SME
General Chemistry II · Problem type
Solubility with a common ion
Use this for “Ksp = 5.4 × 10⁻¹³ for AgBr. Find its molar solubility in 0.020 M NaBr and in pure water.”
Diagram
At equilibrium Q = K = 5.4 × 10⁻¹³: [M⁺] = 2.7 × 10⁻¹¹ M, [X⁻] = 0.02 M.
This is an example. Type your own values to start a new problem.
Ksp
=
s (M)
×
(
c (M)
+
s (M)
)
s₀
Molar solubility in pure water
calculated · M
Formulas
Ksp
=
s
×
(
c
+
s
)
5.4 ×
10
-13
=
2.70 ×
10
-11
×
(0.02
+
(2.70
×
10
-11
))
s₀
=
√
Ksp
7.35 ×
10
-7
=
√
5.4 ×
10
-13
Assumptions
•
A 1:1 salt MX, such as AgCl: each unit that dissolves gives one M⁺ and one X⁻, so [M⁺] = s and [X⁻] = c + s.
•
The common ion comes from a soluble salt, such as NaCl, that is all dissolved: c is its concentration.
•
The solid is left out of Ksp, so Ksp = [M⁺][X⁻] = s(c + s). It is solved exactly, with no small-s guess.
•
One temperature throughout (25 °C unless the problem says), so Ksp stays the same.
Step-by-step
WE KNOW
Ksp = 5.4 × 10⁻¹³, c = 0.02 M
FIND
molar solubility with the common ion (s), molar solubility in pure water (s₀)
Step 1 · Find molar solubility with the common ion (s)
Use Ksp = s × (c + s)
Ksp
=
s
(
c
+
s
)
is
s
2
+
cs
−
Ksp
=
0.
Its
positive
root
is
−
c
+
√
c
2
+ 4Ksp
2,
written
here
as
2K
sp
c
+
√
c
2
+ 4Ksp
so
a
tiny
s
keeps
its
digits.
s
=
2
×
Ksp
c
+
√
c
2
+ 4 × Ksp
s
=
2
×
(5.4
×
10
-13
)
÷
(0.02
+
√(
0.02
2
+
4
×
(5.4
×
10
-13
)))
s
=
2
×
(5.4
×
10
-13
)
÷
(0.02
+
√(0.0004
+
4
×
(5.4
×
10
-13
)))
s
=
2
×
(5.4
×
10
-13
)
÷
(0.02
+
√(0.0004
+
(2.16
×
10
-12
)))
s
=
2
×
5.4 ×
10
-13
0.02 +
√
0.0004
s
=
2
×
5.4 ×
10
-13
0.02 + 0.02
s
=
2
×
5.4 ×
10
-13
0.04
s
=
1.08 ×
10
-12
0.04
s
=
2.70 ×
10
-11
M
Step 2 · Find molar solubility in pure water (s₀)
Use s₀ = √(Ksp)
In
pure
water
c
=
0,
so
Ksp
=
s₀
×
s₀
and
s₀
is
its
square
root.
s₀
=
√
5.4 ×
10
-13
s₀
=
7.35 ×
10
-7
M
Check
Put the numbers back in. Both sides should match.
5.4 ×
10
-13
=
2.70 ×
10
-11
×
(0.02
+
(2.70
×
10
-11
))
✓
7.35 ×
10
-7
=
√
5.4 ×
10
-13
✓
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