Solubility product Ksp — common-ion suppression, selective precipitation: Visual walkthrough

2.6.15 · D2 · Chemistry › Equilibrium › Solubility product Ksp — common-ion suppression, selective p

Ye page parent topic ki poori kahani ko ek single picture se shuru karke rebuild karta hai. Hum ek jar of water se start karte hain, ek chutki si barely-dissolving salt ke saath, aur kuch nahi. Ant tak tum dekh paoge — sirf compute nahi — ki kyon ek aisa ion jo pehle se present hai, usse dalne par solubility hazaron guna kuch nahi reh jaati.

Har symbol pahle earn kiya jaata hai, phir use hota hai. Agar tum count kar sakte ho, aur tumhe pata hai ki "solid woh cheez hai jo tum bottom mein settle hui dekh sakte ho," to tum pehli line se follow kar sakte ho.

Step 1 — "Dissolving" actually hota kya hai (the two-way traffic)

KYA. Ek chutki table-salt-jaisi crystal ka picture karo — crystal ko $MA$ kaho — jo ek beaker of water ke bottom mein baitha hai. Uski surface ke kuch particles toot ke nikal jaate hain aur paani mein alag charged particles ke roop mein float karte hain. Chemistry mein ek charged particle ko ion kehte hain. Positive wale ko hum $M^{+}$ likhte hain, negative wale ko $A^{-}$ .

KYUN. Yahan kuch bhi ek direction mein nahi hai. Jaise free ions paani mein ghumte hain, kuch wapas crystal se takra ke re-stick ho jaate hain. To ek saath do flows ho rahe hain: crystal → ions (dissolving) aur ions → crystal (re-depositing). Figure mein do curved arrows dekho — ek solid se nikal raha hai, ek wapas aa raha hai.

PICTURE.

Half-arrow symbol $⇌$ ka matlab hai "dono directions ek saath, balanced": $MA(s) ⇌ M^{+} (a q) + A^{-} (a q)$

$MA(s)$ — solid crystal (tag " $s$ " matlab solid).
$M^{+} (a q), A^{-} (a q)$ — paani mein float karte free ions (tag " $a q$ " matlab paani mein dissolved).

Step 2 — Balance ko EK number mein banana: $K_{s p}$

KYA. Equilibrium par hum dono ion concentrations ko saath multiply karte hain. "Concentration" ka matlab hai ek litre paani mein us ion ke kitne moles hain; hum ise square brackets se likhte hain, to $[M^{+}]$ = concentration of $M^{+}$ . Yeh product ek given salt ke liye ek given temperature par ek fixed number hota hai.

KYUN multiply karte hain, aur solid ko kyun bahar rakhte hain? Chemical Equilibrium andrium Constant ka equilibrium constant idea kehta hai: freed species ke products lo, har ek ko uski coefficient ki power tak raise karo. Solid ko khud concentration nahi milti — crystal ka ek tukda crystal ka ek tukda hi hai, uska "per litre of water amount" meaningless hai, isliye uski activity $1$ define ki jaati hai. Bas ions bachte hain.

$fixed number K_{s p} = kitna M^{+} [M^{+}] \times kitna A^{-} [A^{-}]$

PICTURE. $K_{s p}$ ko ek rectangle ke area ki tarah socho jisme width $[M^{+}]$ hai aur height $[A^{-}]$ hai. Salt "allow" karta hai sirf ek fixed area ke rectangles ko. Tum ise tall-and-thin ya short-and-wide bana sakte ho, lekin area locked rehta hai.

Step 3 — Solubility $s$ pure water mein (the square rectangle)

KYA. $MA$ ko pure water mein daalo. Maano $s$ = kitne moles per litre actually dissolve hote hain — yahi solubility hai. Kyunki formula mein ek $M^{+}$ per ek $A^{-}$ hai, har dissolved unit ek-ek deta hai. To dono concentrations $s$ ke barabar hain: $[M^{+}] = s, [A^{-}] = s .$

KYUN. Pure water mein koi aur cheez ye ions supply nahi karti — ek maatra source crystal khud hai, aur woh unhe 1-to-1 pair mein release karta hai. To width = height. Locked rectangle ek square hai jiska side $s$ hai.

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

$s^{2}$ — square ka area, side $s$ .
$K_{s p}$ — "kaun si side length is area ko deti hai?" Woh square-root squaring ko undo karta hai.

PICTURE. Step 2 ke rectangle ka ek perfectly square version.

Step 4 — Ab ek common ion daalo (rectangle ko stretch karo)

KYA. Usi beaker mein ek alag, fully-soluble salt dissolve karo jo $A^{-}$ ion share karti hai — AgCl ke liye woh ordinary $NaCl$ hai, jo paani ko $Cl^{-}$ se bhhar deti hai. Extra amount ko $C$ kaho. Ab, AgCl ke kuch bhi dissolve hone se pehle, paani mein pehle se $[A^{-}] = C$ hai.

KYUN ise "common" kehte hain. $A^{-}$ dono salts mein common hai — ek salt woh ion share karta hai jo doosra bhi banata hai. Woh shared ion hi common ion hai.

Yahan key move hai. Locked area $K_{s p}$ nahi badla (same salt, same temperature). Lekin humne abhi height $[A^{-}]$ ko $C$ tak force kar diya. Area constant rakhne ke liye, width $[M^{+}]$ ko shrink karna hoga.

$K_{s p} = shrink karna hoga [M^{+}] \times C tak dhakela gaya [A^{-}]$

PICTURE. Step 3 ke square ko bahut tall stretch karo (height $C$ ). Same area rakhne ke liye, woh razor-thin ho jaata hai. Woh thin width hi naya, tiny solubility hai.

Yeh "area resist karta hai, to doosri side give way karti hai" bilkul wahi hai jo Le Chatelier's Principle kehta hai: products par push karo, balance solid ki taraf wapas jaata hai.

Step 5 — Suppressed-solubility formula (thin width padhna)

KYA. Maano $s^{'}$ = nai solubility jab common ion present hai. Har dissolved AgCl ab bhi ek $A g^{+}$ deta hai, to $[M^{+}] = s^{'}$ . Chloride flood $C$ plus woh thodi si matra jo AgCl add karta hai hai: $[A^{-}] = C + s^{'}$ .

$K_{s p} = [M^{+}] s^{'} \times [A^{-}] (C + s^{'})$

KYUN hum bracket ke andar $s^{'}$ drop kar sakte hain. $s^{'}$ bahut hi fantastically small hai (width almost zero hai), jabki $C$ ek normal lab concentration hai. Ek droplet ko ocean mein add karne se kuch nahi badalta: $C + s^{'} \approx C$ . Figure dekho — extra height ka sliver $C$ ke baad mein invisible hai.

$K_{s p} \approx s^{'} C ⟹ s^{'} \approx \frac{K _{s p}}{C}$

$K_{s p}$ — locked area (fraction ke upar).
$C$ — force ki gayi common-ion concentration (neeche).
Bada $C$ ⇒ chhota $s^{'}$ . Solubility common ion ke saath inversely proportional hai.

PICTURE. Rectangle exact numbers ke saath re-draw kiya gaya, negligible sliver dikhata hua.

Step 6 — 1:2 case (rectangle ki height squared hai)

KYA. Kuch salts do ek ion mein tootte hain, jaise $CaF_{2} ⇌ C a^{2 +} + 2 F^{-}$ . Ab locked shape width×height nahi balki width×height×height hai, kyunki $[F^{-}]$ do baar appear karta hai — ek baar har us fluoride ke liye jo balanced equation banata hai: $K_{s p} = [C a^{2 +}] [F^{-}]^{2}$

$[F^{-}]$ par exponent $2$ literally equation mein coefficient 2 hai — do fluorides per formula unit, isliye concentration khud se multiply hoti hai.

KYUN yeh phir bhi suppress karta hai. Paani ko $F^{-}$ se concentration $C$ par flood karo (NaF se). $[F^{-}]^{2}$ term $C^{2}$ ban jaata hai, aur $K_{s p}$ locked rakhne ke liye $[C a^{2 +}] = s^{'}$ aur bhi zyaada shrink karta hai: $K_{s p} \approx s^{'} C^{2} ⟹ s^{'} \approx \frac{K _{s p}}{C ^{2}}$

Kyunki flood squared hai, choke 1:1 case se zyaada fierce hai.

PICTURE. Ek box jiska ek face $s^{'}$ hai aur jiska square face $C \times C$ hai — volume locked $K_{s p}$ hai.

Step 7 — Edge cases (reader ko kabhi stranded mat chhoddo)

KYA & KYUN — har degenerate scenario:

PICTURE. Sab char cases as char chhote rectangles side by side.

Ek-picture summary

Ek salt = ek locked rectangle area $K_{s p}$ . Pure water square deta hai (side $= s = K_{s p}$ ). Ek common ion flood karna rectangle ko tall stretch karta hai, to doosri side ek sliver $s^{'} = K_{s p} / C$ tak collapse ho jaati hai. Woh collapsing sliver hi suppressed solubility hai — ek hi image mein poora common-ion effect. Yehi $Q$ -vs- $K_{s p}$ reasoning Qualitative Inorganic Analysis aur Ionic Equilibria in Solutions mein bhi kaam aati hai.

Recall Feynman retelling — kisi dost ko bolo

Ek barely-dissolving salt paani mein ek tug-of-war mein baitha hai: kuch bits toot ke nikalte hain, kuch re-stick ho jaate hain, aur balance par dono ion concentrations ka product ek fixed number hota hai, $K_{s p}$ — ise ek aisa rectangle socho jiska area locked hai. Pure water mein dono ions equal amounts mein nikalte hain, isliye rectangle ek square hai aur har side $K_{s p}$ hai. Ab ek doosri salt daalo jo ek ion share karti hai: tumne achanak rectangle ki us side ko bahut bada kar diya. Lekin area locked hai, to doosri side — jo hamari salt ki dissolve hone wali matra hai — almost nothing tak shrink ho jaati hai. Isliye AgCl salty water mein 7500× kam soluble hai: koi magic nahi, bas "ek side stretch karte waqt area constant rakhna." Agar shared ion formula mein do baar aata hai (jaise CaF₂ mein do fluorides), to woh side squared hai, isliye choke aur bhi harsh hai. Aur jab tum do solutions mix karte waqt kuch precipitate hone se pehle, bas current rectangle area $Q$ compute karo aur pucho: kya yeh locked area $K_{s p}$ se bada hai? Agar haan, solid barsata hai jab tak fit nahi ho jaata.

Recall Quick self-test

Solid ko $K_{s p}$ expression se kyun bahar rakha jaata hai? ::: Uski activity 1 define ki jaati hai — ek pure solid ki koi meaningful "concentration per litre of water" nahi hoti, isliye woh 1 ka factor contribute karta hai. Pure water mein rectangle ek ________ shape ka hota hai. ::: Ek square, side $s = K_{s p}$ , kyunki dono ions equal amounts mein aate hain. Common ion ko concentration $C$ par add karne se solubility $s^{'} \approx$ kya ho jaati hai? ::: $K_{s p} / C$ (1:1 salt ke liye) — $C$ ke saath inversely proportional. CaF₂ ke liye flood $C^{2}$ ke roop mein enter karta hai — square kyun? ::: Balanced equation har unit se do F⁻ banata hai, isliye $[F^{-}]$ ko $K_{s p}$ mein power 2 tak raise kiya jaata hai. $s^{'} = K_{s p} / C$ kab FAIL karta hai? ::: Jab $C$ small ho (nahi $≫ s^{'}$ ) — tab poora quadratic $K_{s p} = s^{'} (C + s^{'})$ solve karo; aur yeh kabhi $C = 0$ par apply nahi hota (use karo $K_{s p}$ ).