Buffer solutions — Henderson-Hasselbalch equation: Question bank

2.6.14 · D5 · Chemistry › Equilibrium › Buffer solutions — Henderson-Hasselbalch equation

Yeh Buffer solutions — Henderson-Hasselbalch equation ka companion hai. Jin prerequisites ko open rakhna useful hai: Weak acids and bases, Common ion effect, pH and pOH.

True or false — justify karo

Adding water to a buffer changes its pH significantly

False — dilution se

[A^{-}]

aur

[HA]

dono same factor se scale hote hain, toh

lo g ([A^{-}] / [HA])

ka ratio unchanged rehta hai; pH essentially constant rehta hai (jab tak dilution extreme na ho aur approximation fail na ho).

A buffer at pH = pKₐ has equal moles of acid and conjugate base

True — jab

[A^{-}] = [HA]

hota hai tab ratio 1 hota hai aur

lo g_{10} (1) = 0

, toh pH

=

pKₐ exactly.

A solution of pure weak acid (no conjugate base added) is a buffer

False — ek buffer ko dono partners ke appreciable, comparable amounts chahiye; pure HA mein almost koi

A^{-}

nahi hota, toh woh added OH⁻ ko effectively catch nahi kar sakta.

Mixing a strong acid with excess weak base can produce a buffer

True — strong acid weak base ko partially uske conjugate acid mein convert karta hai, leaving both forms present; yahi woh tarika hai jisse buffers partial neutralization se banaye jaate hain.

Two buffers with the same pKₐ but different total concentrations have the same pH

True (agar unke ratios match karein) — pH

[A^{-}] / [HA]

ke ratio par depend karta hai, absolute concentrations par nahi; lekin zyada concentrated wale ki buffer capacity zyada hoti hai.

Doubling both

[A^{-}]

and

[HA]

raises the buffer capacity but leaves pH unchanged

True — pH ratio se set hota hai (jo unchanged hai), jabki capacity is baat par depend karti hai ki added acid/base ko absorb karne ke liye kitne moles of each reserve available hain.

A buffer works equally well against added acid and added base only when pH = pKₐ

True —

HA

ke equal reserves (jo added OH⁻ se ladte hain) aur

A^{-}

(jo added H⁺ se ladte hain) sirf 1:1 point par exist karte hain.

The Henderson-Hasselbalch equation gives the exact pH of any buffer

False — yeh ek approximation hai jo assume karti hai ki equilibrium concentrations analytical (mixed) amounts ke equal hain aur water se H⁺ negligible hai; dono assumptions bahut dilute ya bahut weak/strong cases mein fail ho jaati hain.

Spot the error

"For an ammonia buffer I use pH = pKₐ + log([NH₄⁺]/[NH₃])."

Galat:

NH_{4}^{+}

acid hai aur

NH_{3}

base hai, toh yeh

lo g ([NH_{3}] / [NH_{4}^{+}])

hona chahiye — base upar, hamesha.

"pKₐ = 4.76, so Kₐ = 4.76 × 10⁻¹."

Galat: pKₐ

= - lo g_{10} K_{a}

hota hai, toh

K_{a} = 1 0^{- p K_{a}} = 1 0^{- 4.76} \approx 1.7 \times 1 0^{- 5}

— aapko antilog lena hoga, directly digits padhne nahi hain.

"Since I added acid, the pH must rise."

Ulta hai: H⁺ add karne se pH girta hai; buffer sirf is girne ko chhota rakhta hai kuch

A^{-}

HA

mein convert karke.

"[A⁻]/[HA] = 100 is fine, the equation still applies, so it's still a good buffer."

Equation ek pH compute karna jaari rakhti hai, lekin 100:1 ratio pKₐ ± 1 range se bahar hai — buffer capacity poor hai kyunki acid reserve almost exhausted ho chuki hai.

"pOH = pKₐ + log([base]/[acid]) for a basic buffer."

Error: Henderson-Hasselbalch pKₐ aur pH ke terms mein likhi jaati hai; agar aap base form use karein toh aapko

pOH = p K_{b} + lo g ([BH^{+}] / [B])

chahiye — alag constant aur flipped ratio.

"I added 0.02 mol strong base, so both forms just shift by 0.02 mol in the same direction."

Error: added OH⁻ HA ko A⁻ mein convert karta hai, toh

[HA]

0.02 se decrease hota hai aur

[A^{-}]

0.02 se increase hota hai — opposite directions mein, total conserve hota hai.

"A 1:1 buffer sits at pH 7."

Sirf tab agar pKₐ exactly 7 ho; 1:1 acetate buffer pH 4.76 par hota hai, aur 1:1 ammonium buffer pH 9.25 par.

Why questions

Why does the ratio set the pH rather than the individual concentrations?

Kyunki

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

mein concentrations sirf ratio ke form mein hain, toh koi bhi common factor (jaise dilution) cancel ho jaata hai.

Why is buffer capacity greatest exactly at pH = pKₐ?

Wahan dono reserves equal aur largest-balanced hote hain, toh added acid ya base ki ek given amount ratio (aur pH) ko kisi bhi direction mein sabse kam shift karti hai.

Why must a buffer contain a weak acid, not a strong one?

Strong acid completely dissociate ho jaata hai, essentially koi undissociated

HA

nahi bachta jo added base ke against H⁺ donate kare — koi reserve pair nahi, hence koi buffering nahi (Common ion effect ko ek aisi equilibrium chahiye jo shift ho sake).

Why does adding a tiny bit of strong acid barely move the pH, while adding it to pure water swings it hugely?

Buffer mein added H⁺ ko

A^{-}

consume kar leta hai (sirf ratio thoda sa nudge hota hai); water mein kuch bhi nahi hai use absorb karne ke liye, toh

[H^{+}]

directly jump kar jaata hai.

Why do we take a logarithm at all in the derivation?

pH ko

- lo g_{10} [H^{+}]

se define kiya jaata hai; dono sides ka

- lo g_{10}

lene par multiplicative

K_{a} \cdot ([HA] / [A^{-}])

additive

p K_{a} + lo g ([A^{-}] / [HA])

mein convert ho jaata hai — padh ne mein aasaan, "pKₐ se start karo, phir shift karo."

Why does flipping the fraction change the sign in front of the log?

Kyunki

lo g (1/ x) = - lo g (x)

hota hai, toh acid-over-acid ki jagah base-over-acid likhne par

-

+

ban jaata hai, standard form milta hai.

Why is the effective range pKₐ ± 1 and not pKₐ ± 3?

pKₐ ± 1 par ratio 10:1 ya 1:10 hota hai; ±2 (100:1) tak jaane par ek component itna scarce hota hai ki almost use up ho jaata hai, toh woh incoming acid ya base ko neutralize nahi kar sakta (Le Chatelier's principle zyada shift nahi kar sakta).

Edge cases

What happens to the Henderson-Hasselbalch prediction when the buffer is diluted a millionfold?

Yeh break down ho jaata hai — jab concentrations water autoionization ke

[H^{+}]

ke kareeb pahunchti hain, toh "equilibrium ≈ analytical" assumption fail ho jaati hai aur pH 7 ki taraf drift karne lagta hai.

If you add enough strong acid to convert all the conjugate base to acid, is it still a buffer?

Nahi — jab

[A^{-}] \to 0

hota hai toh ratio aur log blow up ho jaata hai; ab aapke paas sirf ek weak acid solution hai jo apne buffering point se aage hai (ek titration curve ke equivalence-region behaviour ki tarah).

Can a solution of only NaCl and water act as a buffer?

Nahi —

Na^{+}

aur

Cl^{-}

ek strong base aur strong acid ke ions hain, kisi ke paas usable conjugate equilibrium nahi hai, toh added H⁺ ya OH⁻ ko catch karne ke liye kuch nahi hai.

What is the pH ratio requirement at the extreme edge of the useful range?

pH = pKₐ + 1 par ratio

[A^{-}] / [HA] = 10

hai; pH = pKₐ − 1 par yeh

0.1

hai — ye "still a functional buffer" ke practical cutoffs hain.

For a diprotic acid, why can a buffer form near either pKₐ value?

Har dissociation step ka apna conjugate pair hota hai (jaise

H_{2} A / HA^{-}

aur

HA^{-} / A^{2 -}

), toh buffering har pKₐ ke around alag alag possible hai (Polyprotic acids).

Does blood being buffered near pH 7.4 by a system whose pKₐ is 6.1 violate the pKₐ ± 1 rule?

Yeh thoda stretch karta hai, lekin kaam karta hai kyunki body continuously CO₂ remove karti hai taaki

[HC O_{3}^{-}] / [C O_{2}]

ratio high rahe — ek open system, sealed beaker ki tarah nahi (Biological bufers).

Recall Quick self-check

Which form goes on top of the log, acid or base? ::: Conjugate base $[A^{-}]$ hamesha upar hota hai. pH relative to pKₐ when base exceeds acid? ::: pH pKₐ se upar hota hai (positive log term). Where is buffer capacity maximal? ::: pH $=$ pKₐ par, jahan dono reserves equal hoti hain.