SME
Grade 10 Science · Problem type
Solubility product
Use this for AgCl, a salt that barely dissolves: its Ksp from its molar solubility, or the solubility from Ksp.
Diagram
At equilibrium Q = K = 1.8 × 10⁻¹⁰: [Ag⁺] = 1.34 × 10⁻⁵ M, [Cl⁻] = 1.34 × 10⁻⁵ M.
This is an example. Type your own values to start a new problem.
Ksp
=
s
2
S
Solubility in grams per liter
calculated · g/L
Formulas
Ksp
=
s
2
1.8 ×
10
-10
=
(1.34 ×
10
-5
)
2
S
=
s
×
143.32
0.00192
=
1.34 ×
10
-5
×
143.32
Assumptions
•
AgCl(s) ⇌ Ag⁺ + Cl⁻: each unit that dissolves gives one of each ion, so both are s mol/L.
•
The solid is left out: Ksp = [Ag⁺][Cl⁻] = s². At 25 °C, AgCl’s Ksp is 1.8 × 10⁻¹⁰.
•
Silver chloride’s molar mass is 143.32 g/mol.
Step-by-step
WE KNOW
Ksp = 1.8 × 10⁻¹⁰
FIND
molar solubility (s), solubility in grams per liter (S)
Step 1 · Find molar solubility (s)
Use Ksp = s²
Ksp
is
s
times
s
,
so
s
is
its
square
root.
s
=
√
Ksp
s
=
√
1.8 ×
10
-10
s
=
1.34 ×
10
-5
mol/L
Step 2 · Find solubility in grams per liter (S)
Use S = s × 143.32
Each mole of AgCl that dissolves is 143.32 g.
S
=
1.34 ×
10
-5
×
143.32
S
=
0.00192
g/L
Check
Put the numbers back in. Both sides should match.
1.8 ×
10
-10
≈
(1.34 ×
10
-5
)
2
✓
0.00192
≈
1.34 ×
10
-5
×
143.32
✓
Topic check
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