The document provides examples for calculating the Ksp (solubility product constant) of various ionic compounds given their molar solubility in water. It first explains the general process: writing the dissolution reaction, writing the Ksp expression, inserting concentrations based on molar ratios, and calculating the value. Several examples are then worked through step-by-step, calculating Ksp values from molar solubilities for compounds such as silver bromide, calcium fluoride, and mercury(I) bromide. The document also discusses calculating Ksp from solubilities provided in grams per 100mL by first converting to molar solubility. More examples demonstrate this process for nickel sulfide, magnesium fluoride, and manganese(II) i

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Calculating the Ksp from the Molar Solubility
The molar solubility of a substance is the number of moles that dissolve per liter of
solution. For very soluble substances (like sodium nitrate, NaNO3), this value can be
quite high, exceeding 10.0 moles per liter of solution in some cases.
For insoluble substances like silver bromide (AgBr), the molar solubility can be quite
small. In the case of AgBr, the value is 5.71 x 10¯7
moles per liter.
Given this value, how does one go about calculating the Ksp of the substance? Here is a
skeleton outline of the process:
1) Write the chemical equation for the substance dissolving and dissociating.
2) Write the Ksp expression.
3) Insert the concentration of each ion and multiply out.
Example #1: Determine the Ksp of silver bromide, given that its molar solubility is 5.71
x 10¯7
moles per liter.
Solution:
When AgBr dissolves, it dissociates like this:
AgBr(s) ⇌ Ag+
(aq) + Br¯(aq)
The Ksp expression is:
Ksp = [Ag+
] [Br¯]
There is a 1:1 molar ratio between the AgBr that dissolves and Ag+
that is in solution. In
like manner, there is a 1:1 molar ratio between dissolved AgBr and Br¯ in solution. This
means that, when 5.71 x 10¯7
mole per liter of AgBr dissolves, it produces 5.71 x
10¯7
mole per liter of Ag+
and 5.71 x 10¯7
mole per liter of Br¯ in solution.
Putting the values into the Ksp expression, we obtain:
Ksp = (5.71 x 10¯7
) (5.71 x 10¯7
) = 3.26 x 10¯13
Example #2: Determine the Ksp of calcium fluoride (CaF2), given that its molar
solubility is 2.14 x 10¯4
moles per liter.
Solution:
When CaF2 dissolves, it dissociates like this:
CaF2(s) ⇌ Ca2+
(aq) + 2F¯(aq)
The Ksp expression is:
Ksp = [Ca2+
] [F¯]2
There is a 1:1 molar ratio between CaF2 and Ca2+
, BUT there is a 1:2 molar ratio
between CaF2 and F¯. This means that, when 2.14 x 10¯4
mole per liter of
CaF2 dissolves, it produces 2.14 x 10¯4
mole per liter of Ca2+
and it produces 4.28 x
10¯4
mole per liter of F¯ in solution.
Putting the values into the Ksp expression, we obtain:
Ksp = (2.14 x 10¯4
) (4.28 x 10¯4
)2
= 3.92 x 10¯11
Please note, I DID NOT double the F¯ concentration. The F¯ concentration is TWICE
the value of the amount of CaF2 dissolving.
Example #3: Determine the Ksp of mercury(I) bromide (Hg2Br2), given that its molar
solubility is 2.52 x 10¯8
mole per liter.
Solution:

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When Hg2Br2 dissolves, it dissociates like this:
Hg2Br2(s) ⇌ Hg2
2+
(aq) + 2Br¯(aq)
Important note: it is NOT 2Hg+
. IT IS NOT!!! If you decide that you prefer 2Hg+
, then I
cannot stop you. However, it will give the wrong Ksp expression and the wrong answer
to the problem.
The Ksp expression is:
Ksp = [Hg2
2+
] [Br¯]2
There is a 1:1 ratio between Hg2Br2 and Hg2
2+
, BUT there is a 1:2 ratio between
Hg2Br2 and Br¯. This means that, when 2.52 x 10¯8
mole per liter of Hg2Br2 dissolves, it
produces 2.52 x 10¯8
mole per liter of Hg2
2+
, BUT 5.04 x 10¯8
mole per liter of Br¯ in
solution.
Putting the values into the Ksp expression, we obtain:
Ksp = (2.52 x 10¯8
) (5.04 x 10¯8
)2
= 6.40 x 10¯23
Example #4: Calculate the Ksp for Ce(IO3)4, given that its molar solubility is 1.80 x
10¯4
mol/L
Solution:
The Ksp expression is:
Ksp = [Ce4+
] [IO3¯]4
We know the following:
There is a 1:1 molar ratio between the molar solubility and the cerium(IV) ion's
concentration.
These is a 1:4 molar ratio between the molar solubility and the iodate ion.
Therefore:
Ksp = (1.80 x 10¯4
) (7.20 x 10¯4
)4
Ksp = 4.84 x 10¯17
Example #5: Calculate the Ksp for Mg3(PO4)2, given that its molar solubility is 3.57 x
10-6
mol/L.
Solution:
The Ksp expression is:
Ksp = [Mg2+
]3
[PO4
3
¯]2
We know the following:
These is a 3:1 ratio between the concentration of the magnesium ion and the molar
solubility of the magnesium phosphate.
There is a 2:1 ratio between the concentration of the phosphate ion and the molar
solubility of the magnesium phosphate.
Therefore:
Ksp = (1.071 x 10¯5
)3
(7.14 x 10¯6
)2
Ksp = 6.26 x 10¯26
For what it's worth, my "Handbook of Chemistry and Physics" gives the Ksp as 9.86 x
10¯25
For each compound, the molar solubility is given. Calculate its Ksp.
Example #6: AgCN, 7.73 x 10¯9
M
Example #7: Zn3(AsO4)2, 1.236 x 10¯6
M

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Example #8: Hg2I2, 2.37 x 10¯10
M
Example #9: A saturated solution of magnesium fluoride , MgF2, was prepared by
dissolving solid MgF2 in water. The concentration of Mg2+
ion in the solution was found
to be 2.34 x 10-4
M. Calculate the Ksp for MgF2.
Solution:
1) Here's the chemical equation for the dissolving of MgF2:
MgF2(s) ⇌ Mg2+
(aq) + 2F¯(aq)
2) The Ksp expression is this:
Ksp = [Mg2+
] [F¯]2
3) Based on the stoichiometry of the chemical equation, the [F¯] is this:
4.68 x 10-4
M
4) To calculate the Ksp, do this:
Ksp = (2.34 x 10-4
) (4.68 x 10-4
)2
To three sig figs, the Ksp is 5.12 x 10-11
Calculating the Ksp from the gram per 100 mL solubility
Sometimes, the solubility is given in grams per 100 mL, rather than molar solubility
(which is in mol/L). The Ksp can still be calculated from these data, but indirectly. First,
we have to convert the g/100mL data to mol/L (molar) solubility data. Then the molar
solubility value is used to perform the actual calculation.
Here is how to convert a g/100mL value to molar solubility:
1) multiply the g/100mL value by 10/10. This converts it to grams per 1000 mL or,
better yet, grams per liter. (Sometimes the data is given in g/L. When that happens, this
step is skipped.)
2) divide the grams per liter value by the molar mass of the substance. This gives moles
per liter, which is molar solubility.
After the above conversion, the problem becomes calculate the Ksp from molar solubility
data.
Here are all the problems. The general format is this: "Determine the Ksp of ____, given
the solubility of ___.
1) nickel sulfide (NiS), 2.97 x 10¯10
g/100mL 6) Cu(IO4)2, 0.380 g/300mL
2) magnesium fluoride (MgF2), 1.65 x 10¯3
g/100mL 7) PbBr2, 1.04 x 10¯8
g/100mL
3) manganese(II) iodate [Mn(IO3)2], 1.935 x 10¯1
g/100mL 8) ZnS, 5.28 x 10¯12
g/100mL
4) calcium arsenate [Ca3(AsO4)2], 0.043 g/L 9) Cd3(PO4)2, 6.25 x 10¯6
g/100mL
5) CaF2, 0.00680 g/250mL 10) BaSO4, 0.000245 g/100mL
Note that one problem uses 250mL and another uses 300mL. In each case, a conversion
is made to the equivalent g/L value and from there to the mol/L value.
In these types of problem, the 100mL in the unit never sets the number of significant
figures. Use the amount that dissolves, such as, for example, 2.97 x 10¯10
g/100mL has
three sig. figs.
Example #1: Determine the Ksp of nickel sulfide (NiS), given that its solubility is 2.97 x
10¯10
grams / 100mL.
Solution:

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Convert to grams per 1000 mL, then moles per liter:
2.97 x 10¯10
grams / 100mL x (10/10) = 2.97 x 10¯9
grams / 1000mL
2.97 x 10¯9
grams / L divided by 90.77 g/mol = 3.27 x 10¯11
mol/L
When NiS dissolves, it dissociates like this:
NiS(s) ⇌ Ni2+
(aq) + S2
¯(aq)
The Ksp expression is:
Ksp = [Ni2+
] [S2
¯]
There is a 1:1 ratio between NiS and Ni2+
and there is a 1:1 ratio between NiS and S2
¯.
This means that, when 3.27 x 10¯11
mole per liter of NiS dissolves, it produces 3.27 x
10¯11
mole per liter of Ni2+
and 3.27 x 10¯11
mole per liter of S2
¯ in solution.
Putting the values into the Ksp expression, we obtain:
Ksp = (3.27 x 10¯11
) (3.27 x 10¯11
) = 1.07 x 10¯21
Example #2: Determine the Ksp of magnesium fluoride (MgF2), given that its solubility
is 1.65 x 10¯3
grams per 100mL.
Solution:
Convert to grams per 1000 mL, then moles per liter:
1.65 x 10¯3
grams / 100mL x (10/10) = 1.65 x 10¯2
grams / 1000mL
1.65 x 10¯2
grams / L divided by 62.30 g/mol = 2.65 x 10¯4
mol/L
When MgF2 dissolves, it dissociates like this:
MgF2(s) ⇌ Mg2+
(aq) + 2F¯(aq)
The Ksp expression is:
Ksp = [Mg2+
] [F¯]2
There is a 1:1 ratio between MgF2 and Mg2+
, BUT there is a 1:2 ratio between MgF2 and
F¯. This means that, when 2.65 x 10¯4
mole per liter of MgF2 dissolves, it produces 2.65
x 10¯4
mole per liter of Mg2+
and 5.30 x 10¯4
mole per liter of F¯ in solution.
Putting the values into the Ksp expression, we obtain:
Ksp = (2.65 x 10¯4
) (5.30 x 10¯4
)2
= 7.44 x 10¯11
Please note, I DID NOT double the F¯ concentration. I took the MgF2 concentration (its
molar solubility) and doubled it to get the F¯ concentration. This is because f the 1:2
molar ratio between MgF2 and F¯.
Example #3: Determine the Ksp of manganese(II) iodate [Mn(IO3)2], given that its
solubility is 1.935 x 10¯1
grams per 100mL.
Solution:
Convert to grams per 1000 mL, then moles per liter:
1.935 x 10¯1
grams / 100mL x (10/10) = 1.935 grams / 1000mL
1.935 grams / L divided by 404.746 g/mol = 4.78 x 10¯3
mol/L
When Mn(IO3)2 dissolves, it dissociates like this:
Mn(IO3)2(s) ⇌ Mn2+
(aq) + 2IO3¯(aq)
The Ksp expression is:
Ksp = [Mn2+
] [IO3¯]2
There is a 1:1 ratio between Mn(IO3)2 and Mn2+
, BUT there is a 1:2 ratio between
Mn(IO3)2 and IO3¯. This means that, when 4.78 x 10¯3
mole per liter of
Mn(IO3)2 dissolves, it produces 4.78 x 10¯3
mole per liter of Mn2+
and 9.56 x 10¯3
mole
per liter of IO3¯ in solution.

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Putting the values into the Ksp expression, we obtain:
Ksp = (4.78 x 10¯3
) (9.56 x 10¯3
)2
= 4.37 x 10¯7
Example #4: Determine the Ksp of calcium arsenate [Ca3(AsO4)2], given that its
solubility is 0.13 g/L.
Solution:
Since the solubility is already in g/L, we can proceed directly to calcuating the solubility
in moles per liter:
0.13 grams / L divided by 398.078 g/mol = 3.2657 x 10¯4
mol/L
The Ksp expression is:
Ksp = [Ca2+
]3
[AsO4
2
¯]2
Putting concentration values into the Ksp expression, we obtain:
Ksp = (9.7971 x 10¯4
)3
(6.5314 x 10¯4
)2
= 4.01 x 10¯18
In checking this problem, I found an online listing for the Ksp of calcium arsenate. The
value given was 6.8 x 10¯19
I used the Wikipedia entry for calcium arsenate, where the solubility is given as 0.013
g/100mL.
Example #5: A saturated solution of CaF2 contains 0.00680 g/250mL of solution. Find
the Ksp of CaF2.
Solution:
1) Calculate the grams of CaF2 in one liter:
0.00680 g/250mL x 4 = 0.0272 g/L
Comment: there are four 250mL amounts in 1 liter
2) Convert g/L to mol/L:
0.0272 g/L divided by 78.074 g/mol = 3.48387 x 10¯4
mol/L (a few guard digits)
3) Set up Ksp expression and solve for it:
Ksp = [Ca2+
] [F¯]2
Ksp = (3.48387 x 10¯4
) (6.96774 x 10¯4
)2
Ksp = 1.69 x 10¯10
Example #6: 300. mL of a saturated solution of Cu(IO4)2 contains 0.380 grams of
dissolved salt. Determine the Ksp.
Solution:
1) Calculate the grams of Cu(IO4)2 in one liter:
0.380 g / 0.300 L = x / 1.00 L
x = 1.267 g/L
2) Convert g/L to mol/L
1.267 g/L / 445.338 g/mol = 0.002845 mol/L
3) Plug into Ksp expression for Cu(IO4)2:
Ksp = [0.002845] [0.005690]2
Ksp = 9.21 x 10¯8
Given the solubility in grams per 100 mL, calculate the Ksp
7) PbBr2; 1.04 x 10¯8
g/100mL
8) ZnS; 5.28 x 10¯12
g/100mL

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9) Cd3(PO4)2; 6.25 x 10¯6
g/100mL
Answers only. No detailed solutions
Example #10: A saturated solution of BaSO4 contains 0.000245 g/100mL Calculate its
Ksp.
Solution:
1) Calculate the grams of BaSO4 in one liter:
(0.000245 g/100mL) (10 100mL/L) = 0.00245 g/L
2) Convert g/L to mol/L:
0.00245 g/L / 233.391 g/mol = 0.0000104974 mol/L <--- keep a few guard digits
3) Write the Ksp expression for BaSO4 and calculate the Ksp:
Ksp = [Ba2+
] [SO4
2
¯]
Ksp = (0.0000104974) (0.0000104974)
Ksp = 1.10 x 10¯10
4) By the way, BaSO4 dissociates like this:
BaSO4(s) ⇌ Ba2+
(aq) + SO4
2
¯(aq)
Example #11: The solubility of Ba3(PO4)2 is 7.571 x 10¯4
g/100mL of water at 25 °C.
Determine the Ksp of Ba3(PO4)2 at this temperature.
Solution:
1) Grams in one liter:
(7.571 x 10¯4
g/100mL) (10 100mL/L) = 7.571 x 10¯3
g/L
2) Convert g/L to mol/L:
7.751 x 10¯3
g/L / 601.93 g/mol = 1.25779 x 10¯5
mol/L <--- keep a few guard digits
3) Write the Ksp expression for Ba3(PO4)2 and calculate the Ksp:
Ksp = [Ba2+
]3
[PO4
2
¯]2
<--- I'll skip writing the chemical equation.
[Ba2+
] = (1.25779 x 10¯5
) (3) = 0.0000377337 M
[PO4
2
¯] = (1.25779 x 10¯5
) (2) = 0.0000251558 M
Ksp = (0.0000377337)3
(0.0000251558)2
= 3.40 x 10¯23
This source is where I got the Ksp value and then I back-calculated to get the g/100mL
value to start the problem. Pretty smart, huh?
Calculate the Ksp of a Saturated Solution When Given the pH
Please be aware that this problem does not require a concentration to be given in the
problem. We know the pH and we know the solution is saturated. This information will
be sufficient.
Warning: you need to know about Ksp AND acid base ideas to do this problem type. If
you lack one or the other of these skills, just be aware you just might struggle a little in
your understanding of this problem type.
To solve the problem, we must first calculate the [OH¯]. To do this, we will use the pH
and acid base concepts. Then, we will use the Ksp expression to calculate the Ksp
Final Note: Ksp are almost always given at 25.0 °C in reference sources. All problems in
this tutorial are taken to be at 25.0 °C. If you were to see a problem where the specified
temperature was different, it's probably just that the reference source gave a Ksp that, for
whatever reason, was not at 25.0 °C.

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Example #1: A saturated solution of Mg(OH)2 is prepared. The pH of the solution is
10.17. What is the Ksp for this compound?
Solution:
1) The chemical equation:
Mg(OH)2 ⇌ Mg2+
+ 2OH¯
2) The Ksp expression:
Ksp = [Mg2+
] [OH¯]2
3) Use the pH to get the pOH:
14.00 - 10.17 = 3.83
4) Use the pOH to get the [OH¯]:
[OH¯] = 10¯pOH
= 10¯3.83
= 1.479 x 10¯4
M
5) From the chemical equation, we note that the [Mg2+
] is half the value of the [OH¯],
therefore:
[Mg2+
] = 1.479 x 10¯4
M divided by 2 = 7.395 x 10¯5
M
6) We now have the necessary values to put into the Ksp expression:
Ksp = (7.395 x 10¯5
) (1.479 x 10¯4
)2
Ksp = 1.62 x 10¯12
Note: the book value for the Ksp of Mg(OH)2 is 5.61 x 10¯12
. The reason for the
difference is that teachers like to change the values slightly so as to not allow a student
to simply look up the correct answer and claim they did all their work on the calculator.
Calculating gram per 100 mL solubility, given the Ksp
Basically, this type of problem will (1) calculate the molar solubility from the Ksp. Then,
in an additional step, (2) calculate grams per liter from moles per liter. From there, it is
an easy third step to g/100mL.
Problem #1: Calculate the milligrams of silver carbonate in first 100 mL (then 250 mL)
of a saturated solution of Ag2CO3 (Ksp for silver carbonate = 8.4 x 10¯12
)
Solution:
1) Solve for the molar solubility of Ag2CO3:
Ag2CO3 ⇌ 2Ag+
+ CO3
2
¯
Ksp = [Ag+
]2
[CO3
2
¯]
8.4 x 10¯12
= (2s)2
(s)
s = 1.28 x 10¯4
mol/L
Comment: The value of s is the molar concentration of the carbonate ion. Since there is a
1:1 molar ratio between it and silver carbonate, the value for s is also the molar solubility
of silver carbonate.
2) Convert mol/L to gram/L:
1.28 x 10¯4
mol/L times 275.748 g/mol = 3.53 x 10¯2
g/L
3) Convert from g/L to g/100mL:
3.53 x 10¯2
g/L divided by 10 = 3.53 x 10¯3
g/100mL
Comment: this is done because there are 10 100mL amounts in 1 L.
4) Convert mol/L to g/250mL:
3.53 x 10¯2
g/L divided by 4 = 8.83 x 10¯3
g/250mL
Comment: this is done because there are 4 250mL amounts in 1 L.

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Problem #2: Calculate the milligrams of silver ion that are present in 250 mL of a
saturated solution of silver carbonate.
Solution:
1) Restate the value for s from problem #1, the molar concentration of the carbonate ion:
s = 1.28 x 10¯4
mol/L
2) From the stoichiometry of the chemical equation, the molar concentration of the silver
ion is twice that of the carbonate ion:
[Ag+
] = 2.56 x 10¯4
mol/L
3) Determine g/L of silver ion:
2.56 x 10¯4
mol/L times 107.8682 g/mol = 0.0276 g/L
4) Determine g/250mL and convert to mg/250mL
0.0276 g/L divided by 4 = 0.00690 g/250mL
0.00690 g/250mL = 6.90 mg/250mL
Problem #3: Calculate the mass of Ca5(PO4)3F (Ksp = 1.00 x 10-60
) which will dissolve
in 100 mL of water.
Solution:
1) Determine moles of Ca5(PO4)3F in one liter:
Ca5(PO4)3F(s) ⇌ 5Ca2+
+ 3PO4
3-
+ F¯
Ksp = [Ca2+
]5
[PO4
3-
]3
[F¯]
1.00 x 10-60
= (5s)5
(3s)3
(s)
1.00 x 10-60
= 84375s9
s = 6.11 x 10-8
M
2) Determine moles of Ca5(PO4)3F in 100 mL:
6.11 x 10-8
mole divided by 10 = 6.11 x 10-9
mol / 100mL
3) Determine how many grams this is:
6.11 x 10-9
mol times 504.298 g/mol = 3.08 x 10-6
g / 100mL
Problem #4: Calculate the mass of Ca5(PO4)3OH (Ksp = 6.80 x 10-37
) which will
dissolve in 100 ml of water. (Please ignore any complications which might result as a
consequence of the basicity or acidity of the ions, ion pairing, complex ion formation, or
auto-ionization of water.)
Solution:
1.00 x 10-37
= (5s)5
(3s)3
(s)
1.00 x 10-37
= 84375s9
s = 2.1955 x 10-5
M
This is 2.1955 x 10-6
mol / 100mL
2.1955 x 10-6
mol / 100mL times 502.3069 g/mol = 1.10 x 10-2
g / 100mL
By the way, we can use the molarity to determine the pH of a saturated solution of
hydroxyapatite. It is 9.342.
You may be interested in seeing the above two questions answered on Yahoo
Answers. The question there has an additional part concerning the chemical reasoning
behind fluoridation of water.

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Problem #5: The Ksp for magnesium arsenate, Mg3(AsO4)2, is 2.10 x 10-20
at 25 °C.
What is the solubility of magnesium arsenate in g/L?
Solution:
1) Determine the molar solubility:
Ksp = [Mg2+
]3
[PO4
3-
]2
2.10 x 10-20
= (3s)3
(2s)2
2.10 x 10-20
= 108s5
s = 0.0000454736 M <--- left some extra digits, will round off at end
2) Convert from mol/L to g/L:
0.0000454736 mol/L times 350.751 g/mol = 0.0159 g/L (to three sig figs)