Solubility product Ksp — common-ion suppression, selective precipitation: Question bank

2.6.15 · D5 · Chemistry › Equilibrium › Solubility product Ksp — common-ion suppression, selective p

Yeh page ek misconception minefield walkthrough hai. Neeche diya hua har item ek specific tarike ko target karta hai jisme students solubility equilibria ko galat padhte hain. Prompt padho, apna jawab zyabar se bolne se pehle reveal karo, phir reasoning check karo — answer side hamesha kyun explain karti hai, kabhi sirf "haan/nahi" nahi.

Prerequisites parent note mein hain (Hinglish version yahan) aur yeh Le Chatelier's Principle, Chemical Equilibrium andrium Constant, Ionic Equilibria in Solutions, aur — neeche ke pH traps ke liye — pH and pOH Calculations par lean karte hain.

Definition Paanch symbols jo is page par baar baar aate hain

Traps se pehle, har symbol ko pin down karo taaki koi reveal opaque na lage. Ek salt $M_{x} A_{y} (s) ⇌ x M^{n +} (a q) + y A^{m -} (a q)$ consider karo.

$x, y$ = stoichiometric coefficients: ek formula unit dissolve hone par kitne cations ( $x$ ) aur anions ( $y$ ) release hote hain. AgCl ke liye, $x = y = 1$ ; $CaF_{2}$ ke liye, $x = 1, y = 2$ ; $Ca_{3} (PO_{4})_{2}$ ke liye, $x = 3, y = 2$ .
$s$ = pure water mein solubility, mol/L mein: saturation se pehle ek litre mein kitne formula units salt dissolve hote hain. Har dissolved unit $x$ cations aur $y$ anions deta hai, isliye $[M^{n +}] = x s$ aur $[A^{m -}] = y s$ .
$s^{'}$ = common ion ki presence mein solubility: wahi quantity lekin ek aisi solution mein measure ki gayi jo pehle se hi salt ke ek ion ko contain karti hai. Yeh (almost hamesha) $s$ se bahut chhoti hoti hai.
$C$ = pehle se present common ion ki concentration (jaise added NaCl ya NaF se).
$Q$ = reaction quotient / ionic product: ion concentrations ka wahi product jaise $K_{s p}$ , lekin current, possibly non-equilibrium concentrations par evaluate kiya gaya — $Q = [M^{n +}]_{now}^{x} [A^{m -}]_{now}^{y}$ . Yeh $K_{s p}$ ke barabar sirf saturation par hota hai.

True or false — justify

Chhota

K_{s p}

hamesha chhoti solubility

s

matlab hai.

False. Conversion

s = (K_{s p} / x^{x} y^{y})^{1/ (x + y)}

stoichiometry par depend karta hai, isliye ek 1:2 salt jiska

K_{s p}

chhota hai phir bhi ek 1:1 salt jiska

K_{s p}

bada hai se zyada soluble ho sakta hai — aap

s

ko directly sirf unhi salts mein compare kar sakte hain jinka ion count same ho.

Saturated AgCl solution mein NaCl add karna AgCl ke

K_{s p}

ki value change karta hai.

False.

K_{s p}

sirf temperature par depend karta hai. Common ion add karna equilibrium ki position shift karta hai (AgCl kam dissolve hota hai) lekin product

[Ag^{+}] [Cl^{-}]

abhi bhi same

K_{s p}

ke barabar rehta hai.

Equilibrium par, har solution mein AgCl ke liye

[Ag^{+}] = [Cl^{-}]

hota hai.

False. Yeh equality sirf pure water mein hoti hai. 0.1 M NaCl mein,

[Cl^{-}] \approx 0.1

jabki

[Ag^{+}]

gir ke

\sim 1 0^{- 9}

ho jaata hai — bahut unequal, phir bhi unka product

K_{s p}

hi hai.

Solid

M_{x} A_{y}

ki concentration equilibrium expression ke denominator mein aati hai.

False. Ek pure solid ki activity 1 hoti hai (yeh apni standard state mein hai aur iski "concentration" change nahi hoti), isliye yeh constant mein fold ho jaati hai aur kabhi nahi likhi jaati — isi liye

K_{s p}

mein sirf ion terms hoti hain.

Agar tum neeche baitha undissolved AgCl solid ka amount double kar do, toh solubility double ho jaati hai.

False. Zyada solid koi bhi concentration change nahi karta — solid ki activity 1 rehti hai. Solubility

K_{s p}

aur pehle se present ions par depend karti hai, is par nahi ki kitna solid ikatha hua hai.

CaF_{2}

ke liye,

K_{s p} = [Ca^{2 +}] [F^{-}]

False. Balanced equation hai

CaF_{2} ⇌ Ca^{2 +} + 2 F^{-}

, isliye fluoride coefficient 2 exponent ban jaata hai:

K_{s p} = [Ca^{2 +}] [F^{-}]^{2}

Ek supersaturated solution mein

Q > K_{s p}

hota hai aur yeh thermodynamically stable hai.

False.

Q > K_{s p}

matlab solution tolerated ion product se upar hai aur metastable hai — yeh precipitate karega (kabhi kabhi sirf nucleation trigger ke baad) jab tak

Q

wapas

K_{s p}

par nahi aa jaata.

Common-ion suppression tab bhi kaam karta hai jab added ion salt ka apna ion na ho.

False. Suppression ke liye ek shared (common) ion chahiye taaki Le Chatelier dissolution equilibrium ko backward push kar sake. Ek unrelated spectator ion

K_{s p}

expression mein appear nahi karta aur isliye directly shift nahi kar sakta.

CaF_{2}

ki solubility neutral water aur strong acid mein same hoti hai.

False.

F^{-}

weak acid HF ka conjugate base hai, isliye acid ise protonate karta hai (

F^{-} + H^{+} \to HF

), free fluoride remove karta hai; equilibrium ise replace karne ke liye zyada solid dissolve karta hai, isliye pH girne par solubility badhti hai.

Solid AgCl mein aqueous

NH_{3}

add karna sirf iske dissolution ko decrease kar sakta hai.

False.

NH_{3}

complex

[Ag(NH_{3})_{2}]^{+}

banata hai, free

Ag^{+}

solution se kheench leta hai; dissolution equilibrium respond karta hai zyada AgCl dissolve karke, isliye solubility actually badhti hai.

Spot the error

"AgCl ke liye

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

hai, isliye iski solubility

1.8 \times 1 0^{- 10}

mol/L hai."

Error:

K_{s p}

product

s^{2}

hai,

s

khud nahi. Solubility hai

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

mol/L — raw

K_{s p}

value se paanch orders of magnitude bada.

CaF_{2}

ke liye solubility

s

ke saath, maine

K_{s p} = s \cdot s^{2} = s^{3}

set kiya kyunki teen ions hain."

Error:

[F^{-}] = 2 s

hai,

s

nahi. Correct substitution hai

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

; 2 ka factor bhoolna fluoride ko undercount karta hai aur galat

s

deta hai.

"0.10 M NaCl mein AgCl: maine

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

set kiya aur

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

solve kiya."

Error: chloride NaCl se dominated hai. Correct setup hai

[Cl^{-}] = 0.10 + s^{'} \approx 0.10

aur

[Ag^{+}] = s^{'}

, jo deta hai

K_{s p} = s^{'} (0.10)

— common ion add hota hai,

s^{'}

ke barabar nahi hota.

"100 mL of 0.010 M

AgNO_{3}

ko 100 mL NaCl ke saath mix karte hue, maine

Q

mein

[Ag^{+}] = 0.010

M use kiya."

Error: mixing total volume double kar deti hai, har concentration half kar deti hai.

Q

compute karne se pehle aapko diluted value

[Ag^{+}] = 0.0050

M use karni chahiye; dilution bhoolna

Q

ko overstate karta hai aur galat precipitation predict kar sakta hai.

"Maine check kiya

s^{'} ≪ C

aur yeh fail ho gaya, toh main phir bhi approximate answer use karunga."

Error: agar approximation fail ho toh aapko puri quadratic

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

solve karni chahiye. Failed check ko ignore karna numerically galat solubility deta hai, kabhi kabhi bahut bade factor se.

Q < K_{s p}

matlab precipitate banne wala hai."

Error: yeh ulta hai.

Q < K_{s p}

unsaturated hai — solution abhi bhi zyada salt dissolve kar sakta hai. Precipitation ke liye

Q > K_{s p}

chahiye.

"0.010 M NaF mein

CaF_{2}

ki solubility: maine fluoride ko square karna bhool gaya, isliye

s^{'} = K_{s p} /0.010

Error: fluoride squared enter hoti hai. Correct expression hai

K_{s p} = s^{'} (0.010)^{2}

, isliye

s^{'} = K_{s p} / (0.010)^{2} = K_{s p} /1 0^{- 4}

— square drop karna answer ko 0.010 ke factor se inflate karta hai.

CaCO_{3}

K_{s p}

fixed hai, isliye iski solubility pH par depend nahi kar sakti."

Error:

K_{s p}

fixed zaroor hai, lekin

CO_{3}^{2 -}

acid se protonate hokar

HCO_{3}^{-}

H_{2} CO_{3}

ban jaata hai; free carbonate remove hone se

Q

K_{s p}

se neeche girta hai isliye zyada solid dissolve hota hai — pH solubility change karta hai bina

K_{s p}

change kiye.

Why questions

Pure-solid activity 1 kyun hoti hai jo humein solid ko equilibrium expression se drop karne deti hai?

Kyunki ek pure solid apni standard state mein hai — iski "amount per unit of itself" fixed aur unchanging hai, isliye yeh ek constant factor of 1 contribute karta hai jo

K_{s p}

mein absorb ho jaata hai rather than ek variable ke roop mein track kiya jaaye.

Simple

s^{'} \approx K_{s p} / C

case mein solubility common-ion concentration ke saath inversely proportional kyun hai?

Kyunki

K_{s p}

fixed hai: agar product mein common-ion term

C

badh jaaye, toh doosra ion (jo

s^{'}

ke barabar hai) ko product constant rakhne ke liye kam hona padega, isliye

s^{'} \propto 1/ C

Same

K_{s p}

wale do salts ki solubility alag kyun ho sakti hai?

Kyunki

s

stoichiometry par depend karta hai

s = (K_{s p} / x^{x} y^{y})^{1/ (x + y)}

ke through; exponents aur

x^{x} y^{y}

factor 1:1 aur 1:2 salt ke beech alag hote hain, isliye equal

K_{s p}

equal

s

force nahi karta.

Selective precipitation kyun kaam karta hai — ek ion doosre se pehle kyun drop out hota hai?

Kyunki har target salt apne

K_{s p}

threshold par alag added-reagent concentration par pahunchta hai; aap dhire dhire precipitating ion badhate ho jab tak

Q

chhote

K_{s p}

se pehle exceed na kare, us salt ko precipitate karte hue jabki doosra dissolved rehta hai.

Precipitation decide karne ke liye hum

Q

(na ki

K_{s p}

) kyun use karte hain?

Q

actual current concentrations se compute kiya jaata hai, jo equilibrium se bahut dur ho sakta hai; is instantaneous ion product ko fixed threshold

K_{s p}

se compare karna batata hai ki system ko kis direction mein move karna hai.

Common-ion effect solubility ko "zero tak" kyun nahi girne deta?

Kyunki

K_{s p}

ek nonzero constant hai — chahe

C

kitna bhi bada ho jaaye,

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

positive rehta hai. Kuch salt hamesha dissolve hota hai; suppression sirf

s^{'}

ko bahut chhota karta hai, kabhi exactly zero nahi.

Temperature badhana usually (lekin hamesha nahi)

K_{s p}

kyun badhata hai?

Kyunki zyatartar dissolution endothermic hoti hai, isliye Le Chatelier ke according heat ek reactant ki tarah kaam karti hai aur zyada T equilibrium ko dissolved ions ki taraf shift karta hai; rare exothermic-dissolution salts ke liye ulta hota hai aur

K_{s p}

T ke saath girta hai.

pH kam karna weak-acid anions (jaise

F^{-}

CO_{3}^{2 -}

S^{2 -}

) wale salts ki solubility kyun badhata hai?

Kyunki

H^{+}

anion ko apne weak acid mein protonate kar deta hai, dissolution equilibrium se free anion drain kar deta hai;

[M^{n +}] [A^{m -}]^{y}

K_{s p}

par rakhne ke liye system ko zyada solid dissolve karna padta hai, isliye acid solubility badhata hai.

Complexing agent (jaise

NH_{3}

CN^{-}

) add karna ek "bahut insoluble" salt ko dissolve kyun karta hai?

Ligand free metal ion ko ek complex mein bandh kar leta hai (jaise

[Ag(NH_{3})_{2}]^{+}

), isliye

[M^{n +}]

girta hai,

Q

K_{s p}

se neeche aa jaata hai, aur dissolution jaari rehta hai — apparent solubility pure-water value se hazaaron guna ho sakti hai chahe

K_{s p}

kabhi nahi badla.

Edge cases

Ek solution mein jahan common-ion concentration

C = 0

ho, salt ki solubility kya hai?

Tab koi suppression nahi hai — aap wapas pure water mein hain, isliye

s

full

K_{s p} = x^{x} y^{y} s^{x + y}

relation se aata hai,

s^{'} \approx K_{s p} / C

se nahi (jo blow up ho jaata aur

C = 0

par undefined hai).

Do solutions mix hone ke exact instant par lekin koi solid banne se pehle

Q

ka kya hota hai?

Q

dono ions ki nayi diluted concentrations se compute kiya jaata hai; agar woh

Q > K_{s p}

ho toh system momentarily supersaturated hai aur precipitate karega, agar

Q \leq K_{s p}

ho toh kuch nahi banta.

Agar aap itna common ion add karo ki

C + s^{'}

ko ab

C

se approximate nahi kiya ja sakta, toh kya change karna padega?

Physics mein kuch nahi, lekin algebra mein: aapko full expression

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

(ya non-1:1 salts ke liye exponents ke saath) rakhni hogi aur resulting quadratic/cubic solve karni hogi rather than shortcut use karne ke.

Ek salt jo itni soluble hai ki sparingly soluble nahi hai — kya

K_{s p}

framework cleanly apply hota hai?

Itna nahi. Highly soluble salts ke liye ion concentrations badi hoti hain, activities concentrations se strongly deviate karti hain, aur simple

K_{s p} = [ion]^{x} [ion]^{y}

(jo ideal dilute behaviour assume karta hai) inaccurate ho jaata hai.

Ek saturated AgCl solution mein jisme aap bahut zyada NaCl add karo,

[Ag^{+}]

kya hoga?

Yeh bahut chhota ho jaata hai:

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

, aur jaise

[Cl^{-}]

badhta hai,

[Ag^{+}]

zero ki taraf shrink hota jaata hai (lekin kabhi pahunchta nahi) — isi basis par metal ion ko solution se almost completely drive kiya ja sakta hai.

Lekin agar woh "bahut zyada" chloride sach mein extreme ho, toh AgCl solubility ke saath kya surprising cheez ho sakti hai?

Bahut high

[Cl^{-}]

par, chloride ek ligand ki tarah kaam karta hai aur soluble complexes jaise

[AgCl_{2}]^{-}

banata hai; solubility, jo neeche drive hui thi, phir se badhna shuru ho sakti hai — common-ion suppression aur complex formation ke beech competition.

Jab do salts ek common ion share karte hain aur dono saturation ke paas hain, toh agar aap us shared ion ka aur add karo toh kya hota hai?

Dono equilibria same pressure feel karte hain — shared ion dono

Q

expressions mein appear karta hai, isliye dono salts simultaneously suppress hote hain aur jo bhi pehle apna

K_{s p}

cross karta hai woh precipitate karna shuru karta hai.

Exactly

Q = K_{s p}

ki boundary par, kya kuch precipitate ho raha hai?

Koi net change nahi: solution exactly saturated hai aur equilibrium mein hai, isliye koi bhi solid present ho toh ions ke saath dynamic balance mein coexist karta hai — dissolution aur precipitation rates equal hain, koi net solid nahi banta ya dissolve hota.

Ek metal sulfide ka tiny

K_{s p}

hai; kya yeh akela guarantee karta hai ki yeh kisi bhi acidic sulfide solution se precipitate karega?

Nahi — acid mein,

S^{2 -}

largely

HS^{-}

H_{2} S

mein protonate ho jaata hai, isliye free

[S^{2 -}]

(aur isliye

Q

) pH se throttle hota hai; sulfide ka yeh pH control exactly wahi hai jo qualitative analysis ko metal groups ko selectively separate karne deta hai.

Recall Fast self-test

Precipitation decide karne ke liye kaunsa comparison kiya jaata hai, $s$ vs $K_{s p}$ ya $Q$ vs $K_{s p}$ ? ::: $Q$ vs $K_{s p}$ — $Q$ actual (diluted) concentrations use karta hai, precipitation tab hoti hai jab $Q > K_{s p}$ . $Ca_{3} (PO_{4})_{2} ⇌ 3 Ca^{2 +} + 2 PO_{4}^{3 -}$ ke liye, master formula use karke $s$ mein $K_{s p}$ likho. ::: $x = 3, y = 2$ , isliye $K_{s p} = (3 s)^{3} (2 s)^{2} = 27 \cdot 4 s^{5} = 108 s^{5}$ . Kaun si do side reactions ek salt ko zyada soluble bana sakti hain jitna uska $K_{s p}$ suggest karta hai? ::: Weak-acid anion ka protonation (low pH) aur metal ka complex-ion formation — dono ek free ion drain karte hain aur dissolution equilibrium ko forward pull karte hain.