SME
General Chemistry II
Solubility equilibria
Use this for “Ksp of Ag₂CrO₄ is 1.1 × 10⁻¹². Find its molar solubility and its solubility in g/L.”
Chart
A log scale: each grid line up is 10 times the one below.
Cations per formula unit a = 2
Anions per formula unit b = 1
Solubility product Ksp = 1.1 × 10⁻¹²
Molar mass M = 331.73 g/mol
Solubility in grams per litre Sg = 0.0216 g/L
This is an example. Type your own values to start a new problem.
Ksp
=
(
×
s (M)
)
×
(
×
s (M)
)
[M]
Cation concentration
calculated · M
[X]
Anion concentration
calculated · M
M
Molar mass
example · g/mol
Sg
Solubility in grams per litre
calculated · g/L
Formulas
Ksp
=
a
a
×
b
b
×
s
a
+
b
1.1 ×
10
-12
=
2
2
×
1
1
×
(6.50 ×
10
-5
)
2 + 1
[
M
]
=
a
×
s
1.30 ×
10
-4
=
2
×
(6.50
×
10
-5
)
[X]
=
b
×
s
6.50 ×
10
-5
=
1
×
(6.50
×
10
-5
)
Sg
=
s
×
M
0.0216
=
6.50 ×
10
-5
×
331.73
Assumptions
•
Each formula unit of MₐXᵦ that dissolves gives a cations and b anions, so [M] = as and [X] = bs.
•
The salt dissolves in pure water: no common ion, no hydrolysis and no complex ions. Activities equal concentrations.
•
The solid is left out of Ksp, so Ksp = [M]ᵃ[X]ᵇ = aᵃbᵇs⁽ᵃ⁺ᵇ⁾.
•
One temperature throughout (25 °C unless the problem says), so Ksp stays the same.
Step-by-step
WE KNOW
a = 2, b = 1, Ksp = 1.1 × 10⁻¹², M = 331.73 g/mol
FIND
molar solubility (s), cation concentration ([M]), anion concentration ([X]), solubility in grams per litre (Sg)
Step 1 · Find molar solubility (s)
Use Ksp = aᵃ × bᵇ × sᵃ⁺ᵇ
Ksp
=
(as)ᵃ(bs)ᵇ
=
aᵃbᵇs⁽ᵃ⁺ᵇ⁾.
Divide
Ksp
by
aᵃbᵇ,
then
take
the
(
a
+
b
)th
root.
s
=
∛(
Ksp
a
a
×
b
b
)
s
=
∛(
1.1 ×
10
-12
2
2
×
1
1
)
s
=
∛(
1.1 ×
10
-12
4
)
s
=
∛(2.75
×
10
-13
)
s
=
6.50 ×
10
-5
M
Step 2 · Find cation concentration ([M])
Use [M] = a × s
Multiply
s
by
the
cations
per
formula
unit.
[
M
]
=
2
×
(6.50
×
10
-5
)
[
M
]
=
1.30 ×
10
-4
M
Step 3 · Find anion concentration ([X])
Use [X] = b × s
Multiply
s
by
the
anions
per
formula
unit.
[X]
=
1
×
(6.50
×
10
-5
)
[X]
=
6.50 ×
10
-5
M
Step 4 · Find solubility in grams per litre (Sg)
Use Sg = s × M
Moles per litre times grams per mole gives grams per litre.
Sg
=
6.50 ×
10
-5
×
331.73
Sg = 0.0216 g/L
Check
Put the numbers back in. Both sides should match.
1.1 ×
10
-12
=
2
2
×
1
1
×
(6.50 ×
10
-5
)
2 + 1
✓
1.30 ×
10
-4
=
2
×
(6.50
×
10
-5
)
✓
6.50 ×
10
-5
=
1
×
(6.50
×
10
-5
)
✓
0.0216
=
6.50 ×
10
-5
×
331.73
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.