Buffer solutions — Henderson-Hasselbalch equation

2.6.14 · Chemistry › Equilibrium

Henderson-Hasselbalch equation kya hai?

Henderson-Hasselbalch equation buffer ke pH ko conjugate base aur weak acid ke ratio se relate karti hai:

Ye form kyun? Kyunki jab ratio $[A^{-}] / [HA] = 1$ hoti hai, tab pH bilkul pKₐ ke barabar hoti hai. Log term batata hai ki aap pKₐ se kitne pH units shift karoge ratio ke basis par.

First principles se derivation

Chaliye Henderson-Hasselbalch equation ko acid dissociation equilibrium se derive karte hain.

Step 1: Weak acid ke liye dissociation equilibrium likhte hain

$HA_{(a q)} ⇌ H_{(a q)}^{+} + A_{(a q)}^{-}$

Kyun? Ek weak acid paani mein partially dissociate hota hai. Equilibrium par, forward aur reverse rates balance ho jaati hain.

Step 2: Equilibrium constant expression likhte hain

$K_{a} = \frac{[ H ^{+} ] [ A ^{-} ]}{[ HA ]}$

Kyun? Equilibrium constant equilibrium par concentrations ko relate karta hai. Products over reactants, coefficients as exponents ke saath (yahan sab 1 hain).

Step 3: [H⁺] ke liye solve karte hain

$[H^{+}] = K_{a} \cdot \frac{[ HA ]}{[ A ^{-} ]}$

Ye step kyun? Hum [H⁺] ko acid aur base forms ke ratio se connect karna chahte hain, kyunki pH = -log[H⁺].

Step 4: Dono sides ka negative logarithm lete hain

$- lo g_{10} [H^{+}] = - lo g_{10} (K_{a} \cdot \frac{[ HA ]}{[ A ^{-} ]})$

Kyun? pH ko $- lo g_{10} [H^{+}]$ se define kiya jaata hai, aur pKₐ $- lo g_{10} (K_{a})$ hai. Logarithms multiplicative relationships ko additive mein convert karne dete hain.

Step 5: Logarithm properties apply karte hain

$pH = - lo g_{10} (K_{a}) - lo g_{10} (\frac{[ HA ]}{[ A ^{-} ]})$

Kyun? Kyunki $lo g (x y) = lo g (x) + lo g (y)$ , isliye $- lo g (K_{a} \cdot X) = - lo g (K_{a}) - lo g (X)$ .

Step 6: pKₐ ko recognize karke simplify karte hain

$pH = p K_{a} - lo g_{10} (\frac{[ HA ]}{[ A ^{-} ]})$

Kyun? $p K_{a} = - lo g_{10} (K_{a})$ define karo.

Step 7: Log ke andar fraction flip karte hain

$pH = p K_{a} + lo g_{10} (\frac{[ A ^{-} ]}{[ HA ]})$

Ye final step kyun? Kyunki $lo g (1/ x) = - lo g (x)$ , isliye fraction $[HA] / [A^{-}]$ ko $[A^{-}] / [HA]$ mein flip karne se minus, plus mein badal jaata hai. Ye standard form hai: Henderson-Hasselbalch equation.

Figure — Buffer solutions — Henderson-Hasselbalch equation

Ye equation kab valid hai?

Henderson-Hasselbalch equation ek approximation hai jo tab best kaam karti hai jab:

Acid weak ho ( $K_{a} < 1 0^{- 3}$ ), toh equilibrium concentrations ≈ initial concentrations
[HA] aur [A⁻] dono [H⁺] se bahut zyada ho — buffer components dissociation ya paani ke saath reaction se significantly deplete na hon
Ratio [A⁻]/[HA] 0.1 aur 10 ke beech ho — is range se bahar, buffer capacity poor hoti hai aur activity coefficient effects significant ho jaate hain

Hume ye conditions kyun chahiye? Kyunki humne assume kiya tha ki equilibrium concentrations analytical (initial) concentrations ke barabar hain aur paani ke autoionization se H⁺ ko ignore kiya tha. Dilute ya bahut weak/strong systems ke liye, ye assumptions toot jaate hain.

Worked example Example 1: Acetate buffer pH calculation

Aapne 0.10 M acetic acid (CH₃COOH) aur 0.15 M sodium acetate (CH₃COO⁻) se ek buffer prepare kiya. Acetic acid ka pKₐ 4.76 hai. pH kya hai?

Solution: Step 1: Components identify karo

Weak acid: HA = CH₃COOH, [HA] = 0.10 M
Conjugate base: A⁻ = CH₃COO⁻, [A⁻] = 0.15 M
pKₐ = 4.76

Kyun? Sodium acetate fully dissociate hota hai, acetate ion provide karta hai.

Step 2: Henderson-Hasselbalch apply karo $pH = 4.76 + lo g_{10} (\frac{0.15}{0.10})$ Ye step kyun? Values directly equation mein plug karo.

Step 3: Log term calculate karo $\frac{0.15}{0.10} = 1.5$ $lo g_{10} (1.5) = 0.176$ Kyun? Ratio batata hai ki base zyada hai acid se, isliye pH > pKₐ. Step 4: Final answer $pH = 4.76 + 0.176 = 4.94$

Buffer ka pH 4.94 hai, pKₐ ke relative thoda basic kyunki conjugate base zyada hai.

Worked example Example 2: Target pH ke liye buffer design karna

Aapko pH 9.00 par ek buffer chahiye. Aapke paas ammonia (NH₃, pKₐ = 9.25 for NH₄⁺) aur ammonium chloride (NH₄Cl) hai. Aapko [NH₃]/[NH₄⁺] ka kya ratio chahiye?

Solution:

Step 1: Conjugate pair identify karo

Weak base: NH₃ (lekin hum uske conjugate acid ka pKₐ use karte hain)
Conjugate acid: NH₄⁺ (buffer equilibrium mein weak acid)
Acid form ke liye: HA = NH₄⁺, A⁻ = NH₃
pKₐ(NH₄⁺) = 9.25 Kyun? Henderson-Hasselbalch acid/base pairs ke liye likhi jaati hai. NH₄⁺ acid hai, NH₃ base hai.

Step 2: Equation likhte hain $9.00 = 9.25 + lo g_{10} (\frac{[ NH _{3} ]}{[ NH _{4}^{+} ]})$

Kyun? Target pH = 9.00, hum ratio ke liye solve kar rahe hain.

Step 3: Log term isolate karo $lo g_{10} (\frac{[ NH _{3} ]}{[ NH _{4}^{+} ]}) = 9.00 - 9.25 = - 0.25$

Ye step kyun? Dono sides se pKₐ subtract karo.

Step 4: Ratio ke liye solve karo $\frac{[ NH _{3} ]}{[ NH _{4}^{+} ]} = 1 0^{- 0.25} = 0.562$

Kyun? Logarithm ko undo karne ke liye antilog (10 ki power) lo.

Answer: Aapko NH₃ aur NH₄⁺ ka ratio approximately 0.56:1 chahiye, ya roughly 1 part ammonia aur 1.8 parts ammonium chloride. Kyunki pH < pKₐ, hume acidic form zyada chahiye.

Worked example Example 3: Added acid par buffer response

Ek buffer mein 0.50 M formic acid (HCOH, pKₐ = 3.75) aur 0.50 M formate ion (HCOO⁻) hai. Is buffer ke 1.0 L mein 0.02 mol HCl add karne se pehle aur baad ka pH calculate karo.

Solution:

Initial pH: $pH = 3.75 + lo g_{10} (\frac{0.50}{0.50}) = 3.75 + 0 = 3.75$ Kyun? Jab ratio 1:1 hoti hai, pH = pKₐ exactly.

HCl add karne ke baad: Step 1: Chemically kya hota hai ye determine karo HCl ek strong acid hai, fully dissociate hota hai: HCl → H⁺ + Cl⁻

Added H⁺ conjugate base ke saath react karta hai: $H^{+} + HCOO^{-} \to HCOH$

Kyun? Buffer added H⁺ ko "pakadta" hai base ko acid mein convert karke.

Step 2: Nayi concentrations calculate karo

[HCOO⁻] 0.02 M se decrease hoti hai: 0.50 - 0.02 = 0.48 M
[HCOOH] 0.02 M se increase hoti hai: 0.50 + 0.02 = 0.52 M

Ye step kyun? Stoichiometry: 1 mol H⁺, 1 mol HCOO⁻ consume karta hai aur 1 mol HCOOH produce karta hai.

Step 3: Nayi values ke saath Henderson-Hasselbalch apply karo $pH = 3.75 + lo g_{10} (\frac{0.48}{0.52})$ $= 3.75 + lo g_{10} (0.923)$ $= 3.75 + (- 0.035)$ $= 3.72$

Answer: pH 3.75 se 3.72 tak girti hai, sirf 0.03 pH units ka change. Buffer ke bina, paani mein 0.02 M HCl add karne se pH ≈ 1.7 hota. Buffer ne acid ke effect ka zyaadatar hissa absorb kar liya.

Buffer capacity aur range

Buffer capacity (β) uss amount ki strong acid ya base hai jo ek buffer pH significantly change hone se pehle neutralize kar sakta hai (typically ±1 pH unit ke roop mein define kiya jaata hai).

Maximum capacity tab hoti hai jab [A⁻] = [HA], yaani jab pH = pKₐ. Kyun? Is point par, dono forms ke equal "reserves" hote hain jo added acid ya base ko neutralize kar sakein.

Effective buffer range: pKₐ ± 1

Ye range kyun? Jab pH = pKₐ ± 1, ratio [A⁻]/[HA] 0.1 se 10 ke beech hoti hai:

pH = pKₐ + 1 par: $lo g_{10} (ratio) = 1$ , isliye ratio = 10
pH = pKₐ - 1 par: $lo g_{10} (ratio) = - 1$ , isliye ratio = 0.1

Is range se bahar, ek component itna dilute ho jaata hai ki added H⁺ ya OH⁻ ko effectively neutralize nahi kar paata.

Common mistake Common mistake: Dilution ke baad concentrations ki jagah initial moles use karna

Galat idea: "Main 500 mL mein 0.10 mol HA aur 500 mL mein 0.15 mol A⁻ mix karta hoon, toh main directly 0.10 aur 0.15 plug in karta hoon."

Ye sahi kyun lagta hai: Mole ratio = concentration ratio sirf tab jab dono same volume mein hon.

Fix: Mix karne ke baad, total volume = 1.0 L, isliye:

[HA] = 0.10 mol / 1.0 L = 0.10 M
[A⁻] = 0.15 mol / 1.0 L = 0.15 M

Agar aapne different volumes ya alag initial volumes mix kiye hote, toh aapko final concentrations calculate karni padti. Acchi khabar ye hai ki Henderson-Hasselbalch mein, agar dono species same solution mein hain, toh concentrations ka ratio moles ke ratio ke barabar hai, isliye aap moles directly use kar sakte ho: $pH = p K_{a} + lo g_{10} (\frac{n _{A^{-}}}{n _{HA}})$ Ye shortcut kaam karta hai kyunki volume cancel ho jaata hai. Lekin savdhan raho agar aap volume add kar rahe ho (jaise HCl solution add karna) — tab aapko dilution account karna hoga.

Recall Feynman technique: Ek 12-saal ke bacche ko explain karo

Socho aapke paas ek sponge hai jo paani aur oil dono soak kar sakta hai. Agar koi thoda paani girata hai, sponge use absorb kar leta hai. Agar koi thoda oil girata hai, use bhi absorb kar leta hai. Zameen dono taraf se saaf rehti hai.

Buffer solution usi sponge ki tarah hai, lekin acids aur bases ke liye. Isme do chemicals hote hain: ek jo extra acid (H⁺ ions) "soak up" kar sake aur ek jo extra base (OH⁻ ions) "soak up" kar sake. Jab aap buffer mein thoda acid add karte ho, base part use pakad leta hai. Jab thoda base add karte ho, acid part use pakad leta hai. Toh pH (jo measure karta hai ki koi cheez kitni acidic ya basic hai) barely change hoti hai.

Henderson-Hasselbalch equation sirf ek calculator hai jo batata hai pH kya hogi based on iske paas kitna "acid sponge" aur "base sponge" hai. Agar equal amounts hain, pH bilkul beech mein ek special number pKₐ par baith jaati hai. Agar zyada base hai, pH thodi upar jaati hai. Agar zyada acid hai, thodi neeche jaati hai. "Log" part sirf woh math hai jo ratio ko pH shift mein convert karta hai.

Connections

Acid-base equilibria — Henderson-Hasselbalch, Kₐ expression ka ek rearrangement hai
Weak acids and bases — buffers ke liye weak acids/bases chahiye jo partially dissociate hon
Common ion effect — conjugate base wala salt add karna equilibrium ko shift karta hai
Le Chatelier's principle — buffer action practice mein Le Chatelier's principle hai
pH and pOH — pH scale aur log relationships equation ki neev hain
Titration curves — buffer region curve ka flat part hai half-equivalence ke paas
Biological bufers — blood pH H₂CO₃/HCO₃⁻, proteins, phosphate buffers se maintain hoti hai
Polyprotic acids — har dissociation step ek buffer bana sakti hai (e.g., H₂PO₄⁻/HPO₄²⁻)

#flashcards/chemistry

Buffer solution kya hai? :: Ek aisi solution jo acid ya base ki chhoti matra add karne par pH changes ko resist karti hai, jisme ek weak acid aur uska conjugate base comparable concentrations mein hote hain.

Henderson-Hasselbalch equation state karo.

pH = pKₐ + log₁₀([A⁻]/[HA]), jahaan [A⁻] conjugate base concentration hai aur [HA] weak acid concentration hai.

Jab [A⁻] = [HA] ho tab buffer ka pH kya hoga?

pH = pKₐ exactly, kyunki log₁₀(1) = 0.

Ek weak acid/conjugate base pair ke liye effective buffering range kya hai?

pKₐ ± 1 pH unit, jo [A⁻]/[HA] ke 0.1 aur 10 ke beech ratio se correspond karta hai.

Buffer ki maximum capacity kab hoti hai?

Jab [A⁻] = [HA], yaani jab pH = pKₐ, kyunki dono forms added H⁺ ya OH⁻ neutralize karne ke liye equally abundant hoti hain.

Ek acetate buffer mein [CH₃COOH] = 0.20 M aur [CH₃CO⁻] = 0.10 M hai. pKₐ = 4.76. pH kya hai?

pH = 4.76 + log₁₀(0.10/0.20) = 4.76 - 0.30 = 4.46.

Strong acids ke liye Henderson-Hasselbalch kyun use nahi kar sakte?

Strong acids fully dissociate ho jaate hain, equilibrium par koi appreciable [HA] nahi bachta, isliye ratio [A⁻]/[HA] undefined hoti hai aur koi buffering nahi hoti.

Aap 1 L buffer mein 0.01 mol HCl add karte ho jisme 0.30 M HA aur 0.30 M A⁻ hai. Nayi concentrations kya hongi?

[A⁻] = 0.29 M (decreased), [HA] = 0.31 M (increased), kyunki H⁺, A⁻ ke saath react karke HA banata hai.

Agar thodi matra mein NaOH add karo toh buffer pH ka kya hoga?

pH thodi increase hoti hai kyunki OH⁻, HA ke saath react karke A⁻ banata hai, [A⁻]/[HA] ratio aur isliye log term badh jaata hai.

NH₃/NH₄⁺ buffer ke liye Henderson-Hasselbalch equation mein kaun sa species acid hai?

NH₄⁺ acid hai, NH₃ base hai. Use karo pH = pKₐ(NH₄⁺) + log₁₀([NH₃]/[NH₄⁺]).