SCH4U-C 17
Lesson Solubility of Ionic Compounds Lesson 17 Chemistry SCH4U-C Introduction
In the last lesson, you reviewed solubility guidelines. These guidelines use the term "insoluble,"
but, in reality, every ionic substance will dissolve to some extent. The solubility of some
compounds is so low that it appears as if they do not dissolve. Ionic compounds with low
solubility will likely form precipitates in solutions.
One such ionic compound with low solubility is calcium oxalate. The forming of kidney stones,
which are composed of calcium oxalate, is one of the human health issues that you will study in
this unit. Planning Your Study
You may find this time grid helpful in planning when and how you will work through this
lesson.
Suggested Timing for This Lesson (Hours)
Ionic Compounds and Their Solubility ½ Saturated Solutions and Ksp Expressions ½ Calculating Concentration and Ksp ¾ Predicting Precipitates ½ The Common Ion Effect ¾ Determining Ksp Experimentally ½ Key Questions 1 What You Will Learn
After completing this lesson, you will be able to
• describe and explain equilibrium in saturated solutions
• write Ksp expressions, and calculate molar solubility and Ksp values
• predict whether a precipitate will form during double displacement reactions 1 Lesson 17 Chemistry SCH4U-C Ionic Compounds and Their Solubility
All ionic compounds will dissolve to some extent. Some are very soluble, while others have
extremely low solubility. There is a range of solubility for ionic compounds. A set of solubility
guidelines will help you to determine if an ionic substance will readily dissolve.
In this lesson, you'll need to refer to the solubility guidelines that you saw in Lesson 16: Solubility guidelines for common ionic substances
Ionic substances that contain these positive ions are soluble in water:
• H+
• NH4+
• Alkali metal ions
Ionic substances that contain these negative ions can be either soluble or
insoluble in water:
• ClO3–: All chlorates are soluble.
• NO3–: All nitrates are soluble.
• CH3COO–: All acetates are soluble.
• F
 –: All fluorides are soluble except for Mn2+, Ba2+, Ca2+, Cu2+, Fe2+, Fe3+,
Pb2+, Mg2+, Ni2+, Sr2+, and Zn2+.
• Cl–: All chlorides are soluble except for Hg2+, Ag+, Cr3+, Au+, and Pb2+.
• Br–: All bromides are soluble except for Hg2+, Ag+, Au+, and Pb2+.
• I –: All iodides are soluble except for Bi3+, Cu+, Au+, Au3+, Hg2+, Ag+, Pt3+,
and Pb2+.
• S O42–: All sulfates are soluble except for Sb3+, Ba2+, Ca2+, Fe3+, Pb2+, Mg2+,
Ag+, and Sr2+.
• S 2: All sulfides are insoluble except for NH4+, H+, alkali metal ions, Be2+,
Hg+, Ca2+, Sr2+, Ba2+, and Ra2+.
• C
 O32–, PO43–, and SO32–: All carbonates, phosphates, and sulfites are
insoluble except for NH4+, H+, and alkali metal ions.
• O
 H–: All hydroxides are insoluble except for NH4+, alkali metal ions, Ba2+,
Ca2+, Sr2+, Ra2+, and Tl+. 2 Lesson 17 Chemistry SCH4U-C Consider the compounds NaNO3, Pb(NO3)2, and AgNO3. All will be soluble because, according
to the guidelines, all nitrates are soluble. But if the metals were combined with chloride ions
(that is, NaCl, PbCl2, or AgCl), only NaCl will dissolve. According to the guidelines, most
chlorides are soluble, but there are some exceptions, such as PbCl2 and AgCl.
When ionic compounds dissolve in water, they dissociate (that is, separate) into their ions.
NaCl(s) → Na+(aq) + Cl–(aq)
Pb(NO3)2(s) → Pb2+(aq) + 2NO3–(aq)
In a solution, the water is the solvent that dissolves the solute (here, the ionic compound).
In an unsaturated solution, the solvent can dissolve more solute. But in a saturated solution,
the solvent has dissolved the maximum quantity of solute at that temperature. If you add
more solute, the solute will remain undissolved at the bottom of the container. In a saturated
solution, equilibrium is established. The equilibrium is between the ions that remain associated
(the undissolved solute on the bottom), and the ions that have been dissolved in the solvent. As
a result, equations that show the formation of saturated solutions are correctly written as:
NaCl(s) Na+(aq) + Cl–(aq)
Pb(NO3)2(s) Pb2+(aq) + 2NO3–(aq)
These two ionic equations are examples of dynamic equilibrium, as indicated by the doubleheaded arrows. You will observe that the pile of solid on the bottom of the container, whether
the table salt or the lead (II) nitrate, remains constant. If you could actually see the ions when
the solution has reached equilibrium, you would see that ions are leaving the solid at the same
rate that ions are returning to the solid. The forward rate is equal to the reverse rate. The pile of
solid on the bottom does not change in size. Support Questions
Be sure to try the Support Questions on your own before looking at
the suggested answers provided.
1. Which of the following ionic compounds would have low solubility? Al(OH)3 AgClO3 BaSO4 CaCO3 CuSO4 CuS NH4OH AlPO4 PtI3 NaHCO3
Suggested answer  3 Lesson 17 2. Chemistry SCH4U-C Write ionic equations for saturated solutions of each of the following ionic compounds:
a) Silver chromate Ag2CrO4(s) b) Barium phosphate Ba3(PO4)2(s) c) Aluminum hydroxide Al(OH)3(s)
d) Copper (II) iodate Cu(IO3)2(s) e) Calcium oxalate CaC2O4(s)  Suggested answer Saturated Solutions and Ksp Expressions
A saturated solution is formed when increasing amounts of an ionic compound are dissolved in
water until no more of the solid will dissolve and equilibrium is reached. For instance, you can
add a pinch of table salt to a glass of water and then stir the water to dissolve it, repeating the
process over and over until the salt begins piling up at the bottom of the glass. At this point, the
solution has become saturated, so no additional salt will dissolve.
An equilibrium expression for a saturated solution can be written following the same rules that
were used in previous lessons. Solids are not included in these equilibrium expressions; only the
aqueous or dissolved ions (aq) are included. The solid (concentration) does not form part of the
equilibrium expression because its concentration does not vary.
The equilibrium constant for saturated solutions is known as the Ksp or solubility product
constant. Ksp values, like Keq values, have no units.
Ksp expressions are written from balanced ionic equations, as shown in the following examples:
KOH(s) K+(aq) + OH–(aq)
Ksp = [K+][OH–]
This equilibrium expression is formed in the same manner as all K expressions, [products]/
[reactants]. In this case the [products] is omitted because it does not vary in concentration. You
can see where the terminology "solubility product" originates for this type of equilibrium. The
expression is simply a multiplication (product) of the concentrations.
Al2( SO4)3(s) 2Al3+(aq) + 3SO42–(aq)
Ksp = [Al3+]2[SO42–]3
Complete tables of Ksp values for various ionic compounds are included in chemistry textbooks
and can be found on the Internet. The complete tables only list ionic compounds that have low
solubility. A shortened Ksp table is shown on the next page. The table includes compounds that
will be used in this course. The table specifies that the Ksp values have been measured at 25°C.
As temperature increases, solubility generally increases. This causes the Ksp values to increase
because there will be a higher concentration of ions in the solution. The smaller the Ksp values
are, the less soluble the ionic compounds will be.
4 Lesson 17 Chemistry SCH4U-C Table 17.1: Solubility product constants for common ionic
compounds
Ksp values measured at 25°C
Nickel (II) hydroxide Ni(OH)2 1.6 × 10 –16 Calcium oxalate CaC2O4 2.3 × 10 –9 Iron (III) hydroxide Fe(OH)3 2.6 × 10 –39 Silver iodide AgI 8.5 × 10 –17 Lead (II) iodide PbI2 8.8 × 10 –9 Lead (II) sulfate PbSO4 1.8 × 10 –8 Calcium fluoride CaF2 3.5 × 10 –11 Calcium hydroxide Ca(OH)2 5.0 × 10 –6 Silver iodate AgIO3 3.2 × 10 –8 Silver chromate Ag2CrO4 1.1 × 10 –12 Magnesium hydroxide

Mg(OH)2 5.6 × 10 –12 Lead (II) chloride PbCl2 1.2 × 10 –5 Notice that all of these K values are very small, suggesting that at equilibrium, we will find very
little of the product (dissolved) ions in the system. Ksp values are listed only for chemicals with
very low solubilities. In most situations, these chemicals would be considered "insoluble" for
general purposes. Support Questions
3. Write Ksp expressions for each of the following ionic compounds. Use the balanced ionic
equations that you wrote in the previous set of Support Questions.
a) Ag2CrO4
b) Ba3(PO4)2
c) Al(OH)3
d) Cu(IO3)2
e) CaC2O4
4. Suggested answer How does the Ksp expression differ from most K expressions written earlier? Why is that?
Suggested answer 5 Lesson 17 Chemistry SCH4U-C Calculating Concentration and Ksp
Molar solubility is the term that is used to describe the concentration of the solute in a saturated
solution, which is the number of moles of solute dissolved in one litre of solution. Because
the ionic compound has dissolved into ions in the solution, it is also possible to state the
concentration of the ions as well.
For example, if 0.50 mol of KOH have been dissolved in 1.0 L of water, then the concentration
of KOH in the solution will be 0.50 mol/L. To describe this solution, we write [KOH(aq)] =
0.50 mol/L. Note that this is describing the dissolved KOH, not any KOH that is lying on the
bottom of the container. Even though we refer to the dissolved KOH, in fact that KOH exists in
solution as dissociated ions, not KOH together (associated).
The balanced ionic equation is: KOH(s) K+(aq) + OH–(aq)
The coefficients in the balanced equation are all equal to 1. The concentration of each
individual ion will be the same as the overall compound’s concentration:
[K+] = 0.50 mol/L
[OH–] = 0.50 mol/L
Consider a different ionic compound. If a solution of aluminum sulfate also has the same
concentration, 0.50 mol/L, what are the concentrations of the individual ions? Again, you have
to look at the balanced ionic equation:
Al2(SO4)3(s) 2Al3+(aq) + 3SO42–(aq)
The balanced chemical equation tells us that 1 mole of aluminum sulfate creates 2 moles
of aluminum ion and 3 moles of sulfate ion. Therefore if we have 0.50 mol/L of dissolved
aluminum sulfate, this will create:
[Al3+] = 1.00 mol/L (2 times 0.50 mol/L)
[SO42–] = 1.50 mol/L (3 times 0.50 mol/L)
You are now going to look at four different examples. Each one has a slightly different focus, but
in each case, you will be following a similar strategy to the one in the previous two lessons. The
following steps should be followed when you are asked to find concentration or Ksp. 6 Lesson 17 Chemistry SCH4U-C The initial–change–equilibrium (ICE) strategy
1.  rite the balanced ionic equation for the Ksp expression for the saturated
W
solution given in the question. 2. Record the initial concentrations using an ICE chart. 3.  se the balanced equation and the ICE chart to determine the changes in
U
concentrations of products (dissolved ions). 4. Substitute the equilibrium concentrations into the Ksp expression. 5. Solve for unknown values. Example
The concentration of a saturated solution (molar solubility) of PbI2 is 1.3 × 10–3 mol/L.
Determine the Ksp value of PbI2. Solution
Step 1:
PbI2(s) Pb2+(aq) + 2I–(aq)
Ksp = [Pb2+][I–]2
The equation tells us that 1 mole of PbI2 creates 1 mole of dissolved Pb2+ ion. The consequence
of that is that [Pb2+(aq)] = [PbI2(aq)]. If the molar solubility of lead iodide is given as 1.3 × 10–3
mol/L, this means that the concentration of the dissolved lead iodide in the saturated solution
is [PbI2(aq)] = 1.3 × 10–3 mol/L. We now can say that, in this saturated solution that is at
equilibrium,
[Pb2+(aq)] will be 1.3 × 10–3 mol/L
Note this value in our ICE table below.
Steps 2 and 3:
[Pb2+] in mol/L ICE
Initial 0 Change +x Equilibrium 1.3 × 10 [I–] in mol/L
0
+ 2x –3 2.6 × 10 –3 The initial concentration of the ions is recorded as 0. The ionic compound has just been
added to the water and has yet to start to dissolve. When the maximum amount of this ionic
compound has been dissolved, a saturated solution will have been created. 7 Lesson 17 Chemistry SCH4U-C The x and the 2x are determined by coefficients in the balanced ionic equation. The x and 2x
just reflect the relative concentrations of the lead and iodide ions in the solution as being 1:2
(mole ratio). We know from our previous discussion that the concentration of the lead ion will
be 1.3 × 10–3 mol/L.
Steps 4 and 5:
Ksp = [Pb2+][I–]2
Ksp = (1.3 × 10–3)(2.6 × 10–3)2
Ksp = 8.8 × 10–9
The Ksp for PbI2 is 8.8 × 10–9. Example
What is the molar solubility of nickel (II) hydroxide? Solution
In this example, we will determine the molar solubility using the known Ksp value from
our table. Molar solubility means [Ni(OH)2(aq)]. The balanced equation below shows us that
[Ni(OH)2(aq)] will be the same as [Ni2 +(aq)] (they are 1:1 in the equation). That ion concentration
will be denoted as x on our ICE table.
Step 1:
NI(OH)2(s) Ni2+(aq) + 2OH–(aq)
Ksp = [Ni2+][OH–]2
Steps 2 and 3:
[Ni2+] in mol/L ICE [OH–] in mol/L Initial 0 0 Change +x + 2x Equilibrium x 2x Steps 4 and 5:
Since you are not given the Ksp value, look it up in Table 17.1. You will see that it is
1.6 × 10–16. 8 Lesson 17 Chemistry SCH4U-C Ksp = 1.6 × 10–16 1.6 ×10−16 = (x)(2x)2
1.6 ×10−16 = 4x 3
1.6 ×10−16
4
3
x = 4.0 ×10−17
x3 = x = 3 4.0 ×10−17
x = 3.4 × 10−6
Since x represented the concentration of the nickel ion, then we can conclude that the molar
solubility of Ni(OH)2 is 3.4 × 10–6 mol/L. Example
In a saturated solution of silver carbonate, Ag2CO3, 0.0178 g of solute are dissolved in
500.mL of solution. Calculate the Ksp of silver carbonate. Solution
Step 1:
Ag2CO3(s) 2Ag+(aq) + CO32–(aq)
Ksp = [Ag+]2[CO32–]
Steps 2 and 3:
[Ag+] in mol/L ICE
Initial 0 Change + 2x Equilibrium 2.58 × 10 [CO32–] in mol/L
0
+x –4 1.29 × 10 –4 This problem becomes the same as the first example, where the molar solubility was given.
They tell us that a saturated solution was created here when 0.0178 g was dissolved in 500. mL
(0.500 L). Whatever concentration that creates must be the maximum that can be dissolved (the
molar solubility). So determine what concentration has been created when this solution was
created:
The molar mass of Ag2CO3 is 275.8 g/mol = M (using the periodic table).
Converting 0.0178 g to moles: n =
[Ag2CO3(aq)] = m
M n= 0.0178 g
= 0.0000645 mol
275.8 g / mol n
0.0000645 mol
=
= 1.29 × 10–4 mol/L
V
0.500L 9 Lesson 17 Chemistry SCH4U-C The balanced equation in Step 1 tells us that 1 mole of Ag2CO3 creates 1 mole of CO32– ion. The
CO32– ion created will be denoted as x on our ICE table. Again, the 2x and the x just remind us
that these ions are created in a mole ratio, 2:1. We know that if the [Ag2CO3(aq)] is 1.29 × 10–4
mol/L, then [CO32–] will also be 1.29 × 10–4 mol/L. Study the values in the ICE table to follow
what has been said.
The x and the 2x are determined by coefficients in the balanced ionic equation. The x is always
equal to the molar solubility, in this case, 1.29 × 10–4 mol/L.
Steps 4 and 5:
Ksp = (2.58 × 10–4)2(1.29 × 10–4) = 8.59 × 10–12
The Ksp of silver carbonate is 8.59 × 10–12. Support Questions
Solve the following problems. Your solutions should be in the proper format and your final answers
should have the correct units and number of significant figures.
5. A saturated solution of zinc sulfide has a concentration of 4.47 × 10 –13 mol/L. Determine
Suggested answer
the Ksp of zinc sulfide. 6. Calculate the concentration of a saturated solution (molar solubility) of iron (III)
Suggested answer
hydroxide. 7. If 0.0161 g of strontium fluoride dissolve in 125 mL of water to form a saturated solution,
Suggested answer
determine the Ksp of strontium fluoride.  10 Lesson 17 Chemistry SCH4U-C Predicting Precipitates Figure 17.1: The Brook Bottom Calcareous precipitate formed as a result off the lime-rich water seeping through a
mound of burnt lime that has filled the valley.
Source: http://upload.wikimedia.org/wikipedia/commons/0/03/Brook_Bottom_Calcareous_precipitate_-_geograph.org.uk_-_456623.jpg Precipitates are solids that form when two solutions are mixed together and a double
displacement reaction occurs. The ionic compounds that are found in the original solutions
are soluble. When the solutions are mixed together, a combination of ions creates an ionic
compound that has low solubility. The low solubility compound is the precipitate, which
collects at the bottom of the test tube or reaction vessel. The precipitate forms because
the two ions that make up the low-solubility compound exist in the solution at too high a
concentration. They are too close to each other, so they join together and form a solid.
You can use the solubility guidelines to identify the potential precipitate, but does a precipitate
always form? If the concentrations of the ions that form the precipitate are sufficiently low, then
no precipitate will form. This means that the ions are not too close together in the mixture
created.
Consider the following double displacement reaction and note the location of the two question
marks:
Pb(NO3)2(aq) + 2KI(aq) → 2KNO3(?) + PbI2(?)
This reaction goes to completion so the forward arrow → and not the double-headed arrow is used. The solubility guidelines indicate that PbI2 has low solubility, so the equation is
correctly completed as:
Pb(NO3)2(aq) + 2KI(aq) → 2KNO3(aq) + PbI2(s)
A very small amount of the lead iodide will remain dissolved as a saturated solution of PbI2.
Equilibrium is established between the solid precipitate lying on the bottom and the dissolved
ions in the solution. That equilibrium is described: 11 Lesson 17 Chemistry SCH4U-C PbI2(s) Pb2+(aq) + 2I–(aq)
Ksp = [Pb2+][I–]2
From the tables, Ksp = 8.8 × 10–9
How do you know if a precipitate will form or not? A precipitate forms if the concentrations of
the ions, when substituted into the Ksp expression, produce a value that is equal to or greater
than the Ksp.
In Lesson 15, you used Q as a trial, to see if a reaction had reached equilibrium. You will now
use Q again, to see if a precipitate has formed.
Q = [Pb2+][I–]2
If Q > Ksp, a precipitate forms When Q is greater than the Ksp value, it simply means that the
ions in question are too crowded together or too close together in the solution. The solvent
(water) will be unable to prevent them from joining together and forming the solid precipitate.
If Q < Ksp, there is no precipitate.
You need to know the concentration and volume of the two original solutions in order to
determine the [Pb2+] and the [I–].
Imagine that you mixed 40.0 mL of 0.250 mol/L Pb(NO3)2(aq) with 60.0 mL of 0.300 mol/L KI(aq).
Consider the lead nitrate solution before it was mixed with the other solution. The dissociation
of Pb(NO3)2(aq) is:
Pb(NO3)2(aq) → Pb2+(aq) + 2NO3–(aq)
We are told that in this lead nitrate solution, [Pb(NO3)2(aq)] = 0.250 mol/L. The dissociation
equation shows us that 1 mole of Pb(NO3)2 yields 1 mole of the Pb2+ ion. Therefore, the [Pb2+(aq)]
must also be 0.250 mol/L before this solution is mixed with the second solution. Our thinking
will be restricted to just the Pb2+ ion and the I– ion (from the second solution), since it is these
two ions that could potentially join together and form a precipitate. The NO3– ion and the K+
ions are of no consequence since these always form soluble compounds.
Let’s now look carefully at the second solution, and at the I– ion it contains. The dissociation of
KI is written as:
KI(aq) → K+(aq) + I–(aq)
We are told that in the KI solution, [KI(aq)] is 0.300 mol/L. Therefore the [I–(aq)] will also be 0.300
mol/L.
Summarizing: We will mix two solutions together.
Solution 1 is 40 mL and has [Pb2+(aq)] = 0.250 mol/L
Solution 2 is 60 mL and has [I–(aq)] = 0.300 mol/L
When the two solutions are mixed together, the increase in volume (to 100 mL) causes a dilution
to each ion concentration. The concentration of each ion must be calculated when the two
solutions are combined. 12 Lesson 17 Chemistry SCH4U-C Doing this for the lead ion: the volume changes from V1 = 40 mL to V2 = 100 mL. The
concentration changes from C1 = 0.250 mol/L to C2. Since the number of moles of lead ion
(C×V) does not change when it is diluted,
C1V1 = C2V2
C2 = C1V1
V2 [Pb2+ ]=
and [I– ]= 0.250 mol/L × 40.0 mL
= 0.100 mol/L
100 mL 0.300 mol/L × 60.0 mL
= 0.180 mol/L
100 mL Substitute the concentrations into the Q expression to see if a precipitate forms.
Q = (0.100)(0.180)2 = 3.24 × 10–3.
Q > Ksp, so a precipitate is formed.
The following is a summary of the strategy used to determine if a precipitate will form:
Step 1: Write out the balanced double displacement reaction and identify the potential
precipitate, using solubility guidelines.
Step 2: Write out the equilibrium reaction for the precipitate and the Q expression.
Step 3: Look up the Ksp of the potential precipitate.
Step 4: Calculate the concentrations of the two ions after the two solutions have been mixed,
and substitute the concentrations into the Q expression.
Step 5: Compare Q value and Ksp value. Support Questions
8. How is Q the same as Ksp? Why do we evaluate the Q expression and what does it tell us?
Suggested answer
 9. Solve the following problems. Your solutions should be in the proper format
and your final answers should have the correct units and number of significant figures.
a) Will a precipitate form when 20.0 mL of Pb(NO3)2(aq) with a concentration
of 2.50 × 10–4 mol/L is reacted with 45.0 mL of Na2SO4(aq) with a
concentration of 1.25 × 10 –3 mol/L?
b) Will a precipitate form when 60.0 mL of NaF(aq) with a concentration of
8.50 × 10–5 mol/L is reacted with 85.0 mL of CaCl2(aq) with a concentration
Suggested answer
of 1.25 × 10–4 mol/L? 13 Lesson 17 Chemistry SCH4U-C The Common Ion Effect Figure 17.2: Barbara Askins, NASA chemist
Source: http://upload.wikimedia.org/wikipedia/commons/thumb/0/0c/Barbara_Askins%2C_Chemist_-_GPN-2004-00022.jpg/474px-Barbara_Askins%2C_Chemist_-_GPN-2004-00022.jpg Whenever chemists prepare solutions, they use distilled water, instead of tap water, as the
solvent. Tap water contains ions. Sometimes the ions in tap water are identical to the ions in
the ionic compound being dissolved. Ions like these that are found in both the solute and the
solvent are known as common ions. How does the presence of a common ion, such as chlorine,
affect the molar solubility of an ionic compound? The common ion effect describes what occurs
in a solution of two dissolved solutes that contain the same ion or ions.
Consider the molar solubility of AgCl in distilled water. The Ksp for AgCl is 1.8 × 10–10.
AgCl(s) Ag+(aq) + Cl–(aq)
Ksp = [Ag+][Cl–]
[Ag+] in mol/L ICE [Cl–] in mol/L Initial 0 0 Change +x +x Equilibrium x x 1.8 × 10–10 = (x)(x)
1.8 × 10–10 = x2
x = 1.3 × 10–5
The molar solubility of silver chloride is 1.3 × 10–5 mol/L, when the solution is prepared using
distilled water.
But what if tap water, having a concentration of chloride ions of 1.0 × 10 –4 (0.00010 mol/L),
was used? Would the molar solubility be the same? Would the same amount of silver chloride
dissolve before reaching saturation? 14 Lesson 17 Chemistry SCH4U-C [Ag+] in mol/L ICE [Cl–] in mol/L Initial 0 0.00010 Change +x +x Equilibrium x 0.00010 + x Revised equilibrium x 0.00010 The Ksp expression for this system is
Ksp = [Ag+(aq)][Cl–(aq)]. Using the equilibrium concentrations shown in the ICE table (before the
revision), we can write
1.8 × 10–10 = x(0.00010 + x)
We could solve this equation for x, but because it is a quadratic equation, we should consider
whether we can make a simplifying assumption. The K for this system is very small, so we can
expect that the dissolved silver concentration, x, will be very small. We would expect that x will
be much smaller than 0.00010 in value. If that is so, then 0.00010 + x can be assumed to be just
0.00010. This simplifies our equation to:
1.8 × 10–10 = x(0.00010)
The “100 rule” seen earlier will also confirm this simplification. Applying the 100 rule: [original] 0.00010 mol/L
=
= 55555.6 > 100
K sp
1.8 ×10−10
1.8 × 10–10 = x(0.00010) 1.8 ×10−10
x=
= 1.8 ×10−6
0.00010
Since x represents the concentration of silver ion that was created by the addition of the AgCl,
the concentration of AgCl that must have been added must be the same value. The molar
solubility of silver chloride in tap water is 1.8 × 10–6 mol/L, which is lower than it is in distilled
water. Support Questions
10. Why should solutions, both in laboratory and hospital settings, always be prepared with
Suggested answer
distilled water? 11. Solve the following problems. Your solutions should be in the proper format and your
final answers should have the correct units and number of significant figures.
a) What is the maximum concentration of a silver iodate solution prepared with 1.00 L
of distilled water? (Hint: This is another way of asking, "What is the solubility?") 15 Lesson 17 Chemistry SCH4U-C b) What is the maximum concentration of a silver iodate...

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