Common ion effect: Question bank — Hinglish deep dive

2.6.13 · D5 · Chemistry › Equilibrium › Common ion effect

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

Neeche har jagah use hone wale symbols ka reminder:

$s$ = pure water mein solubility (mol/L jo dissolve hota hai).
$s^{'}$ = common ion ki presence mein solubility.
$c$ = added common ion ki concentration.
$K_{s p}$ = solubility product, ek temperature-fixed constant.
$Q$ = ionic product $[M^{+}] [X^{-}]$ jo abhi jo concentrations hain unse compute kiya (zaruri nahi equilibrium par ho).

True or false — justify karo

Common ion effect

K_{s p}

ki value ko lower karta hai.

False.

K_{s p}

sirf temperature par depend karta hai. Common ion, solubility

s^{'}

ko lower karta hai — ek ion ki concentration raise karke taaki doosri ion ko apni concentration girani pade aur product same

K_{s p}

par bana rahe.

Common ion add karna ek saturated solution se hamesha kuch extra solid precipitate karta hai.

True (already-saturated solution ke liye). Ek ion ko raise karna

Q

K_{s p}

se upar push karta hai, toh system supersaturated ho jaata hai aur tab tak precipitate karta hai jab tak

Q = K_{s p}

na ho jaye — Le Chatelier dissolution equilibrium ko left shift karta hai.

AgCl ke liye,

[Ag^{+}]

har equilibrium par

[Cl^{-}]

ke equal hona chahiye.

False. Ye sirf pure water mein equal hote hain. NaCl add karo toh

[Cl^{-}] ≫ [Ag^{+}]

; AgNO₃ add karo toh

[Ag^{+}] ≫ [Cl^{-}]

. Sirf product

[Ag^{+}] [Cl^{-}] = K_{s p}

force hota hai.

NaNO₃ (koi shared ion nahi) ko saturated AgCl mein add karna idealized calculation mein iske solubility ko bilkul nahi badlata.

Simple model mein True. NaNO₃ na Ag⁺ share karta hai na Cl⁻, isliye ye

K_{s p}

expression mein enter nahi hota. (Reality mein ye ionic-strength/activity effects ke zariye solubility ko thoda increase karta hai — "salt effect", jo common ion effect ka ulta hai.)

Common ion effect aur salt (diverse ion) effect solubility ko same direction mein push karte hain.

False. Common ion (shared ion) solubility ko decrease karta hai; diverse-ion/salt effect (inert electrolyte) ion activities ko lower karke ise thoda increase karta hai. Ye opposite phenomena hain.

Agar aap common ion concentration

c

ko double karo, toh ek MX salt ki solubility half ho jaati hai.

Approximation mein True

s^{'} = K_{s p} / c

, jo

c

ke inversely proportional hai.

c

double karne se

s^{'}

half hoti hai — provided

s^{'} ≪ c

abhi bhi hold kare.

MX_{2}

salt ke liye,

M^{2 +}

common ion ka

c

double karna solubility ko half karta hai.

False. Yahaan

s^{'} = K_{s p} / (4 c) \propto c^{- 1/2}

, toh

c

double karne se

s^{'}

1/ 2 \approx 0.71

se multiply karta hai,

0.5

nahi. Stoichiometry exponent ko change karta hai.

Common ion effect kaam karta hai chahe aap shared cation add karo ya shared anion.

True.

K_{s p}

product mein appear hone wala koi bhi ion equilibrium ko left push karega. AgCl ke liye, extra Cl⁻ (NaCl se) ya extra Ag⁺ (AgNO₃ se) dono dissolution ko suppress karte hain.

Jis exact instant NaCl add hota hai, solution momentarily supersaturated ho jaata hai.

True. Added Cl⁻,

Q

K_{s p}

se upar raise karta hai pehle koi AgCl precipitate ho — tab system solid deposit karke

Q = K_{s p}

par wapas relax karta hai.

Error dhundho

"Kyunki NaCl Cl⁻ add karta hai, AgCl ki solubility girti hai, toh

K_{s p}

bhi gir gaya." — khammi dhundho.

K_{s p}

ke baare mein conclusion galat hai. Solubility giri lekin

K_{s p}

ek constant hai; jo hua wo ye hai ki

[Cl^{-}]

badha aur

[Ag^{+}]

gira toh unka product unchanged hai.

"PbCl₂ ke dissolve hone mein,

[Cl^{-}] = s^{'}

." — error dhundho.

Har formula unit do Cl⁻ release karta hai, toh

[Cl^{-}] = 2 s^{'}

. Stoichiometric coefficient bhoolna

[Cl^{-}]

ko underestimate karta hai aur

K_{s p} = [Pb^{2 +}] (2 s^{'})^{2}

galat likhata hai.

"Common ion present hone ke baad bhi main

K_{s p} = s^{'} \cdot s^{'} = s^{'2}

likhta hoon." — kya galat hai?

Anion concentration sirf

s^{'}

nahi balki total

c + s^{'}

hai. Correct form hai

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

s^{'2}

use karna us ion ko ignore karta hai jo tune deliberately add kiya tha.

s^{'} = K_{s p} / c

exact hai, toh mujhe kabhi kuch check nahi karna." — error?

Ye ek approximation hai jo tab valid hai jab

s^{'} ≪ c

ho (rule of thumb:

s^{'} < 5%

c

). Agar added-ion concentration chhoti hai ya

K_{s p}

bada hai, toh tumhe full equation solve karni padegi — PbCl₂ dekho jahan shortcut ne

s^{'}

overestimate kiya.

"Saturated AgCl mein AgNO₃ add karna solubility increase karta hai kyunki aap zyaada silver salt add kar rahe ho." — trap dhundho.

Ulta hai. Ag⁺ ek common ion hai; extra Ag⁺ equilibrium ko left push karta hai aur AgCl ki dissolved matra ko decrease karta hai, zyaada solid precipitate karta hai.

"PbCl₂ ke liye

K_{s p}

sirf

[Pb^{2 +}] [Cl^{-}]

hai." — kya missing hai?

Cl⁻ term ko square hona chahiye:

K_{s p} = [Pb^{2 +}] [Cl^{-}]^{2}

. Har ion ko equilibrium mein uske stoichiometric coefficient tak raise kiya jaata hai.

"Solid AgCl

K_{s p}

expression mein

[AgCl]

ke roop mein appear hota hai." — khammi?

Pure solids (aur pure liquids) ki activity

= 1

hoti hai aur ye kabhi equilibrium expression mein appear nahi hote.

K_{s p} = [Ag^{+}] [Cl^{-}]

mein sirf aqueous ions hote hain.

"Kyunki kuch Cl⁻ add karne ke baad

Q < K_{s p}

hai, solution precipitate karega." — error?

Direction galat hai. Precipitation ke liye

Q > K_{s p}

chahiye. Agar

Q < K_{s p}

hai toh solution unsaturated hai aur zyaada solid dissolve hoga, precipitate nahi hoga.

Why questions

Common ion effect directly Le Chatelier's Principle se kyun follow karta hai?

Dissolution

MX(s) ⇌ M^{+} + X^{-}

ek equilibrium hai; ek product ion add karna ek stress hai jise system left shift karke relieve karta hai, ions consume karta hai aur solid deposit karta hai — exactly Le Chatelier.

Solid MX

K_{s p}

mein kyun appear nahi hota?

Iske concentration (activity) essentially constant hai aur pure solid ke liye 1 define ki gayi hai, toh ye equilibrium constant se factor out ho jaata hai aur sirf aqueous ion concentrations rehti hain.

s^{'} = K_{s p} / c

ek MX salt ke liye inverse relationship kyun hai?

Kyunki

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

(jab

c + s^{'} \approx c

), toh

s^{'}

hai

K_{s p}

divided by

c

. Bada common-ion load

c

, MX ke dissolve hone ke liye kam room deta hai, isliye chhota

s^{'}

Common ion effect Qualitative Analysis mein ions ko separate kyun kar sakta hai?

Alag salts ke

K_{s p}

values bahut alag hote hain, toh ek fixed common-ion concentration low-

K_{s p}

salt ko precipitate karti hai jabki higher-

K_{s p}

wala dissolved rehta hai — selective precipitation.

Common ion effect buffer ke pH ko kyun stabilize karta hai?

Added common ion (jaise conjugate base A⁻) weak acid HA ki ionization ko suppress karta hai,

[H^{+}]

ko ek set value ke paas pin karta hai toh added acid ya base ise barely shift karta hai.

PbCl₂ se chloride ke liye

s^{'}

ke bajaye

2 s^{'}

kyun use karna padta hai, phir bhi lead ke liye

s^{'}

rakhte ho?

Stoichiometry: ek dissolved unit ek Pb²⁺ deta hai lekin do Cl⁻. Concentrations reflect karti hain ki har dissolving unit kitne ions produce karta hai.

KNO₃ jaisa inert salt add karna solubility ko lower karne ke bajaye thoda raise kyun karta hai?

Iske ions ionic strength badhate hain, Ag⁺ aur Cl⁻ ki effective activities lower karte hain; activity product ko

K_{s p}

par rakhne ke liye, zyaada solid dissolve hona padta hai. Ye salt effect hai, common ion effect nahi.

Edge cases

Agar common-ion concentration

c

zero ke paas aaye toh solubility ka kya hoga?

Formula

s^{'} = K_{s p} / c

blow up karta hai, signal karta hai ki approximation fail ho gayi. Jaise

c \to 0

, tum smoothly pure-water case

s = K_{s p}

recover karte ho — tumhe full expression

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

use karni chahiye.

Agar common-ion concentration

c

, pure-water solubility

s

ke comparable ho toh kya?

Approximation

s^{'} ≪ c

fail hoti hai aur tum

s^{'}

drop nahi kar sakte. Full equation

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

solve karo (MX ke liye ek quadratic) shortcut ke bajaye.

Jab

K_{s p}

bahut large ho (fairly soluble salt) toh common ion effect help karta hai ya hurt?

Bada

K_{s p}

matlab

s^{'}

large rehta hai aur aksar

c

se comparable hota hai, toh neat suppression break down hoti hai aur approximate formula mislead karta hai — tumhe exactly solve karna padega (jaise PbCl₂ mein).

Agar tum common ion add karo lekin solution unsaturated tha (koi solid present nahi), toh kya kuch precipitate hoga?

Tabhi agar addition

Q

K_{s p}

se upar push kare. Agar ion add karne ke baad bhi

Q

K_{s p}

se neeche rahe, toh kuch precipitate nahi hoga — solid par common ion effect ke liye pehle saturation reach karna zaroori hai.

MX_{2}

ke liye pure-water solubility

s = 3 K_{s p} /4

kyun hai na ki

K_{s p}

Pure water mein

[M^{2 +}] = s

aur

[X^{-}] = 2 s

, toh

K_{s p} = s (2 s)^{2} = 4 s^{3}

, jo deta hai

s = (K_{s p} /4)^{1/3}

. Exponent ions ki total number track karta hai, simple square nahi.

MX salt ke liye jaise

c \to \infty

s^{'}

ka limiting behaviour kya hai?

s^{'} = K_{s p} / c \to 0

. Infinite common ion solubility ko zero ke taraf drive karta hai (idealized model mein), though reality mein activity effects iske liye ek floor rakhte hain.

Kya common ion effect kabhi solubility increase kar sakta hai?

Common ion effect khud nahi. Lekin bahut high ligand/common-ion concentrations par, complex-ion formation (jaise

AgCl_{2}^{-}

) solid ko re-dissolve kar sakta hai — ek alag equilibrium simple

K_{s p}

suppression ko override karta hai.

Connections

Le Chatelier's Principle — upar ke har "shift left" answer ka driving principle.
Solubility Product (Ksp) — fixed constant jiske around ye saare traps orbit karte hain.
Buffer Solutions — weak-acid ionization ka common ion suppression.
Qualitative Analysis — common ions ke zariye selective precipitation.
Ionic Equilibrium — coupled equilibria ka wider setting.
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