Solubility product Ksp — common-ion suppression, selective precipitation: Worked examples

2.6.15 · D3 · Chemistry › Equilibrium › Solubility product Ksp — common-ion suppression, selective p

Yeh page parent topic ka drill hall hai. Parent note ne tools banaye the: $K_{s p}$ , solubility $s$ , the common-ion effect, aur ionic product $Q$ . Yahan hum har tarah ke question un tools pe throw karte hain jab tak koi bhi scenario surprise na kar sake.

Ek bhi number se pehle: do symbols jinhe hum poore page bhar use karte hain.

Neeche di gayi figure yeh comparison aapke dimag mein fix karti hai ek bhi $Q$ compute karne se pehle. Ise ek number line ki tarah padhein jo horizontal axis par drawn hai (labelled "ionic product $Q$ , baayi taraf se daayein badh raha hai"): $K_{s p}$ us axis par ek fixed dot hai; $Q$ ek marker hai jo actual ion concentrations ke hisaab se left ya right slide karta hai. Is page par sab kuch yahi decide karna hai ki $Q$ dot ke kis side land karta hai.

Figure — Solubility product Ksp — common-ion suppression, selective precipitation

Figure ek labelled number line hai. Iska horizontal axis $Q$ hai (daayein badhta hua); single dark dot fixed $K_{s p}$ hai. Dot ke left mein band labelled hai "unsaturated — aur dissolve ho sakta hai" ( $Q < K_{s p}$ ); right mein band labelled hai "supersaturated — solid drop out hota hai" ( $Q > K_{s p}$ ); dot khud labelled hai "saturated" ( $Q = K_{s p}$ ). (Colours — left par mint, right par coral, dot par lavender — sirf in teeno written labels ko reinforce karte hain, isliye picture bina colour par rely kiye bhi same padhti hai.) Yeh picture haath mein rakho: har "kya yeh precipitate karega?" question simply yeh pooch raha hai ki $Q$ kis labelled band mein girti hai.

The scenario matrix

Har Ksp problem jo aap kabhi meet karoge woh in cells mein se ek hai. Neeche ke examples us cell ke saath labelled hain jis par woh hit karte hain, taaki aap dekh sako ki poora board covered hai.

Cell	Scenario class	Kya tricky banata hai	Example
A	1:1 salt, pure water	baseline, $K_{s p} = s^{2}$	Ex 1
B	1:2 (ya 2:1) salt, pure water	2 ki power aati hai; $4 s^{3}$ dekho	Ex 2
C	common ion added (1:1)	shift, approximation, badi drop	Ex 3
D	common ion, approximation fails	quadratic honestly solve karni padegi	Ex 4
E	do solutions mix karo → kya precipitate hoga?	dilution on mixing, compute $Q$	Ex 5
F	degenerate / zero input	ek ion absent → $Q = 0$ , kabhi precipitate nahi	Ex 6
G	selective precipitation (pehle kaun girega?)	do salts ek common ion ke liye compete karte hain	Ex 7
H	limiting behaviour (jaise $C \to$ large / small)	ek number nahi, trend check karo	Ex 8
I	real-world word problem	hard water / fluoride — padho, phir model karo	Ex 9
J	exam twist (pH anion ko control karta hai)	anion supply acid–base se set hoti hai, salt se nahi	Ex 10

Prerequisites lean on karte hain Chemical Equilibrium andrium Constant, Le Chatelier's Principle, Ionic Equilibria in Solutions, Stoichiometry and Solution Concentration, aur Ex 10 ke liye, pH and pOH Calculations.

Cell A — 1:1 salt in pure water

Worked example Example 1 — the baseline (Cell A)

Problem: $AgBr$ ka $K_{s p}$ $5.0 \times 1 0^{- 13}$ hai. Pure water mein iska solubility nikalo.

Forecast: Pehle guess karo — kya $s$ $1 0^{- 6}$ ke paas hoga ya $1 0^{- 13}$ ke paas? (Kyunki hum 1:1 salt ka square root lete hain, roughly half-power expect karo, yani $\sim 1 0^{- 6}$ – $1 0^{- 7}$ .)

Dissolution likho. $AgBr(s) ⇌ Ag^{+} + Br^{-}$ . Yeh step kyun? $K_{s p}$ isi reaction ka equilibrium constant hai; jab tak reaction paper par na ho hum ise use nahi kar sakte.
$s$ ko solubility maano. Har formula unit jo dissolve hota hai ek $Ag^{+}$ aur ek $Br^{-}$ deta hai, isliye $[Ag^{+}] = s$ aur $[Br^{-}] = s$ . Yeh step kyun? 1:1 stoichiometry dono ion concentrations ko ek hi unknown $s$ se baandhti hai.
Substitute karo. $K_{s p} = [Ag^{+}] [Br^{-}] = s \cdot s = s^{2}$ , isliye $5.0 \times 1 0^{- 13} = s^{2}$ . Yeh step kyun? Saturation par $Q = K_{s p}$ ; woh equality hi equation hai.
Solve karo. $s = 5.0 \times 1 0^{- 13} = 7.07 \times 1 0^{- 7} mol/L$ .

Verify karo: Plug back karo: $(7.07 \times 1 0^{- 7})^{2} = 5.0 \times 1 0^{- 13}$ ✓. Aur $7 \times 1 0^{- 7}$ exactly forecast band mein aaya. Units: $(mol/L)^{2} = mol/L$ ✓.

Cell B — 1:2 salt in pure water

Worked example Example 2 — the sneaky factor of 4 (Cell B)

Problem: $Ag_{2} CrO_{4}$ ka $K_{s p}$ $1.1 \times 1 0^{- 12}$ hai. Pure water mein $s$ nikalo.

Forecast: Yeh salt 2:1 hai ( $Ag_{2} CrO_{4} \to 2 Ag^{+} + CrO_{4}^{2 -}$ ). Predict karo: kya $s$ AgBr-jaisi $1 0^{- 6}$ se bada hoga ya chhota? (Ek tiny number ka cube root uske square root se bada hota hai, isliye thoda bada expect karo, near $1 0^{- 4}$ .)

Dissolution: $Ag_{2} CrO_{4} (s) ⇌ 2 Ag^{+} + CrO_{4}^{2 -}$ .
Stoichiometry: agar $s$ dissolve hota hai, toh har unit mein do silver aate hain, isliye $[Ag^{+}] = 2 s$ aur $[CrO_{4}^{2 -}] = s$ . Yeh step kyun? Coefficient 2 silver concentration ko double kar deta hai — yeh miss karo toh baad ke saare numbers galat ho jaate hain.
Correct powers ke saath substitute karo: $K_{s p} = [Ag^{+}]^{2} [CrO_{4}^{2 -}] = (2 s)^{2} (s) = 4 s^{3}$ . Yeh step kyun? $[Ag^{+}]$ par exponent uska coefficient hai (2), aur factor $2 s$ uski concentration hai — ek hi "2" ke do alag uses.
Solve karo: $4 s^{3} = 1.1 \times 1 0^{- 12} \Rightarrow s^{3} = 2.75 \times 1 0^{- 13} \Rightarrow s = 6.50 \times 1 0^{- 5} mol/L$ .

Verify karo: $4 (6.50 \times 1 0^{- 5})^{3} = 4 \times 2.746 \times 1 0^{- 13} = 1.10 \times 1 0^{- 12}$ ✓. Forecast ne "near $1 0^{- 4}$ " kaha tha — $6.5 \times 1 0^{- 5}$ wahan hi hai ✓.

Cell C — common ion, approximation holds

Worked example Example 3 — common ion suppresses (Cell C)

Problem: $AgBr$ ( $K_{s p} = 5.0 \times 1 0^{- 13}$ ) ki solubility $0.10 M$ $KBr$ mein nikalo.

Forecast: KBr solution ko $Br^{-}$ se flood kar deta hai. Le Chatelier ke mutabiq, AgBr equilibrium left shift karega. Predict karo: pure water ke $7 \times 1 0^{- 7}$ se zyada soluble hoga ya bahut kam?

Existing $[Br^{-}]$ : KBr fully soluble hai, isliye $[Br^{-}]_{start} = 0.10 M$ .
$s^{'}$ = new solubility maano. Tab $[Ag^{+}] = s^{'}$ aur $[Br^{-}] = 0.10 + s^{'}$ . Yeh step kyun? Bromide ab do sources se aata hai: KBr se 0.10, aur AgBr ke thode se dissolve hone se $s^{'}$ .
Approximate karo: kyunki AgBr bahut kam dissolve hota hai, guess karo $s^{'} ≪ 0.10$ , isliye $[Br^{-}] \approx 0.10$ . Yeh step kyun? Yeh ek quadratic ko one-line algebra mein badal deta hai — bas end mein guess check karo.
Solve karo: $5.0 \times 1 0^{- 13} = s^{'} (0.10) \Rightarrow s^{'} = 5.0 \times 1 0^{- 12} mol/L$ .

Verify karo: Approximation check karo: $5.0 \times 1 0^{- 12} ≪ 0.10$ ✓ (ek factor $2 \times 1 0^{10}$ ). Pure water se compare karo: $\frac{7.07 \times 1 0 ^{- 7}}{5.0 \times 1 0 ^{- 12}} \approx 1.4 \times 1 0^{5}$ — ~140 000× kam soluble. Forecast "bahut kam" ✓.

Cell D — common ion, approximation FAILS

Worked example Example 4 — jab quadratic face karni padti hai (Cell D)

Problem: Ek hypothetical salt $MX$ ka $K_{s p} = 2.0 \times 1 0^{- 3}$ hai — deliberately bada choose kiya gaya hai taaki approximation zaroor break ho (real "slightly insoluble" salts jaise $PbCl_{2}$ , $K_{s p} \approx 1.6 \times 1 0^{- 5}$ , ya $CaSO_{4}$ , $K_{s p} \approx 2.4 \times 1 0^{- 5}$ , chhote hote hain, lekin same quadratic method tab bhi apply hoti hai jab $s^{'}$ nahi hota $≪ C$ ). $0.010 M$ $NaX$ mein $s^{'}$ nikalo.

Forecast: Yahan $K_{s p}$ common-ion concentration squared se sirf ~5× chhota hai. Kya $s^{'} ≪ 0.010$ phir bhi hold karega? Guess karo nahi — aur yahi is cell ka poora point hai.

Setup: $[M^{+}] = s^{'}$ , $[X^{-}] = 0.010 + s^{'}$ , isliye $s^{'} (0.010 + s^{'}) = 2.0 \times 1 0^{- 3}$ .
Naive attempt (trap expose karne ke liye): $s^{'}$ ko 0.010 ke side mein drop karo → $s^{'} = \frac{2.0 \times 1 0 ^{- 3}}{0.010} = 0.20 M$ . Yeh step kyun? Hum shortcut ko test kar rahe hain. Lekin $0.20 \neq ≪ 0.010$ — yeh 20× bada hai. Approximation invalid hai; hum ise use nahi kar sakte.
True quadratic solve karo. Expand karo: $s^{'2} + 0.010 s^{'} - 2.0 \times 1 0^{- 3} = 0$ . Yeh step kyun? Koi shortcut kaam nahi aata, isliye hum poora $(0.010 + s^{'})$ term rakhte hain aur $s^{'} = \frac{- b + b ^{2} - 4 a c}{2 a}$ use karte hain jahan $a = 1, b = 0.010, c = - 2.0 \times 1 0^{- 3}$ .
Compute karo: discriminant $= 0.01 0^{2} + 4 (2.0 \times 1 0^{- 3}) = 1.0 \times 1 0^{- 4} + 8.0 \times 1 0^{- 3} = 8.1 \times 1 0^{- 3}$ ; $= 0.09000$ . $s^{'} = \frac{- 0.010 + 0.09000}{2} = \frac{0.08000}{2} = 0.04000 mol/L .$

Verify karo: Plug back karo: $[X^{-}] = 0.010 + 0.040 = 0.050$ ; $Q = 0.040 \times 0.050 = 2.0 \times 1 0^{- 3} = K_{s p}$ ✓. Note karo $s^{'} = 0.040$ hai, nahi ki naive $0.20$ — shortcut 5× too big hoti. Lesson: hamesha $s^{'} ≪ C$ test karo; agar fail ho, quadratic solve karo.

Cell E — do solutions mix karo, kya precipitate hoga?

Worked example Example 5 — dilution-on-mixing (Cell E)

Problem: $100 mL$ of $0.010 M$ $AgNO_{3}$ ko $100 mL$ of $0.020 M$ $NaCl$ se mix karo. Kya AgCl ( $K_{s p} = 1.8 \times 1 0^{- 10}$ ) precipitate karega?

Forecast: Hume $Q$ ko $K_{s p}$ se compare karna hai. Classic slip hai mixing se pehle dilution bhool jaana. Predict karo: dono concentrations half hone ke baad, kya $Q$ ab bhi $1.8 \times 1 0^{- 10}$ se upar rahega?

Mixing ke baad total volume $= 100 + 100 = 200 mL = 0.200 L$ . Yeh step kyun? Concentration = moles ÷ volume; volume double ho gaya, isliye har concentration drop ho jaati hai.
Har ion ko dilute karo (moles conserved, new volume 0.200 L): $[\text{Cl}^-]=\frac{0.020\times0.100}{0.200}=1.0\times10^{-2}\ \text{M}.$$ *Yeh step kyun?* Sirf **diluted** concentrations actual mixed beaker ko describe karti hain.$
$Q$ compute karo: $Q = [Ag^{+}] [Cl^{-}] = (5.0 \times 1 0^{- 3}) (1.0 \times 1 0^{- 2}) = 5.0 \times 1 0^{- 5}$ . Yeh step kyun? $Q$ hamara "live" marker hai opening figure ki number line par; hum ise actual diluted concentrations se build karte hain taaki dekh sakein woh kis labelled band mein land karta hai.
Compare karo: $Q = 5.0 \times 1 0^{- 5}$ vs $K_{s p} = 1.8 \times 1 0^{- 10}$ . Kyunki $Q ≫ K_{s p}$ hai (karib $2.8 \times 1 0^{5}$ times bada), precipitation hoti hai. Yeh step kyun? Comparison hi decision rule hai: $Q$ supersaturated (right-hand) band mein deep hai, isliye solid crash out hona chahiye jab tak $Q$ wapas $K_{s p}$ par na aa jaaye.

Verify karo: Sanity — dono reactants AgCl ke $\sim 1 0^{- 5}$ saturation level se kaafi upar hain jo parent note mein milaa tha, isliye solid zaroor appear hona chahiye; $Q > K_{s p}$ agree karta hai ✓. $Q$ ki units: $(mol/L)^{2}$ ✓.

Cell F — degenerate / zero input

Worked example Example 6 — ek ion missing hai (Cell F)

Problem: Aapke paas $0.10 M$ $AgNO_{3}$ hai lekin zero chloride ( $[Cl^{-}] = 0$ ). Kya AgCl precipitate kar sakta hai? Kitna $[Cl^{-}]$ chahiye just barely precipitation start karne ke liye?

Forecast: Bilkul bhi chloride nahi hai, guess karo kya koi solid form ho sakta hai — aur nahi, toh kis chloride level par start hoga.

Current $Q$ : $Q = [Ag^{+}] [Cl^{-}] = (0.10) (0) = 0$ . Yeh step kyun? Kisi bhi product mein zero factor ho toh product zero hoga — yeh degenerate case hai.
Compare karo: $Q = 0 < K_{s p} = 1.8 \times 1 0^{- 10}$ . Kyunki $Q < K_{s p}$ hai, solution unsaturated hai: koi precipitate nahi, kabhi bhi, jab tak $[Cl^{-}] = 0$ rahe. Yeh step kyun? Precipitation ke liye $Q \geq K_{s p}$ chahiye; zero product ceiling tak nahi pahunch sakta.
Onset condition: precipitation shuru hoti hai jis instant $Q = K_{s p}$ hota hai: $[Cl^{-}]_{onset} = \frac{K _{s p}}{[ Ag ^{+} ]} = \frac{1.8 \times 1 0 ^{- 10}}{0.10} = 1.8 \times 1 0^{- 9} M .$ Yeh step kyun? Missing ion ke liye $Q = K_{s p}$ solve karo threshold find karne ke liye.

Verify karo: Onset par, $Q = (0.10) (1.8 \times 1 0^{- 9}) = 1.8 \times 1 0^{- 10} = K_{s p}$ ✓. Threshold ek amazing level par chhota hai — even a whiff of $Cl^{-}$ ek silver solution ko cloudy kar deta hai, yahi wajah hai ki Qualitative Inorganic Analysis mein $Cl^{-}$ test itna sensitive hota hai.

Cell G — selective precipitation

Worked example Example 7 — pehle kaun sa salt girega? (Cell G)

Problem: Ek solution mein $0.10 M$ $Cl^{-}$ aur $0.10 M$ $CrO_{4}^{2 -}$ hai. Aap dheere dheere $Ag^{+}$ add karte ho. Diya hai $K_{s p} (AgCl) = 1.8 \times 1 0^{- 10}$ aur $K_{s p} (Ag_{2} CrO_{4}) = 1.1 \times 1 0^{- 12}$ , pehle kaun precipitate karta hai, aur dono kitne alag hote hain?

Forecast: Dono $Ag^{+}$ ke liye compete karte hain. Jo sabse kam $[Ag^{+}]$ par apna $K_{s p}$ reach karta hai, woh pehle precipitate hota hai. Guess karo: AgCl ya $Ag_{2} CrO_{4}$ ?

AgCl ke liye silver chahiye: $[Ag^{+}]_{1} = \frac{K _{s p}}{[ Cl ^{-} ]} = \frac{1.8 \times 1 0 ^{- 10}}{0.10} = 1.8 \times 1 0^{- 9} M$ . Yeh step kyun? AgCl ka precipitation exactly tab shuru hota hai jab $Q = K_{s p}$ ; us equality ko $[Ag^{+}]$ ke liye solve karne se silver level milta hai jo chloride ke liye "ceiling fill" karta hai.
$Ag_{2} CrO_{4}$ ke liye silver chahiye: yahan $K_{s p} = [Ag^{+}]^{2} [CrO_{4}^{2 -}]$ hai, isliye $[Ag^{+}]_{2} = \frac{K _{s p}}{[ CrO _{4}^{2 -} ]} = \frac{1.1 \times 1 0 ^{- 12}}{0.10} = 1.1 \times 1 0^{- 11} = 3.32 \times 1 0^{- 6} M .$ Yeh step kyun? 2:1 salt ko $[Ag^{+}]$ squared chahiye, isliye hum square root lete hain — 1:1 case se alag algebra.
Thresholds compare karo: $1.8 \times 1 0^{- 9} < 3.32 \times 1 0^{- 6}$ , isliye AgCl pehle precipitate karta hai (~1800× lower silver par apni ceiling reach karta hai). Yeh step kyun? Jaise hum silver drip karte hain, $[Ag^{+}]$ chhote threshold ko pehle cross karta hai; jis bhi salt ka threshold pehle cross hota hai woh pehle precipitate hota hai.
Separation ki purity: woh $[Cl^{-}]$ nikalo jo tab bacha hoga jab chromate drop hone waala ho, yani jab $[Ag^{+}] = 3.32 \times 1 0^{- 6}$ : $[Cl^{-}]_{left} = \frac{K _{s p} ( AgCl )}{[ Ag ^{+} ]} = \frac{1.8 \times 1 0 ^{- 10}}{3.32 \times 1 0 ^{- 6}} = 5.4 \times 1 0^{- 5} M .$ Yeh step kyun? Yeh batata hai ki chromate shuru hone par kitna chloride abhi bhi solution mein hai — leftover fraction separation quality measure karta hai.

Verify karo: Chloride ka fraction jo abhi bhi dissolved hai $= \frac{5.4 \times 1 0 ^{- 5}}{0.10} = 5.4 \times 1 0^{- 4} = 0.054%$ . Toh chromate ka pehla particle girne se pehle >99.9% chloride already solid hai — ek excellent selective precipitation, yahi Mohr's titration ka principle hai ✓.

Neeche ki figure dono silver thresholds ko log axis par plot karti hai taaki aap ~1800× ka gap dekh sako jo separation ko clean banata hai — AgCl ki threshold line chromate ki line se kaafi left mein hai.

Picture padhein: horizontal axis added $[Ag^{+}]$ hai log scale par (daayein badhta hua). Do vertical threshold lines labelled hain — "AgCl needs $1.8 \times 1 0^{- 9}$ M" (left) aur " $Ag_{2} CrO_{4}$ needs $3.3 \times 1 0^{- 6}$ M" (right). Jaise aap silver add karte ho (move rightward), aap AgCl line kaafi pehle cross karte ho chromate line se. Dono ke beech ke wide gap mein, essentially sara chloride already precipitate ho chuka hota hai jabki chromate abhi bhi fully dissolved hai — woh gap separation window hai, aur uski width exactly wajah hai ki selective precipitation kaam karti hai. Jo lesson le jaana hai: jab do salts ek common ion share karte hain, separation ka order aur quality poori tarah se compare karne se decide hoti hai ki har ek ko apna $K_{s p}$ hit karne ke liye kitni ion concentration chahiye.

Cell H — limiting behaviour

Worked example Example 8 — ek number nahi, trend (Cell H)

Problem: AgCl ( $K_{s p} = 1.8 \times 1 0^{- 10}$ ) ke liye, solubility $s^{'} \approx K_{s p} / C$ kaise behave karti hai jaise common-ion concentration $C$ bahut chhote se bahut bade hoti jaati hai? $C = 1 0^{- 4}, 1 0^{- 2}, 1 M$ par evaluate karo aur limits describe karo.

Forecast: Shape predict karo: jaise $C$ 100× badhta hai, kya $s^{'}$ 100× shrink hota hai, ya kuch gentle?

Model: $s^{'} \approx \frac{K _{s p}}{C}$ (valid jab tak $s^{'} ≪ C$ ho).
Evaluate karo: $C=10^{-2}:\ s'=1.8\times10^{-8};\quad C=1:\ s'=1.8\times10^{-10}.$$ *Yeh step kyun?* Teen points law reveal karte hain: $C$ mein har 100× rise, $s'$ mein 100× fall deta hai — perfectly inverse.$
Small- $C$ limit: jaise $C \to 0$ , formula $s^{'} = K_{s p} / C$ diverge karta hai — lekin yeh tabhi valid hai jab $C ≫ s^{'}$ ho. Us point se neeche common ion dominate nahi karta aur hum pure-water case par wapas jaate hain, jahan $s \to K_{s p} = 1.34 \times 1 0^{- 5}$ . Interpretation: solubility infinity tak nahi jaati; yeh pure-water value par cap hoti hai. Yeh forecast ki implicit worry ka jawab hai — trend inverse tab hi hota hai valid range mein, phir yeh ek finite ceiling par flatten ho jaata hai. Yeh step kyun? Ek limit meaningless hai agar jo model ise produce karta hai woh already break ho chuka ho; honest limiting analysis batati hai kahan formula apply karna band kar deta hai.
Large- $C$ limit: jaise $C \to \infty$ , $s^{'} = K_{s p} / C \to 0$ . Interpretation: common ion se flood karna solubility ko asymptotically zero ki taraf le jaata hai — sabse strong suppression, aur yahi wajah hai ki common-ion washing precipitate ko almost quantitatively recover karne ke liye use hoti hai. Dono limits milke behaviour bracket karte hain: upar chhote $C$ par $K_{s p}$ se bounded, bade $C$ par $0$ ki taraf squeeze hota hua. Yeh step kyun? Dono endpoints name karna teen isolated numbers ko curve ki ek complete story banata hai.

Verify karo: Log–log axes par, $s^{'}$ vs $C$ valid range mein slope $- 1$ ki straight line hai (inverse proportionality) ✓; product $s^{'} \cdot C = K_{s p} = 1.8 \times 1 0^{- 10}$ teeno points par ✓; aur small- $C$ cap $1.8 \times 1 0^{- 10} = 1.34 \times 1 0^{- 5}$ AgCl ki pure-water solubility se match karta hai ✓.

Cell I — real-world word problem

Worked example Example 9 — fluoridated water (Cell I)

Problem: Tap water mein $[F^{-}] = 1.0 \times 1 0^{- 4} M$ (typical $\approx 1.9 ppm$ ) tak fluoride hoti hai. Ground water $[Ca^{2 +}] = 2.0 \times 1 0^{- 3} M$ carry karta hai. $CaF_{2}$ ke liye, $K_{s p} = 3.9 \times 1 0^{- 11}$ hai. Kya pipes mein $CaF_{2}$ scale form hoga?

Forecast: Padho → model karo → $Q$ compute karo. Guess karo: itne dilute fluoride ke saath, kya $Q$ $K_{s p}$ bar clear karta hai?

Ions mein translate karo. Salt $CaF_{2}$ hai, isliye ( $Q = [cation]^{x} [anion]^{y}$ yaad karte hue) $Q = [Ca^{2 +}] [F^{-}]^{2}$ — fluoride squared hai (do F per formula). Yeh step kyun? Word problem stoichiometry chhupa leta hai; coefficient 2 ko $Q$ mein exponent ki tarah wapas aana chahiye.
$Q$ compute karo: $Q = (2.0 \times 1 0^{- 3}) (1.0 \times 1 0^{- 4})^{2} = (2.0 \times 1 0^{- 3}) (1.0 \times 1 0^{- 8}) = 2.0 \times 1 0^{- 11} .$ Yeh step kyun? $Q$ woh live marker hai jo hum hamesha pehle build karte hain; sirf number milne ke baad hum ise fixed $K_{s p}$ ceiling se compare kar sakte hain.
Compare karo: $Q = 2.0 \times 1 0^{- 11}$ vs $K_{s p} = 3.9 \times 1 0^{- 11}$ . Kyunki $Q < K_{s p}$ hai, water unsaturated hai — koi scale form nahi hoga. Yeh step kyun? $Q < K_{s p}$ (opening figure ka left-hand, unsaturated band) matlab dissolve hone ki jagah hai, precipitate nahi.

Verify karo: $Q / K_{s p} = 2.0/3.9 = 0.51$ — ceiling ka karib aadha, safely neeche ✓. Real check: fluoridation levels exactly $CaF_{2}$ saturation ke neeche rehne ke liye set ki jaati hain, jo agree karta hai ✓.

Cell J — exam twist: pH anion ko control karta hai

Worked example Example 10 — anion ek base hai (Cell J)

Problem: $Mg(OH)_{2}$ ka $K_{s p} = 5.6 \times 1 0^{- 12}$ hai. Ek buffered solution $pH = 9.0$ hold karta hai. Kya $0.010 M$ $Mg^{2 +}$ se $Mg(OH)_{2}$ precipitate hoga?

Forecast: Twist yeh hai: $[OH^{-}]$ pH se fixed hoti hai, hydroxide salt add karne se nahi. Aapko pH and pOH Calculations chahiye hoga. Guess karo ki kya pH 9 itna enough hydroxide supply karta hai ki precipitate ho sake.

pH se $[OH^{-}]$ nikalo. $pOH = 14 - 9 = 5$ , isliye $[OH^{-}] = 1 0^{- 5} M$ . Yeh step kyun? Hydroxide yahan common/participating ion hai, aur iska supply acid–base equilibrium se set hota hai, dissolved salt se nahi — isliye pehle ise pH se extract karna hoga.
Correct power ke saath $Q$ likho. $Mg(OH)_{2} ⇌ Mg^{2 +} + 2 OH^{-}$ , isliye $Q = [Mg^{2 +}] [OH^{-}]^{2}$ . Yeh step kyun? Har formula unit mein do hydroxides hain yani $[OH^{-}]$ $Q$ mein squared enter karta hai; yeh wahi "coefficient becomes exponent" rule hai jaise har doosre cell mein.
Compute karo: $Q = (0.010) (1 0^{- 5})^{2} = (1 0^{- 2}) (1 0^{- 10}) = 1.0 \times 1 0^{- 12}$ . Yeh step kyun? Actual $[Mg^{2 +}]$ aur pH-derived $[OH^{-}]$ se live $Q$ build karo taaki $K_{s p}$ se compare kar sako.
Compare karo: $Q = 1.0 \times 1 0^{- 12} < K_{s p} = 5.6 \times 1 0^{- 12}$ . Kyunki $Q < K_{s p}$ hai, pH 9 par koi precipitate nahi. Yeh step kyun? $Q$ unsaturated (left-hand) band mein hai — pH itna low hai (itna basic nahi) ki $Q$ ceiling ke upar push ho sake.
Bonus — onset pH: precipitation tab shuru hoti hai jab $Q = K_{s p}$ : $[OH^{-}] = \frac{K _{s p}}{[ Mg ^{2 +} ]} = \frac{5.6 \times 1 0 ^{- 12}}{0.010} = 5.6 \times 1 0^{- 10} = 2.37 \times 1 0^{- 5} M,$ jisse $pOH = - lo g (2.37 \times 1 0^{- 5}) = 4.63$ milta hai, isliye $pH_{onset} = 14 - 4.63 = 9.37$ . Yeh step kyun? $Q = K_{s p}$ ko $[OH^{-}]$ ke liye solve karna aur wapas pH mein convert karna exact pH batata hai jahan unsaturated band khatam hota hai aur precipitation shuru hoti hai.

Verify karo: pH 9.37 par, $[OH^{-}] = 2.37 \times 1 0^{- 5}$ aur $Q = (0.010) (2.37 \times 1 0^{- 5})^{2} = 5.6 \times 1 0^{- 12} = K_{s p}$ ✓. Kyunki hamara pH 9.0 < 9.37 hai, koi solid nahi — step 4 ke saath consistent ✓. Yahi exactly woh tarika hai jisme Le Chatelier's Principle aur pH milke Qualitative Inorganic Analysis mein hydroxide precipitations ko control karte hain.

Retrieval

Recall Kaun sa comparison decide karta hai ki precipitate form hoga ya nahi?

Compare $Q$ (live ionic product) to $K_{s p}$ (fixed ceiling) ::: $Q > K_{s p}$ precipitates, $Q = K_{s p}$ saturated, $Q < K_{s p}$ unsaturated.

Recall 2:1 salt jaise

$Ag_{2} CrO_{4}$ ke liye, $K_{s p} = 4 s^{3}$ kyun hai aur $s^{3}$ kyun nahi? Kyunki $[Ag^{+}] = 2 s$ squared enter karta hai: $(2 s)^{2} \cdot s = 4 s^{3}$ ::: coefficient 2 ek baar concentration mein factor ki tarah kaam karta hai aur ek baar exponent ki tarah.

Recall Jab aap

$Q$ compute karne se pehle do solutions mix karte ho, toh pehle kya karna chahiye? Har concentration ko new total volume par dilute karo (moles conserved hain) ::: sirf tab $Q$ compute karo.

Recall Common-ion shortcut

$s^{'} \approx K_{s p} / C$ kab fail karta hai? Jab $s^{'}$ nahi hota $≪ C$ (typically jab $K_{s p}$ bada ho ya $C$ chhota ho) ::: tab full quadratic solve karo.

Recall Selective precipitation mein pehle kaun sa salt precipitate karta hai?

Woh salt jo sabse kam added common ion concentration par apna $K_{s p}$ reach karta hai ::: har threshold check karo aur chhota wala chuno.