Common ion effect

#chemistry/equilibrium #solubility #le-chateliers-principle

What Is the Common Ion Effect?

Example system: AgCl(s) in water.

$\text{AgCl(s)} \rightleftharpoons \text{Ag}^+\text{(aq)} + \text{Cl}^-\text{(aq)}$

Now add NaCl (source of Cl⁻). The equilibrium shifts left to consume the extra Cl⁻, precipitating more AgCl and lowering [Ag⁺].

Derivation from First Principles

Step 1: Pure Water Equilibrium

For a sparingly soluble salt MX:

$\text{MX(s)} \rightleftharpoons \text{M}^+\text{(aq)} + \text{X}^-\text{(aq)}$

The solubility product:

$K_{sp} = [\text{M}^+][\text{X}^-]$

Why this form? Ksp is the equilibrium constant; solid MX activity = 1, so it doesn't appear.

If solubility in pure water = s mol/L, then at equilibrium:

$[\text{M}^+] = s, \quad [\text{X}^-] = s$

$K_{sp} = s \cdot s = s^2 \implies s = \sqrt{K_{sp}}$

Step 2: Add Common Ion (Say, X⁻)

Now dissolve a soluble salt NaX with concentration c mol/L. This instantly adds c mol/L of X⁻ before equilibrium adjusts.

Initial concentrations before MX responds:

[M⁺] ≈ 0 (from MX, negligible initially)
[X⁻] ≈ c (from NaX)

Let new solubility of MX = s' mol/L (amount that dissolves in presence of common ion).

At new equilibrium:

$[M^+] = s', \quad [X^-] = c + s'$

Why c + s'? The X⁻ from NaX (c) plus the X⁻ from dissolved MX (s').

Ksp is constant:

$K_{sp} = s' \cdot (c + s')$

Since s' << c (common ion suppresses solubility):

$c + s' \approx c$

$K_{sp} \approx s' \cdot c \implies s' = \frac{K_{sp}}{c}$

Key result: Solubility drops from $\sqrt{K_{sp}}$ to $\frac{K_{sp}}{c}$ .

Worked Examples

Worked example Example 1: AgCl with Added Cl⁻

Given: $K_{sp}(\text{AgCl}) = 1.8 \times 10^{-10}$ at 25°C. Calculate solubility of AgCl in (a) pure water, (b) 0.1 M NaCl. Solution:

(a) Pure water: $AgCl(s) \rightleftharpoons Ag^+ + Cl^-$

Let solubility = s.

$K_{sp} = s^2 = 1.8 \times 10^{-10}$

$s = \sqrt{1.8 \times 10^{-10}} = 1.34 \times 10^{-5} \text{ M}$

Why this step? In pure water, [Ag⁺] = [Cl⁻] = s, so Ksp = s².

(b) In 0.1 M NaCl:

NaCl provides [Cl⁻] = 0.1 M (common ion).

Let solubility of AgCl = s'.

At equilibrium:

$[\text{Ag}^+] = s', \quad [\text{Cl}^-] = 0.1 + s'$

$K_{sp} = s' (0.1 + s')$

Since $s' \ll 0.1$ (we'll verify):

$1.8 \times 10^{-10} = s' \cdot 0.1$

$s' = \frac{1.8 \times 10^{-10}}{0.1} = 1.8 \times 10^{-9} \text{ M}$

Check assumption: $1.8 \times 10^{-9} \ll 0.1$ ✓

Why this step? Common ion [Cl⁻] from NaCl dominates, so [Cl⁻] ≈ 0.1.

Result: Solubility drops from $1.34 \times 10^{-5}$ M to $1.8 \times 10^{-9}$ M—7500× decrease!

Worked example Example 2: PbCl₂ with Added Pb²⁺

Given: $K_{sp}(\text{PbCl}_2) = 1.7 \times 10^{-5}$ . Find solubility in 0.01 M Pb(NO₃)₂.

Solution:

$\text{PbCl}_2\text{(s)} \rightleftharpoons \text{Pb}^{2+} + 2\text{Cl}^-$

Pb(NO₃)₂ provides [Pb²⁺] = 0.01 M (common ion).

Let solubility = s'.

At equilibrium:

$[\text{Pb}^{2+}] = 0.01 + s', \quad [\text{Cl}^-] = 2s'$

Why 2s'? Each PbCl₂ that dissolves gives 2 Cl⁻.

$K_{sp} = (0.01 + s')(2s')^2$

Assume $s' \ll 0.01$ :

$1.7 \times 10^{-5} = 0.01 \cdot 4(s')^2$

$4(s')^2 = \frac{1.7 \times 10^{-5}}{0.01} = 1.7 \times 10^{-3}$

$(s')^2 = 4.25 \times 10^{-4}$

$s' = 0.0206 \text{ M}$

Wait! $s' = 0.0206$ is not much less than 0.01—assumption fails (0.0206 > 0.01).

Why this step? Must solve exactly:

$1.7 \times 10^{-5} = (0.01 + s') \cdot 4(s')^2$ Let $x = s'$ : $4x^2(0.01 + x) = 1.7 \times 10^{-5}$

$0.04x^2 + 4x^3 = 1.7 \times 10^{-5}$

Trial $x = 0.015$ :

$x^2 = 2.25 \times 10^{-4}$ , so $0.04x^2 = 9.0 \times 10^{-6}$
$x^3 = 3.375 \times 10^{-6}$ , so $4x^3 = 1.35 \times 10^{-5}$
Sum $= 9.0 \times 10^{-6} + 1.35 \times 10^{-5} = 2.25 \times 10^{-5}$ (too high, target is $1.7 \times 10^{-5}$ ).

Trial $x = 0.0135$ :

$x^2 = 1.8225 \times 10^{-4}$ , so $0.04x^2 = 7.29 \times 10^{-6}$
$x^3 = 2.460 \times 10^{-6}$ , so $4x^3 = 9.84 \times 10^{-6}$
Sum $= 7.29 \times 10^{-6} + 9.84 \times 10^{-6} = 1.71 \times 10^{-5}$ ≈ $1.7 \times 10^{-5}$ ✓

Answer: s' ≈ 0.0135 M (≈ $1.35 \times 10^{-2}$ M).

Lesson: Always check your approximation! Here the simple $s' = K_{sp}/(4c)$ shortcut overestimated solubility because s' was comparable to c.

Visualization

What the diagram shows:

Left panel: Solubility of AgCl vs. [Cl⁻] added. Hyperbolic drop as common ion increases.
Right panel: Bar chart comparing solubility in pure water vs. with common ion (Example 1).

Active Recall Challenges

Recall Feynman: Explain to a 12-Year-Old

Imagine you have a bathtub half-full of water, and salt is slowly dissolving into it. The tub reaches a balance—salt dissolves, but some also re-forms as solid at the same rate. Now you dump in a bucket of salty water (already has dissolved salt). What happens? The tub "sees" too much dissolved salt and says, "Whoa, too much!" So more salt turns back into solid to restore balance. The tub can't hold as much new salt anymore because you cheated by adding some dissolved already. That's the common ion effect: adding extra dissolved stuff forces the solid to stay more solid.

Mnemonic

Connections

Le Chatelier's Principle: Common ion effect is a direct application—add product ion, equilibrium shifts left.
Solubility Product (Ksp): Foundation; common ion modifies [ion] terms in Ksp expression.
Buffer Solutions: Common ion effect stabilizes pH by suppressing ionization of weak acids/bases.
Qualitative Analysis: Use common ion to selectively precipitate ions (e.g., add Cl⁻ to precipitate Ag⁺ but not Pb²⁺).
Ionic Equilibrium: Broader context of simultaneous equilibria.

Summary Table

Condition	Solubility Expression	Typical Effect
Pure water	$s = \sqrt{K_{sp}}$ (for MX)	Baseline
Common ion c	$s' = \frac{K_{sp}}{c}$ (if s' << c)	10–1000× decrease

#flashcards/chemistry

What is the common ion effect?

The decrease in solubility of a sparingly soluble salt when a soluble salt containing a common ion (same cation or anion) is added. The equilibrium shifts to reduce the added ion concentration (Le Chatelier).

For AgCl in pure water with

K_{sp} = 1.8 \times 10^{-10}

, what is the solubility?

s = \sqrt{K_{sp}} = \sqrt{1.8 \times 10^{-10}} = 1.34 \times 10^{-5}

M. Each AgCl gives one Ag⁺ and one Cl⁻, so [Ag⁺] = [Cl⁻] = s.

If 0.1 M NaCl is added to AgCl solution, how is solubility calculated?

[Cl⁻] ≈ 0.1 M (from NaCl, common ion). Let solubility = s'. Then

K_{sp} = s' \cdot 0.1

, so

s' = \frac{K_{sp}}{0.1} = \frac{1.8 \times 10^{-10}}{0.1} = 1.8 \times 10^{-9}

M. Solubility drops drastically.

Why does adding a common ion decrease solubility?

Le Chatelier's Principle: adding product ion (e.g., Cl⁻) shifts equilibrium left (toward reactants), precipitating more solid and reducing dissolution.

For

\text{PbCl}_2 \rightleftharpoons \text{Pb}^{2+} + 2\text{Cl}^-

, if [Pb²⁺] from Pb(NO₃)₂ is c and solubility is s', what is [Cl⁻]?

[Cl⁻] = 2s' (not s'!) because each PbCl₂ dissolves to give 2 Cl⁻ ions. Stoichiometry matters!

What approximation is often used in common ion effect problems?

Assume solubility s' << common ion concentration c, so [common ion] ≈ c. Must verify s' < 5% of c after solving.

If the approximation

s' \ll c

fails, what do you do?

Solve the full Ksp expression without approximation—usually a quadratic or cubic equation. Example:

K_{sp} = (c + s')(2s')^2

for PbCl₂.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Common ion effect ka matlab hai ki agar ek sparingly soluble salt (jaise AgCl) already solution mein equilibrium pe hai, aur tum usme ek aur soluble salt add karo jo same ion share karta hai (jaise NaCl, jo Cl⁻ deta hai), toh original salt ki solubility ghat jati hai. Kyun? Kyunki Le Chatelier ka principle kehta hai ki jab tum equilibrium ko disturb karte ho (extra Cl⁻ daalke), toh system ulta shift hota hai—matlab, zyada AgCl precipitate hota hai aur kam dissolve hota hai. Yeh effect common ion effect kehlata hai.

Example dekho: Pure pani mein AgCl ki solubility $1.34 \times 10^{-5}$ M hoti hai. Lekin agar tum 0.1 M NaCl daal do (jo Cl⁻ common ion provide karta hai), toh AgCl ki solubility sirf $1.8 \times 10^{-9}$ M reh jati hai—7500 guna kam! Yeh bahut powerful technique hai chemistry mein, especially qualitative analysis mein jahan selective precipitation karni hoti hai.

Formula bhi simple hai: Agar common ion concentration c hai aur woh solubility se kafi zyada hai, toh naye solubility s' = K $_{sp}$ /c. Bas ek chez dhyan rakhna—stoichiometry! Agar salt PbCl₂ jaisa hai (ek Pb²⁺ aur do Cl⁻ deta hai), toh [Cl⁻] mein 2s' likhna padega, sirf s' nahi. Aur ek baat: agar solubility c ke comparable nikle (jaise PbCl₂ wale example mein), toh approximation chhod ke poora cubic equation solve karna padega. Yeh common mistake hai exams mein, toh hoshiyar rehna!

What Is the Common Ion Effect?

Derivation from First Principles

Step 1: Pure Water Equilibrium

Step 2: Add Common Ion (Say, X⁻)

Worked Examples

Visualization

Common Mistakes

Active Recall Challenges

Mnemonic

Connections

Summary Table

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Equilibrium

Connections