Chemistry 30
Unit 5:
Solutions and
Solubility
TEACHER
Components of a Solution

A solution is composed of two or more pure substances mixed together, one of which is a
solvent and the other a solute.

A cup of instant coffee is the solution, the hot water is the solvent, and the instant coffee is the
solute.

A saturated solution is a solution that contains the maximum amount of solute dissolved in a
solvent. A supersaturated solution is a solution where conditions have been changed to allow
more solute to dissolve than would at room temperature (this is done by heating the solvent
or solution).

When the solvent and solutes are liquids we can use the terms miscible and immiscible.
Miscible means that the two liquids will mix together, and immiscible means that the two
liquids will not dissolve in one another.

Remember, the solvent and solute can be solid, liquid OR gas. There are special names for
solutions where the solute and solvent are both solids and when the solute is a liquid, and the
solvent is a solid.
o Solid-in-solid – alloys (e.g., brass = zinc + copper, sterling silver = silver + copper).
o Liquid-in-solid – amalgams (e.g., older dental fillings = mercury + silver).

Factors that may affect solubility are temperature and pressure:
o Temperature: Gases: Temp ⬆, solubility ⬇
Solids and liquids: Temp ⬆, solubility ⬆
o Pressure: Gases: Pressure ⬆, solubility ⬆
Solids and liquids: unaffected

Like dissolves like. That is polar covalent molecules usually dissolve in other polar covalent
molecules and non-polar covalent molecules dissolve in non-polar covalent molecules.
RECALL: Polarity can be determined by subtracting electronegativities. A polar molecule is one where one
(or more) atom in the molecule has a stronger pull on the electrons than another atom, so the electrons
move closer to the atom with the stronger pull (higher electronegativity). This creates a dipole, where the
atom with the stronger pull has a slightly negative charge and the atom with the lower electronegativity
(less pull) has a slightly positive charge.
Water as a Solvent:

Solutions in which water is the
solvent are called aqueous
solutions.

Most covalent liquids are only
able to dissolve other covalent
compounds; water can dissolve
both ionic and covalent
compounds.
The attraction of water dipoles for ions pulls ions out of a crystalline lattice and into aqueous solution.
Aqueous Solutions:

Sometimes when we mix two dissolved salts together, we end up with a precipitate. This is
because a new ionic compound is formed that is insoluble in water.
We can use solubility charts or solubility tables to determine if a solid will be formed.
Ion
NO3–
ClO4–
Cl–
I–
BrSO42CO32PO43OH–
S2Na+
K+
NH4+
Solubility
soluble
soluble
soluble
soluble
soluble
soluble
insoluble
insoluble
insoluble
insoluble
soluble
soluble
soluble
Exceptions
none
none
except Ag+, Hg22+, *Pb2+
except Ag+, Hg22+, *Pb2+
except Ag+, Hg22+, *Pb2+
except Ca2+, Ba2+, Sr2+, Hg2+, Pb2+, Ag+
except Group IA and NH4+
except Group IA and NH4+
except Group IA, *Ca2+, Ba2+, Sr2+
except Group IA, IIA and NH4+
none
none
none
* = slightly soluble
Ex. If we mix sodium ions with hydroxide ions will a precipitate form?
No
Ex. If we dissolved sodium hydroxide in calcium chloride, will a precipitate form?
2NaOH + CaCl2  2NaCl + Ca(OH)2
NaCl is soluble, Ca(OH)2 is not soluble so a precipitate will form.
Ex. Calcium nitrate with sodium carbonate
Ca(NO3)2 + Na2CO3  CaCO3 + 2NaNO3
Calcium Carbonate is insoluble so a precipitate will form, sodium nitrate is soluble
See Components of Solutions Assign
Concentration

Concentration refers to the amount of solute dissolved in a specific amount of solvent.

Concentrated and dilute are qualitative terms we use to describe concentration. Dilute means
there is not a lot of solute in the solution and concentrated means there is a lot of solute in the
solution.

Common concentrations are molarity, molality, ppm, and ppb.
Molarity
This is the number of moles of solute dissolved in one litre of solution.
The formula for molarity is as follows:
c=
n = # moles of solute
n
V
V = volume of solution in litres
c = concentration in moles per litre (M)
Ex. If a teaspoon (5.0mL) of a 0.50 M solution of NaCl was evaporated, how many moles of sodium
chloride would be left? What mass of NaCl would be left?
𝑛
𝑐=
𝑛 = 𝑐𝑉
𝑉
𝑛 = (0.50𝑀)(0.0050𝐿) = 0.0025 mol
𝑐𝑜𝑛𝑣𝑒𝑟𝑡 𝑡𝑜 𝑚𝑎𝑠𝑠: 0.0025𝑚𝑜𝑙 𝑥
58.4428𝑔
= 0.15𝑔
1𝑚𝑜𝑙
Ex. Antifreeze is a solution of ethylene glycol, C2H6O2, in water. If 4.50 L of antifreeze contains 2.00 kg of
ethylene glycol, what is the concentration of the solution?
Moles of C2H6O2
= 2.00 × 103 g C2H6O2 ×
c=
n
32.2 mol
=
V
4.50 L
1 mol C 2 H 6O 2
62.07g C 2H 6O 2
= 32.2 mol C2H6O2
= 7.16 mol/L
Standard solution:

A standard solution is a solution of a known concentration. This means it has a precise mass of
solute in a specific volume of solution.

Standard solutions are used in experiments where the concentration of a solution must be
known.

We use volumetric flasks to prepare standard solutions as they have a small margin of error
when compared to other pieces of lab equipment. Volumetric flasks come in a variety of sizes
(volumes).
To prepare a standard solution we follow the following steps:
1. Calculate the required mass of solute needed using the volume and concentration you want to
end up with.
2. Weigh out the mass of the solute needed and add it to a volumetric flask of the appropriate size.
3. Dissolve the solid in pure water using less than half of the final solution volume.
4. Once the solute is dissolved, add the rest of the water. Be sure to use a medicine
dropper for the final few milliliters of water. Use the calibration line to set the meniscus
in the appropriate spot.
Ex. Describe the preparation of 2.000 L of a standard aqueous solution containing 0.1000 mol/L
potassium nitrate.
n = Vc
= 2.000 L × 0.1000 mol/L = 0.2000 mol
mass of KNO3 = 0.2000 mol KNO3 ×
101.1g KNO3
= 20.22 g KNO3
1 mol KNO3
To prepare the solution we place 20.22 g of KNO3 in a 2.000 L volumetric flask. About half the required amount of
water is added. When all the potassium nitrate is dissolved, the solution is diluted with the remaining water.
Dilution Calculations:


When making a solution in chemistry laboratories you usually only have access to substances
with high concentration solutions (called stock solutions) and are then required to dilute the
stock solutions. A calculation needs to be completed in order to determine the amount of
distilled water that needs to be added to a certain volume of stock solution in order to create the
desired concentration.
Since the number of moles of solute in a solution does not change when you dilute it, the
equation for dilution is as follows:
V1 C1 = V2 C2
before
after
Ex. Water is added to 200.mL of 2.40M ammonia cleaning solution (NH3), until the final volume is 1.00L.
Find the molar concentration of the final diluted solution.
Vi Ci = Vf Cf
(0.200L)(2.40M) = (1.00L)(Cf)
Cf = 0.480M
Ex . What volume of concentrated sulphuric acid (containing 18.0 M H2SO4) is required to prepare 5.00
L of 0.150 M aqueous sulphuric acid solution by dilution with water?
V1 = ?
V2 = 5.00 L
c1 = 18.0 mol/L
c2 = 0.150 mol/L
V1 =
=
V2 c2
c1
(5.00 L)(0.150
18.0
mol
L
mol
)
L = 0.0417 L = 41.7 mL
Ion Concentration:
Consider a 0.20M aqueous solution of sodium carbonate. The sodium carbonate will be completely
dissociated into ions:
 2 Na+(aq) + CO32-(aq)
Na2CO3(s) 
The concentration of the sodium ions can be calculated using the conversion factors from the balanced
equation (similar to mol- mol stoichiometry). Remember: concentration is represented by using square
brackets.
[Na+]
2 mol Na 
= 0.20 mol/L Na2CO3 ×
= 0.40 mol/L [Na+]
1 mol Na 2 CO3
[CO32-] = 0.20 mol/L Na2CO3 ×
1 mol CO32
= 0.20 mol/L [CO32-]
1 mol Na 2 CO3
Ex. What are the concentrations of the ions in an aqueous solution containing 0.15 mol/L iron(III)
nitrate?
 Fe3+(aq) + 3 NO3−(aq)
Fe(NO3)3(s) 
[Fe3+]
1 mol Fe3
= 0.15 mol/L Fe(NO3)3 ×
= 0.15 mol/L Fe3+
1 mol Fe( NO3 )3
[NO3−] = 0.15 mol/L Fe(NO3)3 ×
3 mol NO3
= 0.45 mol/L NO3−
1 mol Fe( NO3 )3
Ex. 250mL of 0.30M K2SO4 and 250mL of 0.80M MgCl2 are mixed and no reaction results. What is the
concentration of each substance in the final solution, and the concentration of each individual ion?
[K2SO4]: 𝐶𝑓 =
𝐶𝑓 =
𝐶𝑖 𝑉𝑖
[MgCl2]: 𝐶𝑓 =
𝑉𝑓
(0.30𝑀)(250𝑚𝐿)
(500𝑚𝐿)
𝐶𝑓 =
Cf = 0.15M
[K+]=0.30M
𝐶𝑖 𝑉𝑖
𝑉𝑓
(0.80𝑀)(250𝑚𝐿)
(500𝑚𝐿)
Cf = 0.40M
[SO42-]=0.15M
[Mg2+]=0.40M
[Cl-]= 0.80M
Ppm & ppb:

Usually in chemistry we work with molarity when discussing concentration. When working with
very dilute solutions (solutions with low concentrations) it is more convenient to work with
ppm or ppb.

A “part” refers to a gram.

A concentration of 1ppb is equivalent to 1g of solute dissolved in 1,000,000,000g of solvent.

The allowable concentration of toxic chemicals is usually measured in ppm or ppb.

For example the allowable concentration of mercury in air is 82ppm, this means there are
82𝑔 𝐻𝑔
. The allowable concentration of copper in drinking water is 1ppm, this means
1 𝑏𝑖𝑙𝑙𝑖𝑜𝑛 𝑔𝑟𝑎𝑚𝑠 𝑜𝑓 𝑎𝑖𝑟
1 𝑔 𝐶𝑢
1 𝑚𝑖𝑙𝑙𝑖𝑜𝑛 𝑔𝑟𝑎𝑚𝑠 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟

.
To do ppm or ppb calculations we assume that 1g of water has a volume of 1mL.
Ex. Change 2.00M KNO3 to ppm.
2.00𝑀 =
2.0𝑚𝑜𝑙 𝐾𝑁𝑂3
1𝐿 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
change to
𝑔𝑟𝑎𝑚𝑠 𝐾𝑁𝑂3
1 𝑚𝑖𝑙𝑙𝑖𝑜𝑛 𝑔𝑟𝑎𝑚𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
Convert 2.00mol KNO3 to mass: 2.0𝑚𝑜𝑙 𝐾𝑁𝑂3 ×
101.10332 𝑔 𝐾𝑁𝑂3
1 𝑚𝑜𝑙 𝐾𝑁𝑂3
= 202g KNO3
1L solution = 1000mL = 1000g
202𝑔 𝐾𝑁𝑂3
1000
20200𝑔
×
=
= 20200𝑝𝑝𝑚
1000𝑔 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 1000 1000000𝑔
Ex. The concentration of gold in the ocean is 0.005ppb. What would be the molarity of gold in a sample
of ocean water.
0.005𝑝𝑝𝑏 =
0.005𝑔 𝐴𝑢 ÷ 1000000
0.000000005𝑔 𝐴𝑢
=
1000000000𝑔 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 ÷ 1000000
1000𝑔𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
1000g solution = 1000mL solution = 1 L solution
Determine moles Au:
1 𝑚𝑜𝑙 𝐴𝑢
0.000000005𝑔 𝐴𝑢 × 196.9665694 𝑔 𝐴𝑢 = 3 × 10−11 𝑚𝑜𝑙 𝐴𝑢
[Au]=
3×10−11 𝑚𝑜𝑙 𝐴𝑢
1𝐿 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
= 3 x 10-11M
See Concentration and Dilution Assignment
The Solubility Product Constant:

All salts have different solubilities. Even ‘insoluble’ salts will dissolve in water to some extent.

When a salt is dissolved in water, some of the salt dissolves and some may not. The amount of
salt that dissociates into ions is different for every salt.

Ksp is a number that represents how much of the salt will dissociate into ions.

Ksp is equal to the product of the concentration terms each raised to the power of the coefficients
of the substance in the dissociation equation.

Ksp values @ 25oC can be found on the table on the next page.

The equilibrium constant Ksp called the solubility constant indicates how soluble a substance is.
o If the Ksp is low, the solid is not very soluble.
o If the Ksp is high, the solid is soluble.
An expression for the solubility product constant, Ksp, of calcium phosphate would be as follows:
Balanced Equation: Ca3(PO4)2(s) ⇌ 3 Ca2+(aq) + 2 PO43-(aq)
Ksp is equal to the product of the concentrations of the products, each raised to the power of the
coefficient in the balanced equation, over the concentration of the reactant to the power of the
coefficient.
When using Ksp we do not include the concentrations of solids, liquids or gases, as they are not
dissolved. Therefore, the concentration of the reactant should not be included, as it is a solid.
Ksp = [Ca2+]3[PO43-]2
Example: Calculate the solubility product constant for calcium fluoride at 20oC given that it`s solubility
is 1.6 × 10−2 g/L.
First write the chemical equation:
CaF2  Ca2+ + 2FThen convert to molarity (the number of moles of CaF2 that dissolved in 1L):
mm CaF2 = 78.08g/mol
1.6 ×10−2 𝑔
1𝐿
1 𝑚𝑜𝑙
× 78.1𝑔 = 2.0 × 10−4 𝑀 = x
Find Ksp using solubility constant expression:
Ksp = [Ca2+][F-]2 = (x) (2x)2 = (2.0 x 10-4 )(4.0 x 10-4 )2 = 3.2 x 10-11
Solubility Product Constants near 25 oC
Formula
Al(OH)3
AlPO4
BaCO3
BaCrO4
BaF2
Ba(OH)2
BaSO4
BaSO3
BaS2O3
BiOCl
BiOOH
CdCO3
Cd(OH)2
CdC2O4
CdS
CaCO3
CaCrO4
CaF2
CaHPO4
Ca(OH)2
CaC2O4
Ca3(PO4)2
CaSO4
CaSO3
Cr(OH)2
Cr(OH)3
CoCO3
Co(OH)2
Co(OH)3
CoS
CuCl
CuCN
CuI
Cu3(AsO4)2
CuCO3
CuCrO4
Cu[Fe(CN)6]
Ksp
1.3×10–33
6.3×10–19
5.1×10–9
1.2×10–10
1.0×10–6
5×10–3
1.1×10–10
8×10–7
1.6×10–6
1.8×10–31
4×10–10
5.2×10–12
2.5×10–14
1.5×10–8
8×10–28
2.8×10–9
7.1×10–4
5.3×10–9
1×10–7
5.5×10–6
2.7×10–9
2.0×10–29
9.1×10–6
6.8×10–8
2×10–16
6.3×10–31
1.4×10–13
1.6×10–15
1.6×10–44
4×10–21
1.2×10–6
3.2×10–20
1.1×10–12
7.6×10–36
1.4×10–10
3.6×10–6
1.3×10–16
Formula
Cu(OH)2
CuS
FeCO3
Fe(OH)2
FeS
FeAsO4
Fe4[Fe(CN)6]3
Fe(OH)3
FePO4
Pb3(AsO4)2
Pb(N3)2
PbBr2
PbCO3
PbCl2
PbCrO4
PbF2
Pb(OH)2
PbI2
PbSO4
PbS
Li2CO3
LiF
Li3PO4
MgNH4PO4
Mg3(AsO4)2
MgCO3
MgF2
Mg(OH)2
MgC2O4
Mg3(PO4)2
MnCO3
Mn(OH)2
MnS
Hg2Br2
Hg2Cl2
Hg2I2
HgS
Ksp
2.2×10–20
6×10–37
3.2×10–11
8.0×10–16
6×10–19
5.7×10–21
3.3×10–41
4×10–38
1.3×10–22
4×10–36
2.5×10–9
4.0×10–5
7.4×10–14
1.6×10–5
2.8×10–13
2.7×10–8
1.2×10–15
7.1×10–9
1.6×10–8
3×10–28
2.5×10–2
3.8×10–3
3.2×10–9
2.5×10–13
2×10–20
3.5×10–8
3.7×10–8
1.8×10–11
8.5×10–5
1×10–25
1.8×10–11
1.9×10–13
3×10–14
5.6×10–23
1.3×10–18
4.5×10–29
2×10–53
Formula
NiCO3
Ni(OH)2
NiS
ScF3
Sc(OH)3
AgC2H3O2
Ag3AsO4
AgN3
AgBr
AgCl
Ag2CrO4
AgCN
AgIO3
AgI
AgNO2
Ag2SO4
Ag2S
Ag2SO3
AgSCN
SrCO3
SrCrO4
SrF2
SrSO4
TlBr
TlCl
TlI
Tl(OH)3
Sn(OH)2
SnS
ZnCO3
Zn(OH)2
ZnC2O4
Zn3(PO4)2
ZnS
* Sulfide equilibrium of the type:
MS(s) + H2O(l) ⇌ M2+(aq) + HS–(aq) + OH–(aq)
Ksp
6.6×10–9
2.0×10–15
3×10–19
4.2×10–18
8.0×10–31
2.0×10–3
1.0×10–22
2.8×10–9
5.0×10–13
1.8×10–10
1.1×10–12
1.2×10–16
3.0×10–8
8.5×10–17
6.0×10–4
1.4×10–5
6×10–51
1.5×10–14
1.0×10–12
1.1×10–10
2.2×10–5
2.5×10–9
3.2×10–7
3.4×10–6
1.7×10–4
6.5×10–8
6.3×10–46
1.4×10–28
1×10–26
1.4×10–11
1.2×10–17
2.7×10–8
9.0×10–33
2×10–25
Determining Individual Ion Concentration from Ksp:
Ex. Calculate the concentrations of barium ions and sulphate ions in a saturated aqueous solution of
barium sulphate at 25oC.
BaSO4(s) ⇌ Ba2+(aq) + SO42-(aq)
Ksp = [Ba2+][ SO42-] = 1.1 × 10-10  from table
Since [Ba2+] = [SO42-],
then Ksp = [Ba2+][ SO42-] can be written as Ksp = x2
where x = [Ba2+] = [SO42-]
Ksp = 1.1 × 10-10 = x2
x = 1.0 × 10-5 mol/L
[Ba2+] = [SO42-] = 1.0 × 10-5 mol/L
Ex. When 10.0 L of a saturated solution of magnesium carbonate, at 22°C is evaporated to dryness,
0.12g of solid magnesium carbonate is obtained. Calculate the Ksp of magnesium carbonate.
MgCO3(s) ⇌ Mg2+(aq) + CO32-(aq)
Ksp = [Mg2+][ CO32-]
Calculate the concentration of MgCO3:
Moles of MgCO3 = 0.12 g MgCO3 ×
1 mol MgCO 3
84.32g MgCO3
= 1.4 × 10-3 mol MgCO3
c=
1.4 × 10-3 mol
n

 [MgCO3] =
= 1.4 × 10-4 mol/L
V
10.0 L
if [MgCO3] = 1.4 × 10-4 mol/L, then [Mg2+] and [ CO32-] each = 1.4 × 10-4 mol/L
Ksp = (1.4 × 10-4 )(1.4 × 10-4 )
Ksp = (1.4 × 10-4)2 = 2.0 × 10-8
Ex. A saturated solution of ionic calcium hydroxide has a hydroxide ion concentration of 3.0×10-3mol/L.
Calculate the Ksp of calcium hydroxide at the same temperature.
Ca(OH)2(s) ⇌Ca2+(aq) + 2 OH−(aq)
Calculate [Ca2+] :
3.0 × 10-3 mol/L OH- ×
Ksp = [Ca2+][ OH-]2
1 mol Ca 2
= 1.5 × 10-3 mol/L Ca2+

2 mol OH
= (1.5 × 10-3)( 3.0 × 10-3)2 = 1.4 × 10-8
See Solubility Product Constant Assignment
The Common Ion Effect:

Remember that when a salt dissolves in water, some of it dissolves and some may stay as a solid.
There is a point at which the salt dissolving reaches equilibrium. This means that for each
formula unit of salt that dissolves, a pair of ions come together to create a formula unit.

When we dissolve ions in water we know that the product of the concentrations of the ions, to
the power of their coefficients in a balanced equation, is equal to the Ksp.

As long as the product of the concentration of the ions does not exceed the Ksp value, no
precipitate forms.
Let’s look at the following example:
Copper (I) iodide is not very soluble in water (Ksp = 4 × 10−19): CuI(s) ⇌Cu+(aq) + I−(aq)
Since Ksp = [Cu+][I-], if the product of the concentrations of Cu+ and I- stays below 4 × 10−19, no
precipitate will form, however, if we add extra iodine or copper ions (the common ion) things
will change. Let`s say we added some iodide ions; this will increase the concentration of iodide
and the product of the copper ions and iodide ions would be higher than the Ksp value. This is
not possible so the concentrations would need to be lowered. In order to lower the
concentration of iodide ions, some copper (I) iodide would need to precipitate out. This follows
Le Chatelier`s Principle because we increased the concentration of the products (iodide ions) so
the reaction shifts to the reactants.
After adding iodide ions to the solution, copper (I) iodide will precipitate out, but once the
reaction reaches equilibrium again, the concentration of copper ions and iodide will no longer
be equal (there will be more iodide ions in solution than copper ions). This also means the
concentration of Cu+ will be lower than at the original equilibrium and the concentration of Iwill be higher than at the original equilibrium.
Overview: If a slightly soluble ionic compound is dissolved in water, you can force
precipitation of that salt by adding a readily soluble ionic compound that has an ion in common
with the slightly soluble salt. This shift is known as the common ion effect.
Ex. The Ksp of silver iodide is 8.3 1017 . What is the iodide ion concentration of a 1.00 L saturated
solution of AgI to which 0.020 mol of AgNO3 is added?
AgI(aq) ⇌ Ag+(aq) + I-(aq)
Ksp = [ Ag  ][ I  ]  8.3 1017
If the equilibrium concentration of iodide ion from the dissociation is x, then the equilibrium
concentration of silver ion is x + 0.020. Because of the small value of Ksp, x will be small compared to
0.020. Therefore, the [Ag+] at equilibrium equals x + 0.020 ~ 0.020.
Ksp = [ Ag  ][ I  ]  8.3 1017
(0.020)(x) = 8.3 1017
x = 4.2  10 15 M = [I−]
Selective Precipitation

Selective precipitation is when you separate ions from two solutes to create a precipitate.

To calculate selective precipitation we use initial concentrations (Q), instead of the equilibrium
concentrations.

In selective precipitation there are restraints in forming a precipitate; a precipitate doesn't always
form in a solution. It depends on the relationship between K and Q:
o If Q>K a precipitate forms and the reaction proceeds to the left
o If Q<K no precipitate forms and the reaction proceeds to the right
o If Q=K the reaction is at equilibrium
Example: If 40L of 0.0050M KCl is mixed with 60L of 0.0030M Pb(NO3)2. Will the precipitate PbCl2
(Ksp= 1.6 X 10-5) form?
First we calculate the initial concentrations of Pb2+ and Cl-:
[Pb2+]0= (40L)(.005M)/(40L+ 60L)= .002M
[Cl-]0= (60L)(.003M)/(40L+60L) = .0018M
Calculate value of Q:
Q=[Pb2+][Cl-]2
Q=[.002M][.0018M]2
Q= 6.48 X 10-9 M
Q<K no ppt
Example: If 100L of 0.03M Pb(NO3)2 is mixed with 200L of 0.09M KCl. Does a precipitate (PbCl2) form?
(Ksp = 1.6 X 10-5). What are the equilibrium concentrations of Pb2+ and Cl-?
Pb2+
+
2Cl-
----->
PbCl2
[Pb2+]=(100L)(.03M)/(100L+200L) = .01M
[Cl-] = (200L)(.09M)/(100+200L) = .06M
Q=[.01M][.06M]2 -------> Q=3.6 X 10-5
Because Q>K a precipitate forms.
By using selective precipitation, we can calculate the amount of a reagent necessary to react with the
salt.
Example: If a solution contains 1.0 x 10-5 M of Pb2+ and 2.0 x 10-5 M Ag+ and Cl- is slowly added to
the solution, will AgCl (Ksp=1.6 x 10-10) or PbCl2 (Ksp=1.6 x 10-5) precipitate first? Which will be in
excess?
First we plug the given values into the Ksp expression for PbCl2 and solve for the concentration of Cl-.
1.6 x 10-5 = Ksp =[Pb2+][Cl-]2 = [1.0X10-5 M] [Cl-]2
[Cl-]2= 1.6 M
[Cl-] = 1.3 M
Next we plug in the given values to the Ksp expression for AgCl.
Ksp=1.6 X 10-10 = [Ag+][Cl-]
Ksp = 1.6 X 10-10 = [2.0 X 10-5 ] [Cl-]
[Cl-]= 8 X 10-6 M
AgCl will precipitate first because Cl- concentration @ equilibrium is smaller than PbCl2.
See Common Ion and Selective Precipitation Assignment
Lab: Analyzing for Anions
PURPOSE
In this experiment you will observe and record some characteristic chemical reactions of several
negatively charged ions: SO42-, CO32-, and I- using selected reagents. From a study of these observations
you will be able to develop a method for identifying each of these anions in the presence of the other
ions.
Your work will be typical of the work done by chemists who have developed elaborate schemes which
can be used to identify dozens of cations and anions in a given sample. Organize a data table in which
you can record your observations for the results obtained by adding each of the reagents to each of the
solutions of anions to be studied.
MATERIALS
Anion obtained
Cation obtained
Na2SO4, sodium sulfate, 0.1 M SO42-
Ba(NO3)2 , barium nitrate, 0.1 M Ba2+
Na2CO3, sodium carbonate, 0.1 M CO32-
AgNO3, silver nitrate, 0.1 M Ag+
NaCI, sodium chloride, 0.1 M CI-
HNO3, nitric acid, 1.0 M H+
Nal, sodium iodide, 0.1 M IPROCEDURE
a. The tests can easily be made by putting a few drops of a solution on a plastic film and adding to this a
few drops of one of the reagents. DO NOT use the tip of a dropper from a stock bottle for stirring. Doing
so could contaminate the solution contained in the bottle. Use a thin glass rod.
b. Test 2-3 drops of each anion solution separately with 1-2 drops of the 0.1 M Ba(NO3)2 reagent.
Record the results observed. To each mixture that contains a precipitate add 2-3 drops of 1.0 M HNO3
and record any changes you observe.
c. Test 2-3 drops of each anion solution separately with 1-2 drops of the 0.1 M AgNO3 reagent. Record
the results observed. To each mixture that contains a precipitate add 2—3 drops of 1.0 M HNO3 and
record any changes you observe.
d. Prepare new samples of the silver precipitates which formed in step c. Add 2—3 drops of 6 M
aqueous ammonia, NH3(aq), to the precipitates and record any changes you observe.
e. Test 2—3 drops of each anion solution separately with 1—2 drops of 1.0 M HNO3 and record the
results.
f. Examine the pattern of reactions shown by your table of observations. Develop a scheme by which
you could identify either one or a combination of two of the anions in a single solution. Organize your
scheme in such a way that another student could follow it. You may wish to try it out by making up your
own "trial unknowns" containing various combinations of the anions.
g. Obtain an unknown from your teacher and analyze it for the four anions. Report your results. Include
supporting evidence for your conclusions.
ANALYSIS
1) Write net ionic equations for the precipitation reactions which occurred when solutions of the anions
were mixed with the reagent containing Ba2+ ions.
2) Do the same for the precipitation reactions which occurred in step c with the Ag+ ions.
3) Write equations for the reactions which occurred when the precipitates were acidified with 1.0 M
HNO3 .
4) Write equations for the reactions which occurred when the silver precipitate reacted with aqueous
ammonia to form the complex ion, Ag(NH3)2+.