Buffer solutions — Henderson-Hasselbalch equation: Exercises

2.6.14 · D4 · Chemistry › Equilibrium › Buffer solutions — Henderson-Hasselbalch equation

Woh ek tool kahan se aata hai (isko skip mat karo)

Drill karne se pehle, equation ko earn karte hain instead of sirf trust karne ke. Ek weak acid $HA$ paani ke saath ek tug-of-war mein hai: $HA ⇌ H^{+} + A^{-}$ Acid dissociation constant $K_{a}$ measure karta hai ki woh tug-of-war kitna right settle hoti hai: $K_{a} = \frac{[ H ^{+} ] [ A ^{-} ]}{[ HA ]}$ Hume pH ki parwah hai, aur pH bana hai $[H^{+}]$ se, toh $[H^{+}]$ ke liye solve karo: $[H^{+}] = K_{a} \cdot \frac{[ HA ]}{[ A ^{-} ]}$ Yeh ek line poori idea hai disguise mein: free acidity $[H^{+}]$ hoti hai $K_{a}$ jo scale hoti hai acid-to-base ratio se. Base se zyada acid ⇒ zyada free $H^{+}$ ⇒ lower pH.

Ab logarithms se kyun takleef uthayein? Kyunki $[H^{+}]$ absurd ranges mein span karta hai ( $1 0^{- 1}$ se $1 0^{- 14}$ tak), aur multiply karna clumsy hai. $lo g$ "multiply" ko "add" mein convert kar deta hai, toh messy product $K_{a} \cdot (ratio)$ ek tidy sum ban jaata hai. Dono sides ka $- lo g_{10}$ lo: $- lo g_{10} [H^{+}] = - lo g_{10} (K_{a}) - lo g_{10} (\frac{[ HA ]}{[ A ^{-} ]})$ Left side by definition pH hai; $- lo g_{10} (K_{a})$ hai $p K_{a}$ ; aur fraction flip karne se uske log ka sign flip ho jaata hai ( $lo g (1/ x) = - lo g x$ ). Isse milta hai:

Log/antilog ko ek ratio slider ki tarah picture karo

Neeche diya figure is page ke har problem ka mental model hai. Socho ek horizontal slider jiska position hai ratio $[A^{-}] / [HA]$ ek $lo g_{10}$ scale pe, aur jiska readout pH hai. Ek decade slide karna (ratio ×10) pH ko exactly +1 move karta hai; doosri taraf slide karna (ratio ÷10) pH ko −1 move karta hai. Centre notch (ratio = 1) woh jagah hai jahan pH = $p K_{a}$ .

Figure — Buffer solutions — Henderson-Hasselbalch equation

$lo g_{10}$ hai "slider pe main kahan hoon?" wala sawaal: diya gaya ratio, centre se kitne decades door hoon?
Antilog $1 0^{x}$ reverse hai: diya gaya kitna slide kiya, woh ratio kya hai? Agar $lo g_{10} (r) = - 0.25$ toh $r = 1 0^{- 0.25} = 0.562$ — centre se thoda left.

Definition Woh do edges jahan equation kaam nahi karti

Slider figure ke far ends dekho (shaded red):

$[A^{-}] / [HA] \to 0$ (base almost khatam): $lo g_{10}$ $- \infty$ tak gir jaata hai, toh pH crash kar jaata hai. Physically tumhara base khatam ho gaya hai $H^{+}$ pakadne ke liye — ab yeh essentially pure weak acid ka solution hai, aur H-H assumption " $[HA] \approx$ initial" fail ho jaata hai kyunki $HA$ ka dissociation khud $H^{+}$ ka dominant source ban jaata hai.
$[A^{-}] / [HA] \to \infty$ (acid almost khatam): $lo g_{10}$ $+ \infty$ tak soar karta hai, pH shoot up karta hai. Tumhara acid khatam ho gaya hai $H^{+}$ donate karne ke liye; solution weak acid ke salt ki tarah behave karta hai, aur paani ka khud ka $OH^{-}$ matter karta hai.

Lesson: edges ke paas ek partner ka almost koi reserve nahi hota, toh buffer ab change ko resist nahi karta — ek tiny addition pH ko hugely swing kar deta hai. Safe working band hai $0.1 \leq ratio \leq 10$ , yaani $p K_{a} \pm 1$ .

Do undo-buttons jo tum constantly use karoge:

$lo g_{10}$ poochhta hai "10 ko kaunsi power pe raise karein toh yeh number milega?" Toh $lo g_{10} (10) = 1$ , $lo g_{10} (1) = 0$ , $lo g_{10} (0.1) = - 1$ .
Antilog $1 0^{x}$ log ko undo karta hai: agar $lo g_{10} (r) = - 0.25$ , toh $r = 1 0^{- 0.25}$ .

Level 1 — Recognition

L1.1 — Parts ko naam do

Ek solution banaya gaya hai 0.20 M lactic acid ( $H C_{3} H_{5} O_{3}$ ) aur 0.20 M sodium lactate ( $Na C_{3} H_{5} O_{3}$ ) se. Identify karo kaunsi species $HA$ hai, kaunsi $A^{-}$ hai, aur predict karo (no calculator) ki pH, $p K_{a}$ se compare karke kaisi hai.

Recall Solution

$HA$ = lactic acid $H C_{3} H_{5} O_{3}$ — iske paas abhi bhi donatable H hai.
$A^{-}$ = lactate ion $C_{3} H_{5} O_{3}^{-}$ — sodium lactate ek salt hai jo fully dissociate ho jaata hai, conjugate base release karta hai.
Ratio $[A^{-}] / [HA] = 0.20/0.20 = 1$ , toh $lo g_{10} (1) = 0$ .
Isliye $pH = p K_{a}$ exactly. Slider figure pe, hum dead-centre baithe hain.

L1.2 — Kya yeh buffer hai bhi?

Har pair ke liye, YES (buffer) ya NO batao, ek-line reason ke saath: (a) $HCl$ + $NaCl$ · (b) $C H_{3} COOH$ + $C H_{3} COONa$ · (c) $N H_{3}$ + $N H_{4} Cl$ · (d) $NaOH$ + $NaCl$ .

Recall Solution

(a) NO — $HCl$ ek strong acid hai; koi weak-acid/conjugate-base equilibrium nahi hai shift hone ke liye.
(b) YES — weak acid (acetic) + uska conjugate base (acetate). Classic buffer.
(c) YES — weak base $N H_{3}$ + uska conjugate acid $N H_{4}^{+}$ . $p K_{a} (N H_{4}^{+})$ ke aaspaas buffer.
(d) NO — $NaOH$ ek strong base hai aur $NaCl$ ek spectator salt hai; koi conjugate pair nahi.

Rule of thumb: ek buffer ko chahiye ek weak acid ya base plus uske conjugate partner ki comparable amount.

Level 2 — Application

L2.1 — Seedha pH calculation

Ek buffer mein hai 0.30 M benzoic acid ( $p K_{a} = 4.20$ ) aur 0.60 M sodium benzoate. pH find karo.

Recall Solution

$HA$ = benzoic acid, $A^{-}$ = benzoate. $pH = 4.20 + lo g_{10} (\frac{0.60}{0.30}) = 4.20 + lo g_{10} (2)$ $= 4.20 + 0.301 = 4.50$ Ratio 2 slider pe centre se ek chhota step right hai ⇒ pH $p K_{a}$ se upar baithta hai. ✓

L2.2 — Ratio ke liye solve karo

Tum acetic acid ( $p K_{a} = 4.76$ ) use karke pH 5.00 par ek buffer chahte ho. Tumhein kaunsa ratio $[A^{-}] / [HA]$ chahiye?

Recall Solution

$5.00 = 4.76 + lo g_{10} (\frac{[ A ^{-} ]}{[ HA ]})$ Log isolate karo: $lo g_{10} (\frac{[ A ^{-} ]}{[ HA ]}) = 5.00 - 4.76 = 0.24$ Antilog (log undo karo — yeh hai "slider readout → ratio" move): $\frac{[ A ^{-} ]}{[ HA ]} = 1 0^{0.24} = 1.74$ pH $p K_{a}$ se upar hai, toh humein acid se zyada base chahiye — 1 se bada ratio isko confirm karta hai. ✓

L2.3 — $p K_{a}$ ke liye solve karo

Ek buffer pH 7.40 pe measure kiya gaya hai jismein $[HP O_{4}^{2 -}] = 0.080$ M aur $[H_{2} P O_{4}^{-}] = 0.050$ M hai. Relevant dissociation ka $p K_{a}$ kya hai? (Dekho Polyprotic acids.)

Recall Solution

Yahan $HA = H_{2} P O_{4}^{-}$ (donor) aur $A^{-} = HP O_{4}^{2 -}$ (uska conjugate base) hai. $7.40 = p K_{a} + lo g_{10} (\frac{0.080}{0.050}) = p K_{a} + lo g_{10} (1.6)$ $lo g_{10} (1.6) = 0.204 \Rightarrow p K_{a} = 7.40 - 0.204 = 7.20$ Yeh phosphoric acid ke known $p K_{a 2}$ se match karta hai — yehi reason hai ki phosphate buffers physiological pH ke aaspaas kaam karte hain.

Level 3 — Analysis

L3.1 — Strong base add karo

Ek buffer mein 0.40 mol formic acid aur 0.40 mol formate ( $p K_{a} = 3.75$ ) hai 1.0 L mein. 0.10 mol $NaOH$ add karo. Naya pH aur pH change find karo.

Recall Solution

Pehle chemistry: $O H^{-}$ ek strong base hai; yeh acid $\to$ base convert karta hai: $O H^{-} + HCOOH \to HCO O^{-} + H_{2} O$ Nayi amounts (0.10 mol react karta hai, 1 : 1):

$HCOOH$ : $0.40 - 0.10 = 0.30$ mol
$HCO O^{-}$ : $0.40 + 0.10 = 0.50$ mol

Initial pH $= 3.75 + lo g_{10} (1) = 3.75$ . $pH = 3.75 + lo g_{10} (\frac{0.50}{0.30}) = 3.75 + 0.222 = 3.97$ Change $= + 0.22$ pH units. Neeche ke stacked-bar figure mein, orange (acid) block shrink hota hai aur teal (base) block grow karta hai: ratio thoda right nudge hota hai lekin pH barely move karta hai kyunki dono reserves abhi bhi bade hain. (Volume ratio mein cancel ho jaata hai, toh moles theek hain.)

L3.2 — Dilution twist

Original L3.1 buffer lo (0.40 mol / 0.40 mol in 1.0 L) aur pure water se 4.0 L tak dilute karo. Naya pH kya hai?

Recall Solution

Dilution dono concentrations ko same factor se scale karta hai ( $1/4$ ): $[A^{-}] = 0.10 M, [HA] = 0.10 M$ Ratio unchanged rehta hai ( $0.10/0.10 = 1$ ): $pH = 3.75 + lo g_{10} (1) = 3.75$ Kyunki H-H depend karta hai ek ratio pe, absolute concentration pe nahi, dilution pH ko first approximation mein unchanged rehne deta hai. Jo girti hai woh hai buffer capacity — future shocks absorb karne ke liye kum moles.

"First approximation mein" kyun? Do edge effects badhte hain jaise tum aur dilute karte ho:

Water autoionization. H-H silently ignore karta hai woh $H^{+}$ (aur $OH^{-}$ ) jo paani khud supply karta hai ( $K_{w} = 1 0^{- 14}$ ). Jab buffer concentrations $\sim 1 0^{- 6}$ M ke around drop ho jaati hain, paani ka khud ka $1 0^{- 7}$ M $H^{+}$ negligible nahi rehta, aur true pH 7 ki taraf drift karne lagta hai.
Ionic strength. H-H concentrations ko activities ke stand-ins ki tarah use karta hai. Dilute karne se har ion ke aaspaas ionic atmosphere change ho jaata hai, toh effective (active) concentrations thodi shift ho jaati hain; moderate strength pe yeh ek real pH ko naive prediction ke versus $\sim 0.1$ unit tak move kar sakta hai.

Ordinary lab buffers (0.01–1 M) ke liye yeh chhote hain; bahut dilute ya bahut precise kaam ke liye nahi hain.

Level 4 — Synthesis

L4.1 — Acid choose karo, phir recipe banao

pH 4.90 pe ek buffer design karo jisme total concentration $[HA] + [A^{-}] = 0.50$ M ho. In mein se choose karo: formic ( $p K_{a} = 3.75$ ), acetic ( $p K_{a} = 4.76$ ), ya ammonium ( $p K_{a} = 9.25$ ). Acid aur dono concentrations do. (Yaad rakho effective range $p K_{a} \pm 1$ ; dekho Common ion effect for why the salt matters.)

Recall Solution

Acid choose karo: target pH $p K_{a} \pm 1$ ke andar hona chahiye.

Formic: range $2.75$ – $4.75$ → 4.90 (just) bahar hai. ✗
Acetic: range $3.76$ – $5.76$ → 4.90 andar, aur $p K_{a}$ ke karib. ✓ Best.
Ammonium: range $8.25$ – $10.25$ → bahut door. ✗

Ratio find karo acetic ( $p K_{a} = 4.76$ ) ke saath: $lo g_{10} (r) = 4.90 - 4.76 = 0.14 \Rightarrow r = 1 0^{0.14} = 1.38$ Toh $[A^{-}] = 1.38 [HA]$ . Total 0.50 M split karo: $[HA] + 1.38 [HA] = 0.50 \Rightarrow [HA] = \frac{0.50}{2.38} = 0.210 M$ $[A^{-}] = 0.50 - 0.210 = 0.290 M$ Recipe: $\approx$ 0.21 M acetic acid + 0.29 M sodium acetate. Quick check: $4.76 + lo g_{10} (0.290/0.210) = 4.76 + 0.14 = 4.90$ . ✓

L4.2 — Base + strong acid se buffer banana

Tumhare paas sirf 0.50 mol $N H_{3}$ hai 1.0 L mein aur solid $HCl$ . pH 9.05 pe buffer banane ke liye kitne moles $HCl$ add karne honge? ( $p K_{a} (N H_{4}^{+}) = 9.25$ .)

Recall Solution

$HCl$ add karne se base acid mein convert hota hai: $N H_{3} + H^{+} \to N H_{4}^{+}$ . Maano $x$ = mol $HCl$ added. Toh $N H_{4}^{+} = x$ , $N H_{3} = 0.50 - x$ . Target: $9.05 = 9.25 + lo g_{10} (\frac{0.50 - x}{x})$ $lo g_{10} (\frac{0.50 - x}{x}) = - 0.20 \Rightarrow \frac{0.50 - x}{x} = 1 0^{- 0.20} = 0.631$ $0.50 - x = 0.631 x \Rightarrow 0.50 = 1.631 x \Rightarrow x = 0.307 mol$ Check: $N H_{3} = 0.193$ , $N H_{4}^{+} = 0.307$ ; $9.25 + lo g_{10} (0.193/0.307) = 9.25 - 0.20 = 9.05$ . ✓ pH $p K_{a}$ se neeche hai ⇒ acid form dominant hona chahiye, aur indeed $0.307 > 0.193$ . ✓

Level 5 — Mastery

L5.1 — Sequential shocks with a capacity check

Shuru karo 0.60 mol acetic acid + 0.40 mol acetate se 1.0 L mein ( $p K_{a} = 4.76$ ). (a) Starting pH find karo. (b) 0.15 mol $NaOH$ add karo; naya pH find karo. (c) Phir (b) ke result mein 0.35 mol $HCl$ add karo; pH find karo. (d) Comment karo ki (c) ke baad buffer abhi bhi "in range" hai ya nahi.

Recall Solution

(a) Start: $pH = 4.76 + lo g_{10} (\frac{0.40}{0.60}) = 4.76 + (- 0.176) = 4.58$

(b) 0.15 mol $O H^{-}$ add karo (acid → base):

Acid: $0.60 - 0.15 = 0.45$ mol
Base: $0.40 + 0.15 = 0.55$ mol $pH = 4.76 + lo g_{10} (\frac{0.55}{0.45}) = 4.76 + 0.087 = 4.85$

(c) 0.35 mol $H^{+}$ add karo (base → acid), acid = 0.45, base = 0.55 se shuru karke:

Base: $0.55 - 0.35 = 0.20$ mol
Acid: $0.45 + 0.35 = 0.80$ mol $pH = 4.76 + lo g_{10} (\frac{0.20}{0.80}) = 4.76 + (- 0.602) = 4.16$

(d) In range hai? Effective range hai $p K_{a} \pm 1 = 3.76$ se $5.76$ tak. Final pH $4.16$ andar hai, lekin ratio ab $0.20/0.80 = 0.25$ hai — slider pe hum centre se kaafi left drift kar gaye hain, lower edge ( $0.1$ ) ke karib. Buffer abhi bhi kaam karta hai lekin zyada acid ki taraf uski remaining capacity thin ho rahi hai (sirf 0.20 mol base bacha hai H absorb karne ke liye).

Figure teen saare states mein pH ko shaded safe band ke against track karta hai.

L5.2 — Unbuffered water se compare karo

Upar ke step (c) ke baad wale buffer ke liye (pH 4.16), aur 1.0 L pure water (pH 7.00) ke liye, dono mein 0.05 mol $HCl$ add karo. Dono naye pH values find karo aur shocks compare karo.

Recall Solution

Buffer (acid = 0.80, base = 0.20; 0.05 mol base → acid convert karo):

Base: $0.20 - 0.05 = 0.15$ ; Acid: $0.80 + 0.05 = 0.85$ $pH = 4.76 + lo g_{10} (\frac{0.15}{0.85}) = 4.76 + (- 0.753) = 4.01$ Change $= 4.16 \to 4.01 = - 0.15$ pH units.

Pure water: $0.05$ mol $HCl$ $1.0$ L mein $\Rightarrow [H^{+}] = 0.05$ M: $pH = - lo g_{10} (0.05) = 1.30$ Change $= 7.00 \to 1.30 = - 5.70$ pH units.

Comparison: same acid dose water ko 5.7 units move karta hai lekin (already partly spent) buffer ko sirf 0.15 units — ek ~38× chhota swing. Yeh hai Le Chatelier's principle kaam karta hua: add hone wala H consume ho jaata hai $A^{-} \to HA$ shift karke, toh free $[H^{+}]$ barely rise karta hai.

Recall Self-test cloze

Ek buffer pH change ko resist karta hai kyunki iske paas ek weak acid aur uska conjugate base comparable amounts mein hote hain. Jab ratio $[A^{-}] / [HA] = 1$ hota hai, toh $pH$ equals == $p K_{a}$ . Effective buffer range hai $p K_{a} \pm 1$ . Dilution buffer capacity change karta hai lekin (first order tak) pH nahi. H-H actually equation $K_{a}$ == hai jo logarithms ke saath rewrite ki gayi hai.

Log undo karo — zor se bolo:

Agar

lo g_{10} (r) = - 0.25

, toh

r

equals?

1 0^{- 0.25} = 0.562

Dilution par pH kyun fixed rehta hai?

H-H depend karta hai ek ratio pe, aur dilution dono terms ko equally scale karta hai.

Buffer capacity kahan maximum hai?

pH = p K_{a}

pe, jahan

[A^{-}] = [HA]

H-H ratio extremes pe kya break karta hai?

ratio → 0 ya → ∞ ke paas ek partner khatam ho jaata hai, plus water autoionization dominate karne lagta hai.

Related: Titration curves · Biological bufers · pH and pOH