Common ion effect

2.6.13 · Chemistry › Equilibrium

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

Common Ion Effect Kya Hai?

Example system: AgCl(s) paani mein.

$AgCl(s) ⇌ Ag^{+} (aq) + Cl^{-} (aq)$

Ab NaCl (Cl⁻ ka source) add karo. Equilibrium extra Cl⁻ consume karne ke liye left shift hota hai, zyada AgCl precipitate hota hai aur [Ag⁺] kam hoti hai.

First Principles Se Derivation

Step 1: Pure Water Equilibrium

Ek sparingly soluble salt MX ke liye:

$MX(s) ⇌ M^{+} (aq) + X^{-} (aq)$

Solubility product:

$K_{s p} = [M^{+}] [X^{-}]$

Yeh form kyun? Ksp equilibrium constant hai; solid MX ki activity = 1 hoti hai, isliye woh appear nahi hoti.

Agar pure water mein solubility = s mol/L hai, toh equilibrium par:

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

$K_{s p} = s \cdot s = s^{2} ⟹ s = K_{s p}$

Step 2: Common Ion Add Karo (Maano, X⁻)

Ab ek soluble salt NaX ko concentration c mol/L mein dissolve karo. Yeh equilibrium adjust hone se pehle instantly c mol/L X⁻ add karta hai.

MX ke respond karne se pehle initial concentrations:

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

Maano MX ki nai solubility = s' mol/L (common ion ki presence mein jo amount dissolve hoti hai).

Naye equilibrium par:

$[M^{+}] = s^{'}, [X^{-}] = c + s^{'}$

c + s' kyun? NaX se X⁻ (c) plus dissolved MX se X⁻ (s').

Ksp constant hai:

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

Kyunki s' << c (common ion solubility suppress karta hai):

$c + s^{'} \approx c$

$K_{s p} \approx s^{'} \cdot c ⟹ s^{'} = \frac{K _{s p}}{c}$

Key result: Solubility $K_{s p}$ se $\frac{K _{s p}}{c}$ tak drop hoti hai.

Worked Examples

Worked example Example 1: AgCl with Added Cl⁻

Given: $K_{s p} (AgCl) = 1.8 \times 1 0^{- 10}$ at 25°C. AgCl ki solubility calculate karo (a) pure water mein, (b) 0.1 M NaCl mein. Solution:

(a) Pure water: $A g C l (s) ⇌ A g^{+} + C l^{-}$

Maano solubility = s.

$K_{s p} = s^{2} = 1.8 \times 1 0^{- 10}$

$s = 1.8 \times 1 0^{- 10} = 1.34 \times 1 0^{- 5} M$

Yeh step kyun? Pure water mein, [Ag⁺] = [Cl⁻] = s, isliye Ksp = s².

(b) 0.1 M NaCl mein:

NaCl [Cl⁻] = 0.1 M provide karta hai (common ion).

Maano AgCl ki solubility = s'.

Equilibrium par:

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

$K_{s p} = s^{'} (0.1 + s^{'})$

Kyunki $s^{'} ≪ 0.1$ (hum verify karenge):

$1.8 \times 1 0^{- 10} = s^{'} \cdot 0.1$

$s^{'} = \frac{1.8 \times 1 0 ^{- 10}}{0.1} = 1.8 \times 1 0^{- 9} M$

Assumption check karo: $1.8 \times 1 0^{- 9} ≪ 0.1$ ✓

Yeh step kyun? NaCl se common ion [Cl⁻] dominate karta hai, isliye [Cl⁻] ≈ 0.1.

Result: Solubility $1.34 \times 1 0^{- 5}$ M se $1.8 \times 1 0^{- 9}$ M tak drop hoti hai—7500 guna decrease!

Worked example Example 2: PbCl₂ with Added Pb²⁺

Given: $K_{s p} (PbCl_{2}) = 1.7 \times 1 0^{- 5}$ . 0.01 M Pb(NO₃)₂ mein solubility dhundo.

Solution:

$PbCl_{2} (s) ⇌ Pb^{2 +} + 2 Cl^{-}$

Pb(NO₃)₂ [Pb²⁺] = 0.01 M provide karta hai (common ion).

Maano solubility = s'.

Equilibrium par:

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

2s' kyun? Har PbCl₂ jo dissolve hota hai woh 2 Cl⁻ deta hai.

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

Maano $s^{'} ≪ 0.01$ :

$1.7 \times 1 0^{- 5} = 0.01 \cdot 4 (s^{'})^{2}$

$4 (s^{'})^{2} = \frac{1.7 \times 1 0 ^{- 5}}{0.01} = 1.7 \times 1 0^{- 3}$

$(s^{'})^{2} = 4.25 \times 1 0^{- 4}$

$s^{'} = 0.0206 M$

Ruko! $s^{'} = 0.0206$ 0.01 se much less nahi hai—assumption fail ho gayi (0.0206 > 0.01).

Yeh step kyun? Exactly solve karna zaroori hai:

$1.7 \times 1 0^{- 5} = (0.01 + s^{'}) \cdot 4 (s^{'})^{2}$ Maano $x = s^{'}$ : $4 x^{2} (0.01 + x) = 1.7 \times 1 0^{- 5}$

$0.04 x^{2} + 4 x^{3} = 1.7 \times 1 0^{- 5}$

Trial $x = 0.015$ :

$x^{2} = 2.25 \times 1 0^{- 4}$ , isliye $0.04 x^{2} = 9.0 \times 1 0^{- 6}$
$x^{3} = 3.375 \times 1 0^{- 6}$ , isliye $4 x^{3} = 1.35 \times 1 0^{- 5}$
Sum $= 9.0 \times 1 0^{- 6} + 1.35 \times 1 0^{- 5} = 2.25 \times 1 0^{- 5}$ (bahut zyada, target $1.7 \times 1 0^{- 5}$ hai).

Trial $x = 0.0135$ :

$x^{2} = 1.8225 \times 1 0^{- 4}$ , isliye $0.04 x^{2} = 7.29 \times 1 0^{- 6}$
$x^{3} = 2.460 \times 1 0^{- 6}$ , isliye $4 x^{3} = 9.84 \times 1 0^{- 6}$
Sum $= 7.29 \times 1 0^{- 6} + 9.84 \times 1 0^{- 6} = 1.71 \times 1 0^{- 5}$ ≈ $1.7 \times 1 0^{- 5}$ ✓

Answer: s' ≈ 0.0135 M (≈ $1.35 \times 1 0^{- 2}$ M).

Lesson: Apni approximation hamesha check karo! Yahan simple $s^{'} = K_{s p} / (4 c)$ shortcut ne solubility overestimate ki kyunki s' aur c comparable the.

Visualization

Diagram kya dikhata hai:

Left panel: AgCl ki solubility vs. added [Cl⁻]. Common ion badhne par hyperbolic drop.
Right panel: Bar chart jo pure water aur common ion wali solubility compare karta hai (Example 1).

Active Recall Challenges

Recall Feynman: Ek 12-Saal Ke Bacche Ko Explain Karo

Socho tumhare paas ek bathtub hai jo aadha paani se bhara hai, aur namak dheere dheere dissolve ho raha hai. Tub ek balance reach karta hai—namak dissolve hota hai, lekin utna hi solid bhi wapas banta hai. Ab tum ek bucket bhari salty water daal do (pehle se dissolved namak hai). Kya hota hai? Tub "dekhta" hai ki zyada dissolved namak hai aur kehta hai, "Arre yaar, bahut zyada!" Toh zyada namak wapas solid ban jaata hai balance restore karne ke liye. Tub utna naya namak nahi le sakta kyunki tumne pehle se kuch dissolved daala tha. Yahi hai common ion effect: pehle se dissolved cheez add karna solid ko aur solid rehne par majboor karta hai.

Mnemonic

Connections

Le Chatelier's Principle: Common ion effect iska direct application hai—product ion add karo, equilibrium left shift hota hai.
Solubility Product (Ksp): Foundation; common ion Ksp expression mein [ion] terms ko modify karta hai.
Buffer Solutions: Common ion effect weak acids/bases ki ionization suppress karke pH stabilize karta hai.
Qualitative Analysis: Common ion use karke ions selectively precipitate karo (jaise, Ag⁺ precipitate karne ke liye Cl⁻ add karo lekin Pb²⁺ nahi).
Ionic Equilibrium: Simultaneous equilibria ka broader context.

Summary Table

Condition	Solubility Expression	Typical Effect
Pure water	$s = K_{s p}$ (MX ke liye)	Baseline
Common ion c	$s^{'} = \frac{K _{s p}}{c}$ (agar s' << c)	10–1000 guna decrease

#flashcards/chemistry

Common ion effect kya hai?

Ek sparingly soluble salt ki solubility mein decrease jab ek common ion (same cation ya anion) wala soluble salt add kiya jaata hai. Equilibrium added ion concentration reduce karne ke liye shift hota hai (Le Chatelier).

AgCl ke liye pure water mein

K_{s p} = 1.8 \times 1 0^{- 10}

ke saath solubility kya hai?

s = K_{s p} = 1.8 \times 1 0^{- 10} = 1.34 \times 1 0^{- 5}

M. Har AgCl ek Ag⁺ aur ek Cl⁻ deta hai, isliye [Ag⁺] = [Cl⁻] = s.

Agar AgCl solution mein 0.1 M NaCl add kiya jaaye, toh solubility kaise calculate hogi?

[Cl⁻] ≈ 0.1 M (NaCl se, common ion). Maano solubility = s'. Tab

K_{s p} = s^{'} \cdot 0.1

, isliye

s^{'} = \frac{K _{s p}}{0.1} = \frac{1.8 \times 1 0 ^{- 10}}{0.1} = 1.8 \times 1 0^{- 9}

M. Solubility drastically drop hoti hai.

Common ion add karne se solubility kyun decrease hoti hai?

Le Chatelier's Principle: product ion (jaise Cl⁻) add karne se equilibrium left shift hota hai (reactants ki taraf), zyada solid precipitate hota hai aur dissolution kam hoti hai.

PbCl_{2} ⇌ Pb^{2 +} + 2 Cl^{-}

ke liye, agar Pb(NO₃)₂ se [Pb²⁺] = c hai aur solubility s' hai, toh [Cl⁻] kya hai?

[Cl⁻] = 2s' (s' nahi!) kyunki har PbCl₂ dissolve hokar 2 Cl⁻ ions deta hai. Stoichiometry matter karta hai!

Common ion effect problems mein commonly kaun si approximation use hoti hai?

Assume karo solubility s' << common ion concentration c, isliye [common ion] ≈ c. Solve karne ke baad verify karna zaroori hai ki s' < 5% of c.

Agar approximation

s^{'} ≪ c

fail ho jaaye, toh kya karo?

Full Ksp expression ko approximation ke bina solve karo—usually quadratic ya cubic equation. Example: PbCl₂ ke liye

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