Buffer solutions — Henderson-Hasselbalch equation: Worked examples

2.6.14 · D3 · Chemistry › Equilibrium › Buffer solutions — Henderson-Hasselbalch equation

Yeh page buffer questions ke har tarah ke cases dhundh ke nikalti hai jo parent topic exam mein throw kar sakta hai. Pehle hum saare cases ka ek map banate hain, phir map ke har region ke liye ek example solve karte hain taaki koi bhi scenario exam mein surprise na kar sake.

Sab kuch ek hi equation par tikaa hai jo tumne pehle se dekhi hai:

Agar "pH", "log", "conjugate base", ya " $K_{a}$ " mein se kuch bhi shaky lagta hai, toh pehle pH and pOH, Weak acids and bases, aur Acid-base equilibria ko revisit karo.

The scenario matrix

Har buffer problem bilkul inhi cells mein se ek mein land karta hai. Last column us example ka naam batata hai jo use cover karta hai.

Cell	Kya special hai	Log term kya karta hai	Example
A. Base-rich	$[A^{-}] > [HA]$	ratio $> 1$ , log $> 0$ → pH above $p K_{a}$	Ex 1
B. Acid-rich	$[A^{-}] < [HA]$	ratio $< 1$ , log $< 0$ → pH below $p K_{a}$	Ex 2
C. Exact centre	$[A^{-}] = [HA]$	ratio $= 1$ , log $= 0$ → pH $= p K_{a}$	Ex 3
D. Degenerate: pure acid	$[A^{-}] \to 0$	log $\to - \infty$ → equation breaks, buffer nahi hai	Ex 4
E. Base-buffer (roles flip)	weak base + uska salt diya ho	acid form sahi se identify karni padti hai	Ex 5
F. Add strong acid	H⁺ base form ko kha jaata hai	ratio thoda neeche shift hota hai	Ex 6
G. Add strong base	OH⁻ acid form ko kha jaata hai	ratio thoda upar shift hota hai	Ex 7
H. Design for target pH	pH diya hai, ratio unknown hai	antilog $1 0^{x}$ se invert karo	Ex 8
I. Real-world / edge of range	ratio 0.1–10 ke bahar hai	buffer weak hai, reader ko warn karo	Ex 9

Figure — Buffer solutions — Henderson-Hasselbalch equation

Upar wali figure poore page ke liye ek "number line" hai: log term simply tumhe $p K_{a}$ se left ya right slide karta hai. Har example ke liye ise apne dimaag mein rakho.

Base-rich, acid-rich, aur dead centre (cells A, B, C)

The degenerate case — jab yeh buffer NAHI hai (cell D)

Worked example Example 4 — Cell D (pure acid,

$[A^{-}] \to 0$ ) Tum $0.10 M$ acetic acid akele dissolve karte ho (koi acetate nahi). Kya Henderson–Hasselbalch use kar sakte ho?

Forecast: zero ki taraf jaane wale number ka $lo g$ kya karta hai?

Yahan $[A^{-}] \to 0$ hai, toh ratio $\frac{[ A ^{-} ]}{[ HA ]} \to 0$ . Yeh step kyun? Blindly plug karne se pehle hume limit test karni chahiye.
$lo g_{10} (0^{+}) \to - \infty$ , isliye formula predict karta hai $pH \to - \infty$ — physically impossible hai. Kyun? Derivation ne maana tha ki $[A^{-}]$ zyaadatar ek added salt se aaya, na ki acid ki apni chhoti si dissociation se. Bina salt ke, woh assumption collapse ho jaati hai.
Sahi raasta: ise ek plain weak acid ki tarah equilibrium se treat karo: $K_{a} = \frac{x ^{2}}{0.10 - x} \approx \frac{x ^{2}}{0.10}, x = [H^{+}] .$ $K_{a} = 1 0^{- 4.76} = 1.74 \times 1 0^{- 5}$ ke saath: $x = 1.74 \times 1 0^{- 5} \times 0.10 = 1.32 \times 1 0^{- 3} .$
$pH = - lo g_{10} (1.32 \times 1 0^{- 3}) = 2.88$ .

Verify: Ek single weak acid clearly acidic honi chahiye aur uski pH $p K_{a}$ se neeche honi chahiye; $2.88 < 4.76$ . ✓ Lesson: koi conjugate base nahi ⇒ koi buffer nahi ⇒ equation use mat karo. ✓

Roles flip karna — ek base buffer (cell E)

Worked example Example 5 — Cell E (weak base + uska salt)

Ek buffer mein $0.30 M$ methylamine ( $C H_{3} N H_{2}$ ) aur $0.30 M$ methylammonium chloride ( $C H_{3} N H_{3}^{+}$ ) hai. Conjugate acid $C H_{3} N H_{3}^{+}$ ka $p K_{a} = 10.66$ hai. pH find karo.

Forecast: kaun sa species " $HA$ " acid hai aur kaun sa " $A^{-}$ " base?

Proton ke point of view se roles assign karo:
- $HA = C H_{3} N H_{3}^{+}$ (yeh proton de sakta hai) $= 0.30 M$ .
- $A^{-} = C H_{3} N H_{2}$ (yeh proton pakad sakta hai) $= 0.30 M$ . Yeh step kyun? Henderson–Hasselbalch hamesha ek acid aur uske apne conjugate base ke around likha jaata hai; yahan neutral amine base ka role play karta hai.
Ratio $= \frac{0.30}{0.30} = 1$ , isliye $lo g = 0$ .
$pH = 10.66 + 0 = 10.66$ .

Verify: Ek base buffer basic hona chahiye (pH $> 7$ ); $10.66 > 7$ . ✓ Aur kyunki concentrations equal hain hum sahi tarah $p K_{a}$ par land karte hain. ✓

Recall Roles cheat-sheet

Kaun sa species $HA$ hai? ::: Woh jo proton donate kar sake (amine buffers ke liye charged/protonated form). Kaun sa species $A^{-}$ hai? ::: Woh jo proton accept kar sake (neutral amine, ya acid buffers ke liye salt ka anion).

Buffer under attack (cells F aur G)

Jab tum strong acid ya base add karo, pehle stoichiometry karo (ek one-way reaction jo completion tak chalti hai), tab bache hue par Henderson–Hasselbalch apply karo.

Buffer design karna aur range ka edge (cells H, I)

Worked example Example 8 — Cell H (target pH, ratio ke liye solve karo)

Tumhe $pH = 7.00$ par ek phosphate buffer chahiye jisme $H_{2} P O_{4}^{-}$ / $HP O_{4}^{2 -}$ ho, jahan relevant $p K_{a} = 7.21$ hai. Tumhe kaun sa ratio $[HP O_{4}^{2 -}] / [H_{2} P O_{4}^{-}]$ chahiye? (Phosphate ke kaafi saare $p K_{a}$ values kyun hain iske liye Polyprotic acids dekho.)

Forecast: pH, $p K_{a}$ se neeche hai — tumhe zyada acid chahiye ya zyada base?

$HA = H_{2} P O_{4}^{-}$ , $A^{-} = HP O_{4}^{2 -}$ ke saath set up karo: $7.00 = 7.21 + lo g_{10} (r), r = \frac{[ A ^{-} ]}{[ HA ]} .$
Log isolate karo: $lo g_{10} (r) = 7.00 - 7.21 = - 0.21$ . Yeh step kyun? Fixed $p K_{a}$ subtract karo taaki unknown ratio expose ho.
Log ko antilog $1 0^{x}$ se undo karo: $r = 1 0^{- 0.21} = 0.617.$ Kyun? $lo g$ aur $1 0^{()}$ inverse operations hain — $1 0^{x}$ ka jawab hai "kaun se ratio ka yeh log hai?"
Isliye $[HP O_{4}^{2 -}] : [H_{2} P O_{4}^{-}] \approx 0.62 : 1$ , matlab zyada acid form chahiye.

Verify: pH $p K_{a}$ se neeche ⇒ ratio $< 1$ ⇒ acid-rich; $0.62 < 1$ . ✓ Sanity check: $7.21 + lo g_{10} (0.617) = 7.21 - 0.21 = 7.00$ . ✓

Worked example Example 9 — Cell I (real-world, buffer range ka edge)

Blood plasma mein $C O_{2} /HC O_{3}^{-}$ system use hota hai jiska $p K_{a} = 6.1$ hai. Healthy blood ka $pH = 7.4$ hota hai. Body kaun sa ratio $[HC O_{3}^{-}] / [C O_{2}]$ maintain karti hai, aur kya yeh ek "textbook-safe" buffer hai? (Zyada ke liye Biological bufers dekho.)

Forecast: pH, $p K_{a}$ se puri $1.3$ units upar hai — kya ratio comfy $0.1$ – $10$ window ke andar hai?

$7.4 = 6.1 + lo g_{10} (r)$ isliye $lo g_{10} (r) = 1.3$ .
$r = 1 0^{1.3} = 20.0$ . Yeh step kyun? Antilog actual concentration ratio recover karta hai.
Interpretation: $[HC O_{3}^{-}] : [C O_{2}] \approx 20 : 1$ — ideal $0.1$ – $10$ range se bahar hai, isliye paper par yeh ek poor buffer lagta hai. Kyun? Ratio $20$ par, added base se ladne ke liye bahut kam dissolved $C O_{2}$ reserve bachta hai.
Body phir bhi kyun jeetti hai: yeh ek open system hai — lungs $C O_{2}$ bahar nikaalt ya retain karte hain, acid form ko continuously resupply karte hain, isliye effective capacity bahut badi hai.

Verify: Reverse-check: $6.1 + lo g_{10} (20) = 6.1 + 1.301 = 7.40$ . ✓ Aur $20 > 10$ confirm karta hai ki yeh classical range se bahar hai — "open lungs" caveat zaroori hai. ✓

Recall Full-matrix self-test

Ratio $> 1$ ⇒ pH $p K_{a}$ ke compare mein kahan hai? ::: Usse upar (log positive). Ratio $= 1$ ⇒ pH? ::: Bilkul $p K_{a}$ (log $= 0$ ). Buffer mein strong acid add karo — kaun sa form badhta hai? ::: Acid form $HA$ (base kha jaata hai). Pure weak acid, koi salt nahi — Henderson–Hasselbalch use karo? ::: Nahi, $[A^{-}] \to 0$ se toot jaata hai; $K_{a} = x^{2} / C$ use karo. Target pH hit karne ke liye tum log ko kis se invert karte ho? ::: Antilog $1 0^{x}$ se.