Solubility product Ksp — common-ion suppression, selective precipitation

2.6.15 · Chemistry › Equilibrium

Overview

Yeh note explore karta hai ki solubility equilibria ko $K_{s p}$ ke through kaise quantify kiya jaata hai, aur solution mein already present ions ko add karne se (common ions) solubility dramatically kaise suppress ho jaati hai. Hum $K_{s p}$ ko equilibrium principles se derive karenge, predict karenge ki precipitates kab bante hain, aur selective precipitation schemes design karenge taaki mixed ions ko separate kiya ja sake—yeh qualitative analysis aur industrial purification mein ek cornerstone technique hai.

Core Concepts

Formula Equilibrium Principles Se

$K_{s p}$ Derive Karna Step 1: Dissolution equilibrium likho. $M_{x} A_{y} (s) ⟷ x M^{n +} (a q) + y A^{m -} (a q)$

Step 2: Equilibrium constant expression apply karo. Kisi bhi equilibrium ke liye, $K_{c} = \frac{[ products ]}{[ reactants ]}$ jahan har species apne stoichiometric coefficient tak raise ki gayi ho. $K_{c} = \frac{[ M ^{n +} ] ^{x} [ A ^{m -} ] ^{y}}{[ M _{x} A _{y} ( s )]}$

Step 3: Pure solids ki unit activity hoti hai. Thermodynamics mein, pure solid ki activity 1 define ki jaati hai (uski concentration change nahi hoti—woh apni standard state mein hai). Toh: $K_{c} = [M^{n +}]^{x} [A^{m -}]^{y} \equiv K_{s p}$

Yeh KYUN matter karta hai? Yeh derivation dikhata hai ki $K_{s p}$ arbitrary nahi hai—yeh dissolution process ka equilibrium constant hai. Koi bhi ion concentrations jo is equality ko satisfy karti hain, woh ek saturated solution represent karti hain.

Solubility vs. Solubility Product

Common-Ion Effect

Worked example Example 3: AgCl ke Saath Common-Ion Effect

Problem: 0.10 M NaCl solution mein AgCl ki solubility calculate karo. ( $K_{s p} = 1.8 \times 1 0^{- 10}$ )

Solution:

Equilibrium: $A g C l (s) ⇌ A g^{+} + C l^{-}$
Initial conditions: NaCl se $[Cl^{-}] = 0.10$ M (fully dissociated), $[Ag^{+}] = 0$ .
Equilibrium pe: $[Ag^{+}] = s^{'}$ , $[Cl^{-}] = 0.10 + s^{'}$ .
Substitute karo: $K_{s p} = s^{'} (0.10 + s^{'})$
Assume $s^{'} ≪ 0.10$ : Toh $0.10 + s^{'} \approx 0.10$ . $1.8 \times 1 0^{- 10} = s^{'} (0.10)$ $s^{'} = \frac{1.8 \times 1 0 ^{- 10}}{0.10} = 1.8 \times 1 0^{- 9} mol/L$
Assumption check karo: $1.8 \times 1 0^{- 9} ≪ 0.10$ ✓ Yeh step KYUN? Approximation $0.10 + s^{'} \approx 0.10$ valid hai kyunki AgCl itna kam dissolve hota hai ki woh $[Cl^{-}]$ ko appreciably change nahi karta. Yeh algebra simplify karta hai aur standard practice hai.

Compare karo: Pure water mein, $s = 1.34 \times 1 0^{- 5}$ mol/L. Yahan, $s^{'} = 1.8 \times 1 0^{- 9}$ mol/L—7500 guna kam soluble!

Predicting Precipitation: The Ionic Product Q

Selective Precipitation

Worked example Example 6: Ag⁺ aur Pb²⁺ ko Chloride Se Separate Karna

Problem: Ek solution mein 0.010 M Ag⁺ aur 0.010 M Pb²⁺ hai. Kya tum slowly Cl⁻ add karke inhe separate kar sakte ho? $K_{s p} (AgCl) = 1.8 \times 1 0^{- 10}$ , $K_{s p} (PbCl_{2}) = 1.7 \times 1 0^{- 5}$ .

Solution:

$[Cl^{-}]$ find karo jo har ek ko precipitate karna shuru kare.

AgCl ke liye: $K_{s p} = [Ag^{+}] [Cl^{-}]$ $[Cl^{-}]_{AgCl} = \frac{K _{s p}}{[ Ag ^{+} ]} = \frac{1.8 \times 1 0 ^{- 10}}{0.010} = 1.8 \times 1 0^{- 8} M$
PbCl₂ ke liye: $K_{s p} = [Pb^{2 +}] [Cl^{-}]^{2}$ $[Cl^{-}]_{PbCl_{2}} = \frac{K _{s p}}{[ Pb ^{2 +} ]} = \frac{1.7 \times 1 0 ^{- 5}}{0.010} = 1.7 \times 1 0^{- 3} = 0.041 M$

Compare karo: AgCl $1.8 \times 1 0^{- 8}$ M Cl⁻ pe precipitate hota hai, PbCl₂ 0.041 M pe.
Strategy: Slowly Cl⁻ add karo. Jab $[Cl^{-}] = 1.8 \times 1 0^{- 8}$ M ho, AgCl precipitate hona shuru ho jaayega. Tab tak add karo jab tak $[Cl^{-}]$ 0.041 M ke paas pahunche (lekin neeche rakho). AgCl essentially completely precipitate ho jaata hai; Pb²⁺ dissolved rehta hai.
AgCl filter off karo. Phir PbCl₂ precipitate karne ke liye 0.041 M se zyada Cl⁻ add karo. Yeh step KYUN? Hum har salt ke liye threshold $[Cl^{-}]$ calculate karte hain. Bahut bada gap (~2 million ka factor) ka matlab hai ki hum process control kar sakte hain. Practice mein, hum monitor karte hain aur Cl⁻ add karna tab band karte hain jab $[Ag^{+}]$ negligible levels tak drop ho jaaye (maano, $1 0^{- 5}$ M), yeh ensure karte hue ki PbCl₂ abhi precipitate na ho.

Quantitative check: Jab $[Cl^{-}] = 0.041$ M ho (PbCl₂ precipitate hone se just pehle), bacha hua $[Ag^{+}]$ kya hai? $[Ag^{+}] = \frac{K _{sp}}{[ Cl ^{-} ]} = \frac{1.8 \times 1 0 ^{- 10}}{0.041} = 4.4 \times 1 0^{- 9} M$ Essentially saara Ag⁺ (originally 0.010 M) remove ho gaya. Separation achieved!

Worked example Example 7: Sulfides ka Fractional Precipitation

Problem: Ek solution mein 0.10 M Cu²⁺ aur 0.10 M Mn²⁺ hai. pH adjust karke CuS precipitate karo lekin MnS nahi, $[S^{2 -}]$ ko pH ke through control karke. $K_{s p} (CuS) = 6 \times 1 0^{- 37}$ , $K_{s p} (MnS) = 3 \times 1 0^{- 14}$ .

Solution:

$[S^{2 -}]$ find karo jo zaruri hai:

CuS: $K_{s p} = [Cu^{2 +}] [S^{2 -}]$ $[S^{2 -}]_{CuS} = \frac{6 \times 1 0 ^{- 37}}{0.10} = 6 \times 1 0^{- 36} M$
MnS: $[S^{2 -}]_{MnS} = \frac{3 \times 1 0 ^{- 14}}{0.10} = 3 \times 1 0^{- 13} M$

Range: $6 \times 1 0^{- 36} < [S^{2 -}] < 3 \times 1 0^{- 13}$ M rakho.
Mechanism: $[\ce S^{2 -}]$ ko $\ce H 2 S$ solution mein pH se control kiya jaata hai ( $\ce H 2 S <=> 2 H + + S^{2 -}$ , low pH pe suppress hota hai). Acidity adjust karke, hum $[\ce S^{2 -}]$ ko tune kar sakte hain taaki selectively CuS precipitate ho.
Conclusion: CuS bahut low $[\ce S^{2 -}]$ pe precipitate hota hai, jabki MnS ko $1 0^{23}$ guna zyada chahiye. Aasaan separation.

Yeh step KYUN? $K_{s p}$ ka colossal difference (23 orders of magnitude!) ka matlab hai ki CuS vastly less soluble hai. Hum isse exploit karte hain precipitating ion ki concentration ko pH ke through control karke. Yeh ek classic qualitative analysis technique hai (Group II cations).

Common Mistakes

Advanced: Temperature Dependence aur Dissolution Thermodynamics

Mnemonic & Recall

Recall 12-Saal Ke Bacche Ko Feynman Explanation

Imagine karo tum lemonade bana rahe ho. Tum sugar dalte rehte ho jab tak aur dissolve na ho—yeh "saturated" hai. Kitna dissolve hota hai woh solubility ki tarah hai. Ab, kya hoga agar tumhare lemonade mein pehle se bahut saari sugar ho kisi aur cheez se (jaise tumne pre-sweetened water add kiya)? Agar tum aur sugar daloge, bahut kam dissolve hogi kyunki paani mein pehle se sugar particles ki "bheed" hai. Yahi common-ion effect hai!

Aur maano tumhare paas do tarah ki sugar hai: ek jo bahut aasaani se dissolve hoti hai (jaise regular sugar) aur ek jo mushkil se dissolve hoti hai (jaise rock candy). Agar tum slowly paani add karo, toh rock candy pehle "gir jaayegi" (precipitate ho jaayegi) kyunki woh apni limit sooner hit karti hai. Yeh control karke ki tum kitna paani add karte ho, tum dono sugars ko separate kar sakte ho. Yahi selective precipitation hai—pehle kam-soluble cheez ko bahar nikalna!

$K_{s p}$ number har tarah ki sugar ki "crowding limit" ki tarah hai. Chhota $K_{s p}$ matlab limit hit karne aur precipitate hone mein zyada nahi lagta. Bada $K_{s p}$ matlab bahut kuch dissolve ho sakta hai limit hit hone se pehle.

Connections

Le Chatelier's Principle – common ion add hone pe equilibrium shift explain karta hai
Chemical Equilibrium andrium Constant – $K_{s p}$ , $K_{c}$ ka ek special case hai
Ionic Equilibria in Solutions – acids, bases, aur buffers bhi equilibrium expressions involve karte hain
Qualitative Inorganic Analysis – selective precipitation cation groups separate karne ka basis hai
Gibs Free Energy and Spontaneity – $Δ G^{\circ} = - R T ln K_{s p}$ thermodynamics ko solubility se jodta hai
pH and pOH Calculations – pH ke through $[\ce S^{2 -}]$ ya $[OH^{-}]$ control karna selective precipitation enable karta hai
Stoichiometry and Solution Concentration – dilution aur mole calculations mixing problems ki neenv hain

Flashcards

#flashcards/chemistry

$M_{x} A_{y} (s) ⟷ x M^{n +} + y A^{m -}$ ke liye solubility product constant $K_{s p}$ kya hai? :: $K_{s p} = [M^{n +}]^{x} [A^{m -}]^{y}$ . Yeh dissolution ka equilibrium constant hai; pure solid appear nahi hota kyunki uski activity 1 hai.

Common ion add karne se solubility suppress kyun hoti hai?

Le Chatelier: product ion badhane se equilibrium reactants (solid) ki taraf shift hota hai. Mathematically,

K_{s p}

constant rakhne ke liye, agar ek ion ka

[\cdot]

badhta hai, doosre ko girna padta hai—kam dissolve hota hai.

0.10 M NaCl mein

AgCl

ki solubility

s^{'}

estimate kaise karo agar

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

Assume karo

[Cl^{-}] \approx 0.10

(common ion dominate karta hai). Toh

K_{s p} = s^{'} \cdot 0.10 ⟹ s^{'} = K_{s p} /0.10 = 1.8 \times 1 0^{- 9}

M. Baad mein check karo

s^{'} ≪ 0.10

Ionic product Q kya hai aur yeh precipitation kaise predict karta hai?

Q current ion concentrations ka product hai (stoichiometric powers tak raised), equilibrium pe nahi. Agar

Q > K_{s p}

, precipitation hoti hai; agar

Q < K_{s p}

, aur dissolve ho sakta hai; agar

Q = K_{s p}

, saturated equilibrium hai.

Solutions mix karna: 50 mL of 0.002 M Ag⁺ + 50 mL of 0.004 M Cl⁻. Mix karne ke baad concentrations kya hain?

Total volume 100 mL.

[Ag^{+}] = 0.002 \times (50/100) = 0.001

[Cl^{-}] = 0.004 \times (50/100) = 0.002

M. Hamesha dilution account karo.

CaF_{2}

ke liye

K_{s p} = 3.9 \times 1 0^{- 11}

ke saath, pure water mein solubility kya hai?

C a F_{2} ⇌ C a^{2 +} + 2 F^{-}

. Agar solubility

= s

, toh

[Ca^{2 +}] = s

[F^{-}] = 2 s

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

. Solve karo:

s = (K_{s p} /4)^{1/3} = (9.75 \times 1 0^{- 12})^{1/3} \approx 2.14 \times 1 0^{- 4}

M (toh

[F^{-}] = 2 s \approx 4.3 \times 1 0^{- 4}

M).