Common ion effect: Worked examples — Hinglish deep dive

2.6.13 · D3 · Chemistry › Equilibrium › Common ion effect

#chemistry/equilibrium #solubility

Intuition Yeh page kyun hai

Parent note ne tumhe machinery di thi: $K_{s p}$ , "common ion add karo, solubility giregi" wali story, aur do examples. Lekin real problems alag-alag flavours mein aate hain — alag salt shapes, common ion cation ho ya anion, approximation kaam kare ya na kare, ajeeb edge cases jaise "agar koi common ion hi nahi hai" ya "agar common ion concentration itni tiny hai ki kuch karta hi nahi". Yeh page un sabhi cells ko walk karta hai taaki tum kabhi koi aisa scenario na dekho jo tumne pehle nahi dekha.

Agar koi symbol yahaan unfamiliar lage, toh woh parent note mein aur Solubility Product (Ksp) mein build kiya gaya tha — neeche har quantity ya toh $K_{s p}$ hai (woh fixed number jo nature har solid ko assign karta hai), ya concentration $[\cdot]$ mol/L mein likhi hai, ya solubility $s$ hai = "ek litre mein kitne moles solid dissolve hote hain".

Is page par use ki gayi Notation

Definition Teen symbols jo neeche har jagah milenge

$K_{s p}$ — solubility product: ek solid ke dissolve hone ka fixed equilibrium constant, jo saturation par ion concentrations ke product ke barabar hota hai (Solubility Product (Ksp) mein build hua).
$s$ — pure water mein solubility: jab solution mein kuch aur nahi hota tab ek litre mein kitne moles solid dissolve hote hain.
$s^{'}$ — common ion ki presence mein solubility: jab ek common ion pehle se maujood ho tab ek litre mein usi solid ke kitne moles dissolve hote hain. Yeh $s$ se alag number hai, isliye hum ise apna prime dete hain.
$c$ — initial common-ion concentration: shared ion ki matra (mol/L) jo ek soluble salt solution mein dalta hai pehle ki sparingly soluble solid re-adjust kare. Generic 1:1 shortcut $s^{'} = K_{s p} / c$ mein, yeh $c$ wahi supplied concentration hai.

Scenario matrix

Kisi bhi number se pehle, aao har tarah ki problem list karein jo yeh topic tumpe phenk sakta hai. Table ko yun padho: "yeh sab doors hain; neeche har worked example ek door kholega." (Recall: $c$ = common ion ki initial concentration jo ek soluble salt supply karta hai; $s^{'}$ = uski presence mein sparingly soluble solid ki solubility.)

Cell	Kya alag banata hai ise	Kaunsa example
A. 1:1 salt, koi common ion nahi	Baseline $s = K_{s p}$ case	Ex 1
B. 1:1 salt, common ion (anion)	X⁻ add karo, $s^{'} = K_{s p} / c$ use karo	Ex 2
C. 1:1 salt, common ion (cation)	Uski jagah M⁺ add karo — symmetry check	Ex 3
D. 1:2 salt, koi common ion nahi	Stoichiometry ⇒ $s = 3 K_{s p} /4$	Ex 4
E. 1:2 salt, common ion, approx VALID	$s^{'} ≪ c$ genuinely holds	Ex 5
F. 1:2 salt, common ion, approx FAILS	Cubic exactly solve karni padegi	Ex 6
G. Degenerate: common ion itna chhota ki negligible	Limiting behaviour $c \to 0$ se cell A wapas milta hai	Ex 7
H. Real-world word problem	Kya precipitate banta bhi hai? ( $Q$ vs $K_{s p}$ )	Ex 8
I. Exam twist: percent-suppression	"Solubility kis factor se giri?"	Ex 9

Is matrix ke do axes hain: salt shape (1:1 jaise AgCl, ya 1:2 jaise PbCl₂) aur kaunsa ion common hai / approximation survive karti hai ya nahi. Har row cover karo aur topic cover ho jayega.

Figure s01 — AgCl ki solubility $s^{'}$ (vertical axis, log scale) initial common-ion concentration $c$ (horizontal axis, log scale) ke against plot ki gayi hai. Blue curve ek hyperbola hai: chhote $c$ par yeh pure-water value $s = 1.34 \times 1 0^{- 5}$ M par flat rehta hai (green arrow, cell G), phir steep-drop region mein gir jaata hai (orange arrow, cells B aur C), $c = 0.02$ M par ~1500× suppression tak pahunchta hai (red arrow, cell I). Ek picture mein matrix ka zyaadatar hissa hai.

Upar ki figure AgCl ki solubility ko added common-ion concentration $c$ ke against plot karti hai. Shape dekho: ek hyperbola — jaise hi $c$ pure-water value se upar charhta hai, solubility gir jaati hai. Flat left region (tiny $c$ ) cell G hai; steep-drop region cells B/C hai; far-right cell I ka "huge suppression" hai.

Cell A — 1:1 salt, koi common ion nahi (baseline)

Worked example Ex 1: AgCl pure water mein

Diya gaya $K_{s p} (AgCl) = 1.8 \times 1 0^{- 10}$ . Solubility $s$ nikalo.

Forecast: order of magnitude guess karo — $s$ $1 0^{- 5}$ ke kareeb hai ya $1 0^{- 10}$ ? (Dimaag mein square root lo: $1 0^{- 10} = 1 0^{- 5}$ .)

Dissolving equilibrium likho. $AgCl(s) ⇌ Ag^{+} + Cl^{-}$ Yeh step kyun? Har solubility problem is cheez se shuru hoti hai ki kya kya mein toot raha hai, taaki pata chale kitne ions aate hain.
Maano $s$ = moles of AgCl dissolving per litre. Tab $[Ag^{+}] = s$ aur $[Cl^{-}] = s$ . Yeh step kyun? Ek formula unit se ek Ag⁺ aur ek Cl⁻ milta hai, isliye dono concentrations $s$ ke barabar hain.
$K_{s p} = [Ag^{+}] [Cl^{-}]$ mein plug karo: $K_{s p} = s \cdot s = s^{2} ⟹ s = K_{s p} = 1.8 \times 1 0^{- 10} = 1.34 \times 1 0^{- 5} M .$ Yeh step kyun? Jab koi aur source nahi kisi bhi ion ka, toh $K_{s p}$ bas $s^{2}$ hai — sabse simple case.

Verify: plug back karo: $(1.34 \times 1 0^{- 5})^{2} = 1.80 \times 1 0^{- 10}$ ✓. Units: (mol/L)² = M², 1:1 salt ke liye sahi.

Cell B — 1:1 salt, common anion

Worked example Ex 2: AgCl, 0.10 M NaCl mein

Diya gaya wahi $K_{s p} = 1.8 \times 1 0^{- 10}$ , ab 0.10 M NaCl mein (Cl⁻ supply karta hai, toh yahaan $c = 0.10$ M).

Forecast: common ion Cl⁻ hai 0.10 M par — woh pure water ke $1.34 \times 1 0^{- 5}$ se bahut zyaada hai. Guess: kya $s^{'}$ hazaron ke factor se giregi?

Total $[Cl^{-}] = 0.10 + s^{'}$ , aur $[Ag^{+}] = s^{'}$ . Yeh step kyun? Chloride ke ab do sources hain: NaCl se 0.10 M aur AgCl dissolve hone se $s^{'}$ .
Approximate karo $0.10 + s^{'} \approx 0.10$ (bet lagate hain ki $s^{'}$ tiny hai). Yeh step kyun? Agar guess sahi nikla toh ek quadratic ek line ki algebra ban jaati hai.
$K_{s p} = s^{'} (0.10) ⟹ s^{'} = \frac{1.8 \times 1 0 ^{- 10}}{0.10} = 1.8 \times 1 0^{- 9} M .$ Yeh step kyun? Jab $[Cl^{-}]$ common-ion value $c = 0.10$ par pin ho, toh $K_{s p}$ expression $s^{'} \cdot c$ directly divide karke $s^{'}$ solve ho jaata hai — yeh hai generic shortcut $s^{'} = K_{s p} / c$ action mein.

Verify: kya $s^{'} ≪ 0.10$ hai? $1.8 \times 1 0^{- 9}$ 0.10 ka $1.8 \times 1 0^{- 6} %$ hai — haan. Suppression factor $= \frac{1.34 \times 1 0 ^{- 5}}{1.8 \times 1 0 ^{- 9}} \approx 7.4 \times 1 0^{3}$ . Lagbhag 7500× kam soluble ✓.

Cell C — 1:1 salt, common cation (symmetry check)

Worked example Ex 3: AgCl, 0.050 M AgNO₃ mein

Diya gaya wahi $K_{s p}$ , ab common ion Ag⁺ hai 0.050 M par (AgNO₃ se, toh $c = 0.050$ M).

Forecast: kya farq padta hai agar cation add karein ya anion? $K_{s p} = [Ag^{+}] [Cl^{-}]$ symmetric hai — same form ka answer predict karo.

$[Ag^{+}] = 0.050 + s^{'}$ , $[Cl^{-}] = s^{'}$ . Yeh step kyun? Ab Ag⁺ ka extra source hai; Cl⁻ sirf AgCl dissolve hone se aata hai.
Approximate karo $0.050 + s^{'} \approx 0.050$ , phir $K_{s p} = (0.050) s^{'} ⟹ s^{'} = \frac{1.8 \times 1 0 ^{- 10}}{0.050} = 3.6 \times 1 0^{- 9} M .$ Yeh step kyun? Math bilkul Ex 2 jaisi hai $c = 0.050$ ke saath — yeh prove karta hai ki effect cation vs anion mein symmetric hai. Le Chatelier's Principle ko parwah nahi ki tum kaunsa product ion add karte ho.

Verify: $s^{'} = 3.6 \times 1 0^{- 9} ≪ 0.050$ ✓. Note karo ki $s^{'}$ yahaan Ex 2 se double hai kyunki $c$ aadha hai — $s^{'} \propto 1/ c$ ke consistent.

Cell D — 1:2 salt, koi common ion nahi

Cell E — 1:2 salt, common ion, approximation VALID

Worked example Ex 5: PbCl₂, 0.50 M NaCl mein

Diya gaya wahi $K_{s p} = 1.7 \times 1 0^{- 5}$ ; ab 0.50 M NaCl Cl⁻ supply karta hai (toh $c = 0.50$ M).

Forecast: common ion $K_{s p}$ mein squared hai, toh ek bada $c$ solubility ko bahut crush karega — guess karo $s^{'}$ pure water ke $1.62 \times 1 0^{- 2}$ se bahut neeche hogi.

$[Pb^{2 +}] = s^{'}$ , $[Cl^{-}] = 0.50 + 2 s^{'}$ . Yeh step kyun? Chloride ab = 0.50 M NaCl se aur $2 s^{'}$ PbCl₂ dissolve hone se.
Approximate karo $0.50 + 2 s^{'} \approx 0.50$ (bet karo $s^{'}$ tiny hai), jisse milta hai $K_{s p} = s^{'} (0.50)^{2} = 0.25 s^{'} ⟹ s^{'} = \frac{1.7 \times 1 0 ^{- 5}}{0.25} = 6.8 \times 1 0^{- 5} M .$ Yeh step kyun? Common ion squared hai, isliye woh $(0.50)^{2} = 0.25$ ke roop mein appear hota hai — yahan 1:2 salts 1:1 se alag hoti hain.

Verify: kya $2 s^{'} ≪ 0.50$ hai? $2 s^{'} = 1.36 \times 1 0^{- 4}$ , jo 0.50 ka $0.027%$ hai — approximation holds ✓. Suppression: $1.62 \times 1 0^{- 2}$ se $6.8 \times 1 0^{- 5}$ tak, lagbhag 240×.

Cell F — 1:2 salt, common ion, approximation FAILS

Worked example Ex 6: PbCl₂, 0.010 M Pb(NO₃)₂ mein

Diya gaya wahi $K_{s p}$ ; common ion Pb²⁺ hai 0.010 M par (toh $c = 0.010$ M).

Forecast: yahaan common ion unsquared wala hai aur chhota hai (0.010). Guess: kya $s^{'}$ 0.010 se comparable nikal sakta hai, shortcut tod ke?

$[Pb^{2 +}] = 0.010 + s^{'}$ , $[Cl^{-}] = 2 s^{'}$ . Yeh step kyun? Ab lead ka extra source hai; chloride sirf PbCl₂ dissolve hone se aata hai, isliye $2 s^{'}$ .
Naive approximation $0.010 + s^{'} \approx 0.010$ : $K_{s p} = (0.010) (2 s^{'})^{2} = 0.010 \cdot 4 s^{'2} ⟹ s^{'2} = \frac{1.7 \times 1 0 ^{- 5}}{0.04} = 4.25 \times 1 0^{- 4}, s^{'} = 0.0206 M .$ Yeh step kyun? Hum jaanbujhkar usual shortcut chalate hain ( $s^{'}$ ko $c = 0.010$ ke saath drop karte hain) taaki uski failure visible ho sake — result 0.0206 $c$ se bada nikla, jo alarm bell hai ki $s^{'} ≪ c$ assumption galat hai.
Exact cubic solve karo: $4 s^{'2} (0.010 + s^{'}) = 1.7 \times 1 0^{- 5} .$ Iterate karo (try $s^{'} = 0.0135$ ): $4 (0.0135)^{2} (0.010 + 0.0135) = 4 (1.8225 \times 1 0^{- 4}) (0.0235) = 1.71 \times 1 0^{- 5}$ ✓. Yeh step kyun? Jab approximation fail ho toh pura $(c + s^{'})$ factor rakho aur numerically root dhundho.

Answer: $s^{'} \approx 1.35 \times 1 0^{- 2}$ M.

Verify: $s^{'} = 0.0135$ plug karo: $4 (0.0135)^{2} (0.0235) = 1.71 \times 1 0^{- 5} \approx K_{s p}$ ✓. Lesson: hamesha test karo $s^{'} < 5%$ of $c$ shortcut trust karne se pehle.

Cell G — Degenerate limit: common ion → 0

Worked example Ex 7: AgCl, 1.0×10⁻⁸ M NaCl mein (essentially kuch nahi)

Diya gaya $K_{s p} = 1.8 \times 1 0^{- 10}$ ; common ion ki whisper, $c = 1.0 \times 1 0^{- 8}$ M.

Forecast: common ion pure-water solubility ( $1.34 \times 1 0^{- 5}$ ) se bahut chhota hai. Guess: kya answer basically Ex 1 par wapas aa jaata hai?

Total $[Cl^{-}] = c + s^{'} = 1.0 \times 1 0^{- 8} + s^{'}$ , $[Ag^{+}] = s^{'}$ . Yeh step kyun? Hum assume nahi kar sakte ki $c$ dominate karta hai — yahaan karta nahi.
Kyunki $c ≪ s^{'}$ hai, hum $c$ drop karte hain: $c + s^{'} \approx s^{'}$ , jisse milta hai $K_{s p} = s^{'} \cdot s^{'} = s^{'2} ⟹ s^{'} = 1.8 \times 1 0^{- 10} = 1.34 \times 1 0^{- 5} M .$ Yeh step kyun? Limit $c \to 0$ mein formula zaroor pure-water result recover karna chahiye — ek sanity check ki hamara poora framework continuous hai.

Verify: $s^{'} = 1.34 \times 1 0^{- 5}$ , aur indeed $c = 1 0^{- 8}$ uska $0.075%$ hai — negligible ✓. Toh common-ion effect switch off ho jaata hai jab $c$ tiny ho; figure mein hyperbola wahaan flat hai.

Cell H — Real-world word problem: kya precipitate bhi banega?

Worked example Ex 8: Precipitate test ke liye mixing

Diya gaya $2.0 \times 1 0^{- 4}$ M AgNO₃ aur $2.0 \times 1 0^{- 4}$ M NaCl ke equal volumes mix karo. Kya AgCl precipitate hoga? ( $K_{s p} = 1.8 \times 1 0^{- 10}$ .)

Forecast: equal volumes mix karne se har concentration half ho jaati hai. Guess: kya product phir bhi $K_{s p}$ se upar hai?

Mix karne ke baad (volume double ho gaya), har ion half ho jaata hai: $[Ag^{+}] = [Cl^{-}] = \frac{2.0 \times 1 0 ^{- 4}}{2} = 1.0 \times 1 0^{- 4} M .$ Yeh step kyun? Dilution: same moles, double litres ⇒ half concentration. Yeh bhool jaana classic word-problem error hai.
Reaction quotient $Q = [Ag^{+}] [Cl^{-}]$ compute karo: $Q = (1.0 \times 1 0^{- 4}) (1.0 \times 1 0^{- 4}) = 1.0 \times 1 0^{- 8} .$ Yeh step kyun? $Q$ hai " $K_{s p}$ current concentrations par evaluate kiya gaya". $Q$ ko $K_{s p}$ se compare karna direction batata hai (yeh Le Chatelier's Principle quantitative banaya gaya — dekho Qualitative Analysis).
Compare karo: $Q = 1.0 \times 1 0^{- 8} > K_{s p} = 1.8 \times 1 0^{- 10}$ . Yeh step kyun? $Q > K_{s p}$ matlab solution supersaturated hai ⇒ precipitation hoti hai jab tak $Q$ girke $K_{s p}$ na ho jaaye.

Verify: ratio $Q / K_{s p} = 1.0 \times 1 0^{- 8} /1.8 \times 1 0^{- 10} = 55.6 > 1$ ✓ — haan, AgCl precipitate karta hai.

Cell I — Exam twist: suppression ka percent / factor

Worked example Ex 9: "Solubility kis factor se giri?"

Diya gaya AgCl: pure water mein vs 0.020 M NaCl mein solubility. Suppression factor batao. ( $K_{s p} = 1.8 \times 1 0^{- 10}$ .)

Forecast: tum dono formulas jaante ho — $s = K_{s p}$ aur $s^{'} = K_{s p} / c$ . Guess: factor hoga $K_{s p} / (K_{s p} / c) = c / K_{s p}$ .

Pure water: $s = 1.8 \times 1 0^{- 10} = 1.34 \times 1 0^{- 5}$ M. Yeh step kyun? Cell A se baseline.
NaCl mein ( $c = 0.020$ M): $s^{'} = \frac{K _{s p}}{c} = \frac{1.8 \times 1 0 ^{- 10}}{0.020} = 9.0 \times 1 0^{- 9}$ M. Yeh step kyun? Cell B shortcut, valid hai kyunki $9.0 \times 1 0^{- 9} ≪ 0.020$ .
Suppression factor: $\frac{s}{s ^{'}} = \frac{1.34 \times 1 0 ^{- 5}}{9.0 \times 1 0 ^{- 9}} \approx 1.49 \times 1 0^{3} .$ Yeh step kyun? Exam ratio maangta hai, raw solubilities nahi — question precisely padhna zaroori hai.

Verify: derived shortcut se cross-check karo $c / K_{s p} = 0.020/1.34 \times 1 0^{- 5} = 1.49 \times 1 0^{3}$ ✓. Solubility lagbhag 1500 guna girती hai.

Recall

Recall Main

$s^{'} = K_{s p} / c$ kab use kar sakta hoon poore quadratic ki jagah? Sirf tab jab dissolved amount common ion ke saath negligible ho ::: jab $s^{'} < 5%$ of $c$ ho; warna pura $(c + s^{'})$ (ya $(c + 2 s^{'})$ ) factor rakho aur exactly solve karo (Ex 6).

Recall Pure water mein 1:2 salt

$MX_{2}$ ke liye, $K_{s p}$ ke terms mein $s$ kya hoga? $s = 3 K_{s p} /4$ ::: kyunki $K_{s p} = s (2 s)^{2} = 4 s^{3}$ .

Recall Do solutions mix karne par kya precipitate banta hai yeh kaise decide karoon?

$Q$ compute karo diluted concentrations par aur compare karo ::: agar $Q > K_{s p}$ toh precipitate hoga; agar $Q < K_{s p}$ toh dissolved rehega; $Q = K_{s p}$ exactly saturated hai.

Connections

Solubility Product (Ksp) — har $Q$ -vs- $K_{s p}$ comparison yahaan rehta hai.
Le Chatelier's Principle — leftward shift ke kyun ka explanation.
Buffer Solutions — wahi algebra weak-acid ionization par apply hoti hai.
Qualitative Analysis — common ions ke zariye selective precipitation (Ex 8 ki logic).
Ionic Equilibrium — in sabhi simultaneous equilibria ka broader ghar.