(aq) + 2NO3-1(aq) + 2K+1(aq) + 2I-1(aq)
---- PbI2(s) + 2K+1(aq) + 2NO3-1(aq)
 This is called a TOTAL IONIC EQUATION
 Any ion (atom, molecule, etc.) present in a reaction system that
does not change or become involved during a chemical reaction is
called a spectator.
 If we rewrite an equation for the reaction, showing only the
entities that change we have what’s called a net ionic equation.
 In other words;
Pb2+(aq) + 2NO3-1(aq) + 2K+1(aq) + 2I-1(aq)
---- PbI2(s) + 2K+1(aq) + 2NO3-1(aq)
would be re-written as:
Pb2+(aq) + 2I-1(aq) ---- PbI2(s) ……..our net ionic equation
Writing Net Ionic Equations
1. Write the balanced chemical equation with full chemical
formulas for all the reactants and products.
2. Using the solubility table (see p. 324), rewrite the formulas for
all the high solubility ionic compounds as dissociated ions, to
show the total ionic equation.
3. Cancel identical amounts of identical entities appearing on both
sides of the equation.
4. Write the net ionic equation reducing the coefficients if
necessary.
Sample Problem 1
Q: Write the net ionic equation for the reaction for the reaction of
aqueous barium chloride and aqueous sodium sulphate.
A: step 1: BaCl2(aq) + Na2SO4(aq) --- BaSO4(s) + 2NaCl(aq)
step 2 & 3: Ba2+(aq) + 2Cl-1(aq) + 2Na+1(aq) + SO4-2(aq)
---- BaSO4(s) + 2Na+1(aq) + 2Cl-1(aq)
step 4: Ba2+(aq) + SO4-2(aq) ---- BaSO4(s)
…..net ionic equation
Sample problem 2
Q: Write the net ionic equation for the reaction of zinc metal and
aqueous copper (II) sulphate and then write a statement to
communicate the meaning of the net ionic equation..
A: step 1: Zn(s) + CuSO4(aq) --- Cu(s) + ZnSO4(aq)
steps 2 & 3: Zn(s) + Cu2+(aq) + SO42-(aq)
--- Cu(s) + Zn2+(aq) + SO42-(aq)
step 4: Zn(s) + Cu2+(aq) ---Cu(s) + Zn2-(aq) ….net ionic equation
Solid zinc in an aqueous solution containing copper (II) ions will
produce solid copper and aqueous zinc ions.
Do questions 3-7 on Pages 335 & 336 in your textbooks.
7.4 Waste Water Treatment
Read Section 7.4 Waste Water Treatment (pages 337-340) &
make your own notes!
Answer Practice Questions 1-4 on page 340.
7.5 Qualitative Chemical Analysis
1. Qualitative Analysis – identification of the specific
substances present.
2. Quantitative Analysis – measurement of the quantity of a
substance present.
1-(a) Qualitative analysis – can be done by colour
ION
Group 1,2,17
Cr2+(aq)
Cr3+(aq)
Co2+(aq)
Cu+(aq)
Cu2+(aq)
Fe2+(aq)
Fe3+(aq)
Mn2+(aq)
Ni2+(aq)
CrO4-(aq)
Cr2O72-(aq)
MnO4-1(aq)
SOLUTION COLOUR
colourless
blue
green
pink
green
blue
pale green
yellow-brown
pale pink
green
yellow
orange
purple
 In conjunction with colour solution tests, we can also use flame
tests to determine qualitatively what substances are present.
(see Fig. 2 on p. 342)
1-(b)-Sequential Qualitative Chemical Analysis
 this is a type of test where we set up a double displacement
reaction using one unknown solution and one known solution
 in this test, we would predict that if a precipitate forms then a
certain ion must have been present in the unknown solution
 Book example: if you were given a solution that contains lead (II)
ions or strontium ions (or neither or both) what would we do?
To Complete A Sequential Analysis
F
O
R
cations
C
A
T
I
use
O
N
A
N
A
L
Y
S
I
S
1. Locate the possible cations on the solubility table.
2. Determine which anions would precipitate the possible
cations.
3. Plan a sequence of precipitation reactions that would
use anions to precipitate a single cation at a time.
4. Use filtration between the steps to remove cation
precipitates that might interfere with subsequent
additions of anions.
5. Draw a flow chart to assist testing & communication.
Solution known to contain Pb2+(aq) and/or Sr2+(aq)
Add NaCl(aq)
White precipitate
Solution contained
Lead(II) ions,
precipitated as PbCl2
ions
no precipitate
No Lead(II) ions were present
Filter
Add Na2SO4(aq)
White precipitate
Solution containing
Strontium ions,
precipitated as
SrSO4(s)
no precipitate
No strontium ions present
Finish questions 5-7 on page 346, as well
as, questions 1 & 2 on page 346 and
handout.
Quantitative Analysis
 often quantitative analysis will follow a qualitative analysis.
e.g. a police officer sees erratic driving, stops the car, smells
alcohol on the breath of the driver, and then does a breathalyzer
to get an actual reading of the diver’s blood alcohol.
Orange solution
containing dichromate
ions
Exhaled Air
Alcohol present
Solution turns
pale green
indicating
chromate (III)
ions formed
No alcohol present
No colour
change
 light is passed through a solution detects colour change (if any)
and light that passes through is converted by a photocell into an
electric current which indicates blood alcohol content
 a blood test may be done but it takes longer
 in Ontario, the legal limit is 0.080g/100mL (800ppm)
[0.08]
Solution Stoichiometry
 most solutions take place in aqueous conditions
 we need to know how to find the concentration of reactants in
solutions
Stoichiometry Calculations
1. Write a balanced equation for the reaction, to obtain the mole
ratios.
2. Convert the given value to an amount in moles using the
appropriate conversion factor.
3. Convert the given amount in moles to the required amount in
moles, using the mole ratio from the balanced equation.
4. Convert the required amount in moles to the required value
again using the appropriate conversion factor.
Molar Concentration Problems
 many well known fertilizers include ammonium hydrogen
phosphate
 made commercially by reacting concentrated aqueous solutions of
ammonia and phosphoric acid
Q: What volume of 14.8mol/L NH3(aq) would be needed to react
completely with each 1.00kL (1.00m3) of 12.9mol/L H3PO4(aq) to
produce fertilizer in a commercial operation?
A: step 1:
2NH3(aq) +
v=?
C=14.8mol/L
H3PO4(aq) ---
v=1.00kL
C=12.9mol/L
(NH4)2HPO4(aq)
step 2:
to find the number of moles of H3PO4, we use the formula C=n/v
so, n=v x C--- nH3PO4=1.00kL x 12.9mol/L =12.9kmol
step 3:
using the mole ratio of NH3 : H3PO4
2 : 1
nNH3 = 12.9kmol x 2/1 = 25.8kmol
step 4: again using C=n/v, to find the volume of NH3
v=n/C
vNH3=25.8kmol
14.8mol/L
v=1.74kL
Therefore the volume of NH3 needed would be 1.75kL.
Sample Problem 1
Q: A 10.00mL sample of sulfuric acid reacts completely with 15.9mL
of a 0.150mol/L potassium hydroxide solution. Calculate the molar
concentration of the sulfuric acid.
A:
step 1: H2SO4(aq) +
v=10.00mL
C=?
2KOH----2H2O(l) + K2SO4(aq)
v=15.9mL
C=0.150mol/L
step 2: CKOH =nKOH
vKOH
therefore
nKOH =vKOH x CKOH
=15.9mL x 0.150mol/L
=2.39mmol
step 3: nH2SO4 =2.39mmol x 1
2
nH2SO4 =1.19mmol
step 4: to find C of H2SO4
CH2SO4 = n/v = 1.19mmol
10.00mL
= 0.119mol/L
Therefore the molar concentration of the sulfuric acid is 0.119mol/L.
Answer questions 1-3 & #4 on Page 353.
For #4 do only (a), (b) and (c).
(c) =% error