Strong vs weak acids - bases; degree of dissociation α

2.6.11 · Chemistry › Equilibrium

Overview

Saare acids aur bases ek jaise nahi hote. Kuch completely ionize ho jaate hain paani mein (strong), jabki doosre sirf partially dissociate karte hain (weak). Degree of dissociation (α) is behavior ko quantify karta hai aur solution pH, buffer capacity, aur reaction dynamics determine karta hai.

Figure — Strong vs weak acids - bases; degree of dissociation α

Core Concepts

[!intuition] Kuch Acids "Stronger" Kyun Hote Hain

Ek acid ko ek company ki tarah socho jo hydrogen ions (H⁺) ko apne "employees" ki tarah pakad ke rakhti hai. Ek strong acid ek bura employer hai—jaise hi woh paani mein dissolve hota hai, saare H⁺ "employees" quit karke chale jaate hain (100% dissociation). Ek weak acid ek achha employer hai—zyaadatar H⁺ ions bonded rehte hain, sirf kuch hi kisi bhi waqt jaate hain.

Aisa kyun hota hai?

Strong acids: Conjugate base (A⁻) H⁺ kho dene ke baad bahut zyaada stable hota hai. Paani ke saath ion-dipole interactions strongly dissociated state ko favor karte hain. Reverse reaction (recombination) ki rate negligible hoti hai.
Weak acids: Conjugate base kam stable hoti hai ya acid form relatively stable hoti hai. Ek equilibrium form hota hai jahan HA aur H⁺ + A⁻ dono saath exist karte hain.

Physical basis: Bond polarity, conjugate base mein charge delocalization, aur solvation energy strength determine karte hain.

[!definition] Strong vs Weak Acids aur Bases

Strong Acid: Ek acid jo aqueous solution mein completely dissociate karta hai. $HA H_{2} O H^{+} + A^{-} (100% dissociation)$

Common strong acids: HCl, HBr, HI, HNO₃, H₂SO₄, HClO₄

Weak Acid: Ek acid jo solution mein partially dissociate karta hai, equilibrium establish karta hai. $HA ⇌ H^{+} + A^{-}$

Common weak acids: CH₃COOH, HF, H₂CO₃, H₃PO₄

Strong Base: Ek base jo solution mein completely dissociate karta hai. $BOH H_{2} O B^{+} + OH^{-}$

Common strong bases: NaOH, KOH, Ba(OH)₂, Ca(OH)₂

Weak Base: Ek base jo partially protons accept karta hai, equilibrium establish karta hai. $B + H_{2} O ⇌ BH^{+} + OH^{-}$

Common weak bases: NH₃, CH₃NH₂, pyridine

[!definition] Degree of Dissociation (α)

Degree of dissociation (α) woh fraction hai jo original acid ya base molecules ka equilibrium par ions mein dissociate hota hai.

$α = \frac{amount dissociated}{initial amount} = \frac{c - c _{eq}}{c}$

Jahaan:

$c$ = initial concentration
$c_{eq}$ = undissociated species ki equilibrium concentration

Alternative form (aksar zyaada useful): $α = \frac{[ H ^{+} ]}{c _{0}} (for acids)$ $α = \frac{[ OH ^{-} ]}{c _{0}} (for bases)$

Key insights:

Strong acids/bases ke liye: $α \approx 1$ (essentially complete dissociation)
Weak acids/bases ke liye: $α ≪ 1$ (moderate concentrations ke liye typically 0.01 se 0.1)
$α$ concentration par depend karta hai—kam concentrated solutions mein α zyaada hota hai (Le Chatelier's principle)

[!formula] α aur Ka ke Beech Relationship (Derivation)

Starting point: Ek weak acid HA consider karo jiska initial concentration $c$ hai.

Step 1: Dissociation equilibrium likho $HA ⇌ H^{+} + A^{-}$

Step 2: ICE table banao (Initial, Change, Equilibrium)

	HA	H⁺	A⁻
Initial	$c$	0	0
Change	$- c α$	$+ c α$	$+ c α$
Equilibrium	$c (1 - α)$	$c α$	$c α$

Yeh kyun kaam karta hai: Agar α dissociated fraction hai, toh $c α$ moles dissociate hote hain, aur $c (1 - α)$ undissociated bachte hain.

Step 3: Equilibrium constant expression likho $K_{a} = \frac{[ H ^{+} ] [ A ^{-} ]}{[ HA ]} = \frac{( c α ) ( c α )}{c ( 1 - α )} = \frac{c α ^{2}}{1 - α}$

Step 4: Weak acids ke liye jahan $α ≪ 1$ hai, hum approximate kar sakte hain $1 - α \approx 1$ : $K_{a} \approx c α^{2}$

α ke liye solve karo: $α \approx \frac{K _{a}}{c}$

α concentration ke saath kyun decrease karta hai? Le Chatelier's principle ke according, zyaada HA add karne se equilibrium left shift hota hai (undissociated form ki taraf). System zyaada concentration par complete dissociation ko "resist" karta hai.

Exact formula (koi approximation nahi): $α = \frac{- K _{a} + K _{a}^{2} + 4 c K _{a}}{2 c}$

Derivation: $K_{a} = \frac{c α ^{2}}{1 - α}$ se, multiply through karo: $K_{a} (1 - α) = c α^{2}$ $K_{a} - K_{a} α = c α^{2}$ $c α^{2} + K_{a} α - K_{a} = 0$

Quadratic formula use karo jahan $α$ variable hai: $α = \frac{- K _{a} \pm K _{a}^{2} + 4 c K _{a}}{2 c}$

Hum positive root lete hain kyunki α positive hona chahiye.

[!example] Worked Examples

Example 1: Acetic Acid ke liye α Nikalna

Problem: 0.1 M acetic acid (CH₃COOH) ka degree of dissociation calculate karo. Given: $K_{a} = 1.8 \times 1 0^{- 5}$ .

Solution:

Step 1: Check karo ki approximation valid hai ya nahi $\frac{c}{K _{a}} = \frac{0.1}{1.8 \times 1 0 ^{- 5}} = 5555 ≫ 100$

"Rule of 100" satisfy hota hai, toh hum approximation use kar sakte hain.

Yeh step kyun? Approximation $1 - α \approx 1$ tabhi valid hai jab α small ho (typically < 5%). Ratio $c / K_{a}$ batata hai ki dissociation minimal hogi ya nahi.

Step 2: Approximate formula apply karo $α \approx \frac{K _{a}}{c} = \frac{1.8 \times 1 0 ^{- 5}}{0.1} = 1.8 \times 1 0^{- 4} = 0.0134$

Step 3: Percentage mein convert karo $α = 1.34%$

Interpretation: Sirf 1.34% acetic acid molecules dissociate karte hain. Equilibrium par:

$[H^{+}] = c α = 0.1 \times 0.0134 = 1.34 \times 1 0^{- 3}$ M
pH = $- lo g (1.34 \times 1 0^{- 3}) = 2.87$

Yeh kyun important hai: Isse explain hota hai ki vinegar (5% acetic acid) HCl ki tarah tumhari skin kyun nahi jalaata—zyaadatar molecules undissociated rehte hain.

Example 2: α ki Concentration Dependence

Problem: Acetic acid ke liye 0.01 M aur 1.0 M par α calculate karo. Example 1 ke 0.1 M se compare karo.

Solution:

0.01 M ke liye: $α = \frac{1.8 \times 1 0 ^{- 5}}{0.01} = 1.8 \times 1 0^{- 3} = 0.0424 = 4.24%$

1.0 M ke liye: $α = \frac{1.8 \times 1 0 ^{- 5}}{1.0} = 1.8 \times 1 0^{- 5} = 0.00424 = 0.424%$

Summary table:

Concentration	α	[H⁺]
0.01 M	4.24%	$4.24 \times 1 0^{- 4}$ M
0.1 M	1.34%	$1.34 \times 1 0^{- 3}$ M
1.0 M	0.424%	$4.24 \times 1 0^{- 3}$ M

Aisa kyun hota hai?

Kam concentration → Zyaada dilution → Paani ke molecules ions ko better stabilize kar sakte hain → Equilibrium right shift → Higher α
Zyaada concentration → Ions zyaada frequently collide karte hain → Zyaada recombination → Lower α

Important: Bhale hi α decrease karta hai, [H⁺] concentration ke saath badhta hi rehta hai (sirf linearly nahi).

Example 3: Strong vs Weak Acid Comparison

Problem: 0.1 M HCl (strong) aur 0.1 M acetic acid (weak) ko compare karo.

HCl (strong acid):

$α = 1$ (complete dissociation)
$[H^{+}] = 0.1$ M
pH = 1.0

CH₃COOH (weak acid):

$α = 0.0134$
$[H^{+}] = 1.34 \times 1 0^{- 3}$ M
pH = 2.87

Key insight: Same concentration par, strong acid 75 times zyaada H⁺ ions produce karta hai acetic acid se. Iseelie HCl kaafi zyaada corrosive aur dangerous hai.

[!mistake] Common Errors

Mistake 1: α aur Ka ko Confuse Karna

Galat soch: "Zyaada Ka matlab zyaada α hai, toh sirf Ka values compare karo α compare karne ke liye."

Yeh sahi kyun lagta hai: Ka actually acid strength se related hai, aur stronger acids mein higher α hota hi hai.

Fix: α dono Ka aur concentration par depend karta hai: $α = \frac{K _{a}}{c}$

Do acids ka same Ka ho sakta hai lekin alag α agar concentrations alag hoon. Ya same α ho sakta hai lekin alag Ka.

Example:

0.1 M acetic acid ( $K_{a} = 1.8 \times 1 0^{- 5}$ ): α = 1.34%
0.01 M formic acid ( $K_{a} = 1.8 \times 1 0^{- 4}$ ): α = $\frac{1.8 \times 1 0 ^{- 4}}{0.01}$ = 13.4%

Formic ek stronger acid hai (higher Ka), lekin sahi concentrations par, dono arbitrary α values le sakte hain.

Mistake 2: Weak Acids ko Strong ki Tarah Treat Karna

Galat soch: "CH₃COOH ka formula CH₃COOH hai, toh woh 1 H⁺ per molecule release karta hai. 0.1 M solution ke liye, [H⁺] = 0.1 M."

Yeh sahi kyun lagta hai: Stoichiometry kehti hai ek H⁺ per molecule, aur humne early chemistry mein stoichiometry use karna seekha tha.

Fix: Equilibrium stoichiometry ko override karta hai. Sirf α fraction dissociate karta hai: $[H^{+}] = c \cdot α, not c$

0.1 M acetic acid ke liye, [H⁺] = 0.00134 M hai, na ki 0.1 M.

Is mistake ka consequence: Tum pH = 1 calculate karte jabki sahi pH = 2.87 hai—ek badi mistake jo tumhe lagwati ki vinegar utna hi acidic hai jitna stomach acid.

Mistake 3: Water Autoionization ko Ignore Karna

Galat soch: "Bahut weak acid ya bahut dilute solution ke liye, main [H⁺] sirf Ka se calculate kar sakta hoon."

Fix: Jab acid dissociation se calculated [H⁺] $1 0^{- 7}$ M ke kareeb aaye, tab tumhe water ki autoionization zaroor account karni chahiye: $[H^{+}]_{total} = [H^{+}]_{acid} + 1 0^{- 7}$

Yeh kab matter karta hai?

Bahut weak acids ( $K_{a} < 1 0^{- 7}$ )
Bahut dilute solutions (< $1 0^{- 6}$ M)

Example: 0.001 M acetic acid se sirf acid ke basis par calculated [H⁺] ≈ $4.2 \times 1 0^{- 7}$ M milta hai, lekin paani $1 0^{- 7}$ M contribute karta hai, toh actual [H⁺] ≈ $5.2 \times 1 0^{- 7}$ M hoga. Paani ko ignore karne par pH = 6.37 aata hai jabki sahi pH = 6.28 hai.

[!recall]- Ek 12-Saal ke Bacche ko Explain Karo

Socho tumhare paas ek box Legos hain jo saath mein jude hue hain. Kuch Lego sets super glue se jude hote hain (weak acids)—jab tum unhe paani mein daalte ho, toh sirf kuch pieces alag hote hain. Doosre sets sirf loosely snap kiye hue hote hain (strong acids)—jaise hi paani unhe touch karta hai, SAARE pieces turant alag ho jaate hain.

"Degree of dissociation" (α) aise poochh raha hai: "Mere kitne percent Lego pieces alag hue?" Super-glued sets ke liye, shayad sirf 1-5% alag hote hain. Loose sets ke liye, 100% alag ho jaate hain.

Yahan cool part hai: agar tumhare paas paani mein ZYAADA Lego sets hain, toh jo pieces alag hote hain unke apne partners se takraane ki zyaada chance hoti hai aur woh wapas snap ho jaate hain. Toh Legos ka bada dher → kam percentage alag hona (bhale hi loose pieces ki total number abhi bhi zyaada ho).

Strong acids matlab Legos WITHOUT glue. Weak acids matlab Legos with varying amounts of glue. Ka number batata hai ki kitna glue hai—chhota Ka matlab strong glue, toh kam pieces alag hote hain.

[!mnemonic] Memory Aids

Strong Acids: "Hary Clarke Brought Ice Nightly, So Please Clap Often"

HCl, Bromide (HBr), Iodide (HI), Nitric (HNO₃), Sulfuric (H₂SO₄), Percloric (*HClO₄)

α Formula: "Square root isliye kyunki equilibrium expression mein squared term hai"

$K_{a} = c α^{2} / (1 - α)$ → jab solve karo, α mein square root aata hai

Concentration Effect: "Dilute → Dissociates zyaada" (Kam c par Higher α)

Connections

Acid-base equilibria and Ka, Kb—α directly Ka se calculate hota hai
pH and pOH calculations—α determine karta hai [H⁺], jo pH determine karta hai
Buffer solutions—buffers isliye kaam karte hain kyunki weak acids mein α < 1 hota hai
Common ion effect—A⁻ add karne se HA ka α suppress hota hai
Ostwald's dilution law— $α = K_{a} / c$ relationship ko formalize karta hai
Polyprotic acids—har dissociation step ke liye multiple α values
Hydrolysis of salts—related concept jab ions acids/bases ki tarah act karte hain

Key Takeaways

Strong acids/bases completely dissociate hote hain (α ≈ 1); weak acids/bases partially dissociate karte hain (α ≪ 1)
==Degree of dissociation α = (amount dissociated)/(initial amount)==
Weak acids ke liye: $α \approx K_{a} / c$ (jab $c / K_{a} > 100$ )
α concentration badhne ke saath decrease karta hai Le Chatelier's principle ki wajah se
Kabhi complete dissociation assume mat karo jab tak acid/base strong na ho
Relationship $K_{a} = c α^{2} / (1 - α)$ thermodynamics (Ka) ko kinetics/extent (α) se connect karta hai

#flashcards/chemistry

Degree of dissociation (α) kya hota hai? :: Woh fraction jo acid ya base molecules ka equilibrium par ions mein dissociate hota hai. Formula: α = (amount dissociated)/(initial amount).

Ek weak acid ke liye, α ka Ka aur concentration se kya relation hai?

α ≈ √(Ka/c) jab acid sufficiently weak ho (c/Ka > 100). Exact: α = [−Ka + √(Ka² + 4cKa)]/(2c).

Concentration badhne par degree of dissociation α kyun decrease karta hai?

Le Chatelier's principle: zyaada concentration equilibrium ko undissociated form ki taraf shift karta hai. Zyaada frequent ion collisions se zyaada recombination hoti hai.

Teen strong acids ke naam batao.

HCl, HNO₃, H₂SO₄ (yeh bhi accept: HBr, HI, HClO₄).

Strong acid ke liye degree of dissociation kya hoti hai?

α ≈ 1 (essentially 100% dissociation).

Agar Ka = 1.8×10⁻⁵ aur c = 0.1 M ho, toh α calculate karo.

α ≈ √(1.8×10⁻⁵/0.1) = √(1.8×10⁻⁴) ≈ 0.0134 ya 1.34%.

Weak acids ko calculations mein strong acids ki tarah kyun treat nahi kar sakte?

Weak acids sirf partially dissociate karte hain (α ≪ 1), isliye [H⁺] = c·α hota hai, na ki c. Complete dissociation assume karne par drastically galat pH values milti hain.

Weak acid solution ko dilute karne par α kaise change hota hai?

Dilution par α increase hota hai. Kam concentration equilibrium ko right shift karta hai (zyaada dissociation) Le Chatelier's principle ke according.

Kaunsi approximation α ≈ √(Ka/c) allow karti hai?

Approximation 1 − α ≈ 1, jo valid hoti hai jab α ≪ 1 ho (typically jab c/Ka > 100).

pH calculations mein water autoionization kab account karni chahiye?

Jab acid dissociation se [H⁺] 10⁻⁷ M ke kareeb ho (bahut weak acids ya bahut dilute solutions mein).

First principles se Ka = cα²/(1−α) relationship derive karo.

HA ⇌ H⁺ + A⁻ se shuru karo. Equilibrium par: [HA] = c(1−α), [H⁺] = [A⁻] = cα. Ka = [H⁺][A⁻]/[HA] = (cα)(cα)/[c(1−α)] = cα²/(1−α).

Agar do acids ka same Ka ho lekin alag concentrations hoon, toh unke α values ke baare mein kya keh sakte ho? :: Zyaada dilute solution mein higher α hoga, kyunki α = √(Ka/c) aur α inversely related hai √c se.