Buffer solutions — Henderson-Hasselbalch equation

What is the Henderson-Hasselbalch equation?

The Henderson-Hasselbalch equation relates the pH of a buffer to the ratio of conjugate base to weak acid:

Why this form? Because when the ratio $[\mathrm{A}^-]/[\mathrm{HA}] = 1$ , the pH equals the pKₐ exactly. The log term tells you how many pH units you shift from pKₐ based on the ratio.

Derivation from first principles

Let's derive the Henderson-Hasselbalch equation starting from the acid dissociation equilibrium.

Step 1: Write the dissociation equilibrium for a weak acid

$\mathrm{HA}_{(aq)} \rightleftharpoons \mathrm{H}^+_{(aq)} + \mathrm{A}^-_{(aq)}$

Why? A weak acid partially dissociates in water. At equilibrium, forward and reverse rates balance.

Step 2: Write the equilibrium constant expression

$K_a = \frac{[\mathrm{H}^+][\mathrm{A}^-]}{[\mathrm{HA}]}$

Why? The equilibrium constant relates concentrations at equilibrium. Products over reactants, with coefficients as exponents (all 1 here).

Step 3: Solve for [H⁺]

$[\mathrm{H}^+] = K_a \cdot \frac{[\mathrm{HA}]}{[\mathrm{A}^-]}$

Why this step? We want to connect [H⁺] to the ratio of acid and base forms, because pH = -log[H⁺].

Step 4: Take the negative logarithm of both sides

$-\log_{10}[\mathrm{H}^+] = -\log_{10}\left(K_a \cdot \frac{[\mathrm{HA}]}{[\mathrm{A}^-]}\right)$

Why? The pH is defined as $-\log_{10}[\mathrm{H}^+]$ , and pKₐ is $-\log_{10}(K_a)$ . Logarithms let us turn multiplicative relationships into additive ones.

Step 5: Apply logarithm properties

$\mathrm{pH} = -\log_{10}(K_a) - \log_{10}\left(\frac{[\mathrm{HA}]}{[\mathrm{A}^-]}\right)$

Why? Because $\log(xy) = \log(x) + \log(y)$ , so $-\log(K_a \cdot X) = -\log(K_a) - \log(X)$ .

Step 6: Recognize pKₐ and simplify

$\mathrm{pH} = \mathrm{p}K_a - \log_{10}\left(\frac{[\mathrm{HA}]}{[\mathrm{A}^-]}\right)$

Why? Define $\mathrm{p}K_a = -\log_{10}(K_a)$ .

Step 7: Flip the fraction inside the log

$\mathrm{pH} = \mathrm{p}K_a + \log_{10}\left(\frac{[\mathrm{A}^-]}{[\mathrm{HA}]}\right)$

Why this final step? Because $\log(1/x) = -\log(x)$ , so flipping the fraction $[\mathrm{HA}]/[\mathrm{A}^-]$ to $[\mathrm{A}^-]/[\mathrm{HA}]$ changes the minus to a plus. This is the standard form: Henderson-Hasselbalch equation.

Figure — Buffer solutions — Henderson-Hasselbalch equation

When is this equation valid?

The Henderson-Hasselbalch equation is an approximation that works best when:

The acid is weak ( $K_a < 10^{-3}$ ), so equilibrium concentrations ≈ initial concentrations
Both [HA] and [A⁻] are much larger than [H⁺] — the buffer components don't get significantly depleted by dissociation or reaction with water
The ratio [A⁻]/[HA] is between 0.1 and 10 — outside this range, the buffer capacity is poor and activity coefficient effects become significant

Why do we need these conditions? Because we assumed that the equilibrium concentrations equal the analytical (initial) concentrations and ignored the H⁺ from water autoionization. For dilute or very weak/strong systems, these assumptions break down.

Worked example Example 1: Acetate buffer pH calculation

You prepare a buffer with 0.10 M acetic acid (CH₃COOH) and 0.15 M sodium acetate (CH₃COO⁻). The pKₐ of acetic acid is 4.76. What is the pH?

Solution: Step 1: Identify the components

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

Why? Sodium acetate fully dissociates, providing the acetate ion.

Step 2: Apply Henderson-Hasselbalch $\mathrm{pH} = 4.76 + \log_{10}\left(\frac{0.15}{0.10}\right)$ Why this step? Plug values directly into the equation.

Step 3: Calculate the log term $\frac{0.15}{0.10} = 1.5$ $\log_{10}(1.5) = 0.176$ Why? The ratio tells us we have more base than acid, so pH > pKₐ. Step 4: Final answer $\mathrm{pH} = 4.76 + 0.176 = 4.94$

The buffer has a pH of 4.94, slightly basic relative to the pKₐ because we have more conjugate base.

Worked example Example 2: Designing a buffer for target pH

You need a buffer at pH 9.00 You have ammonia (NH₃, pKₐ = 9.25 for NH₄⁺) and amonium chloride (NH₄Cl). What ratio [NH₃]/[NH₄⁺] do you need?

Solution:

Step 1: Identify the conjugate pair

Weak base: NH₃ (but we use its conjugate acid's pKₐ)
Conjugate acid: NH₄⁺ (the weak acid in the buffer equilibrium)
For the acid form: HA = NH₄⁺, A⁻ = NH₃
pKₐ(NH₄⁺) = 9.25 Why? Henderson-Hasselbalch is written for acid/base pairs. NH₄⁺ is the acid, NH₃ is the base.

Step 2: Write the equation $9.00 = 9.25 + \log_{10}\left(\frac{[\mathrm{NH}_3]}{[\mathrm{NH}_4^+]}\right)$

Why? Target pH = 9.00, we're solving for the ratio.

Step 3: Isolate the log term $\log_{10}\left(\frac{[\mathrm{NH}_3]}{[\mathrm{NH}_4^+]}\right) = 9.00 - 9.25 = -0.25$

Why this step? Subtract pKₐ from both sides.

Step 4: Solve for the ratio $\frac{[\mathrm{NH}_3]}{[\mathrm{NH}_4^+]} = 10^{-0.25} = 0.562$

Why? Take the antilog (10 to the power) toundo the logarithm.

Answer: You need a ratio of NH₃ to NH₄⁺ of approximately 0.56:1, or roughly 1part ammonia to 1.8 parts ammonium chloride. Since pH< pKₐ, we need more of the acidic form.

Worked example Example 3: Buffer response to added acid

A buffer contains 0.50 M formic acid (HCOH, pKₐ = 3.75) and 0.50 M formate ion (HCOO⁻). Calculate the pH before and after adding 0.02 mol of HCl to 1.0 L of this buffer.

Solution:

Initial pH: $\mathrm{pH} = 3.75 + \log_{10}\left(\frac{0.50}{0.50}\right) = 3.75 + 0 = 3.75$ Why? When the ratio is 1:1, pH = pKₐ exactly.

After adding HCl: Step 1: Determine what happens chemically HCl is a strong acid, fully dissociates: HCl → H⁺ + Cl⁻

The added H⁺ reacts with the conjugate base: $\mathrm{H}^+ + \mathrm{HCOO}^- \rightarrow \mathrm{HCOH}$

Why? The buffer "catches" the added H⁺ by converting base to acid.

Step 2: Calculate new concentrations

[HCOO⁻] decreases by 0.02 M:0.50 - 0.02 = 0.48 M
[HCOOH] increases by 0.02 M: 0.50 + 0.02 = 0.52 M

Why this step? Stoichiometry: 1 mol H⁺ consumes 1 mol HCOO⁻ and produces 1 mol HCOOH.

Step 3: Apply Henderson-Hasselbalch with new values $\mathrm{pH} = 3.75 + \log_{10}\left(\frac{0.48}{0.52}\right)$ $= 3.75 + \log_{10}(0.923)$ $= 3.75 + (-0.035)$ $= 3.72$

Answer: pH drops from 3.75 to 3.72, a change of only 0.03 pH units. Without the buffer, adding 0.02 M HCl to water would give pH≈ 1.7 The buffer absorbed most of the acid's effect.

Buffer capacity and range

Buffer capacity (β) is the amount of strong acid or base a buffer can neutralize before the pH changes significantly (typically defined as ±1 pH unit).

Maximum capacity occurs when [A⁻] = [HA], i.e., when pH = pKₐ. Why? At this point, you have equal "reserves" of both forms to neutralize added acid or base.

Effective buffer range: pKₐ ± 1

Why this range? When pH = pKₐ ± 1, the ratio [A⁻]/[HA] ranges from 0.1 to 10:

At pH = pKₐ + 1: $\log_{10}(\text{ratio}) = 1$ , so ratio = 10
At pH = pKₐ - 1: $\log_{10}(\text{ratio}) = -1$ , so ratio = 0.1

Outside this range, one component becomes too dilute to effectively neutralize added H⁺ or OH⁻.

Common mistake Common mistake: Forgetting which form is acid vs base

The wrong idea: "For an ammonia buffer at pH 9, I should use pH = pKₐ + log([NH₄⁺]/[NH₃])."

Why it feels right: Ammonia is the starting compound, so students think it goes in the denominator as "the acid."

The fix: NH₃ is the base, NH₄⁺ is the acid. Henderson-Hasselbalch is always: $\mathrm{pH} = \mathrm{p}K_a + \log_{10}\left(\frac{[\text{base form}]}{[\text{acid form}]}\right)$ For NH₃/NH₄⁺, the acid is NH₄⁺, so: $\mathrm{pH} = \mathrm{p}K_a(\mathrm{NH}_4^+) + \log_{10}\left(\frac{[\mathrm{NH}_3]}{[\mathrm{NH}_4^+]}\right)$

Memory aid: The conjugate base (missing an H⁺) goes on top.

Common mistake Common mistake: Using initial moles instead of concentrations after dilution

The wrong idea: "I mix 0.10 molHA in 500 mL with 0.15 mol A⁻ in 500 mL, so I plug in 0.10 and 0.15 directly."

Why it feels right: The mole ratio equals the concentration ratio only if they're in the same volume.

The fix: After mixing, total volume = 1.0 L, so:

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

If you had mixed different volumes or different initial volumes, you must calculate the final concentrations. The good news: in Henderson-Hasselbalch, if both species are in the same solution, the ratio of concentrations equals the ratio of moles, so you can use moles directly: $\mathrm{pH} = \mathrm{p}K_a + \log_{10}\left(\frac{n_{\mathrm{A}^-}}{n_{\mathrm{HA}}}\right)$ This shortcut works because the volume cancels. But be careful if you add volume (like adding HCl solution) — then you must account for dilution.

Recall Feynman technique: Explain to a 12-year-old

Imagine you have a sponge that can soak up both water and oil. If someone spills a little water, the sponge absorbs it. If someone spills a little oil, it absorbs that too. The floor stays clean either way.

A buffer solution is like that sponge, but for acids and bases. It has two chemicals: one that can "soak up" extra acid (H⁺ ions) and one that can "soak up" extra base (OH⁻ ions). When you add a little acid to the buffer, the base part grabs it. When you add a little base, the acid part grabs it. So the pH (which measures how acidic or basic something is) barely changes.

The Henderson-Hasselbalch equation is just a calculator that tells you what the pH will be based on how much "acid sponge" and "base sponge" you have. If you have equal amounts, the pH sits right in the middle at a special number called pKₐ. If you have more base, the pH goes up a bit. If you have more acid, it goes down a bit. The "log" part is just math that turns the ratio into a pH shift.

Connections

Acid-base equilibria — Henderson-Hasselbalch is a rearangement of the Kₐ expression
Weak acids and bases — bufers require weak acids/bases that partially dissociate
Common ion effect — adding a salt with the conjugate base shifts equilibrium
Le Chatelier's principle — buffer action is Le Chatelier's principle in practice
pH and pOH — pH scale and log relationships underlie the equation
Titration curves — buffer region is the flat part of the curve near half-equivalence
Biological bufers — blood pH maintained by H₂CO₃/HCO₃⁻, proteins, phosphate buffers
Polyprotic acids — each dissociation step can form a buffer (e.g., H₂PO₄⁻/HPO₄²⁻)

#flashcards/chemistry

What is a buffer solution? :: A solution that resists pH changes when small amounts of acid or base are added, containing a weak acid and its conjugate base in comparable concentrations.

State the Henderson-Hasselbalch equation.

pH = pKₐ + log₁₀([A⁻]/[HA]), where [A⁻] is the conjugate base concentration and [HA] is the weak acid concentration.

What is the pH of a buffer when [A⁻] = [HA]?

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

What is the effective buffering range for a weak acid/conjugate base pair?

pKₐ ± 1 pH unit, corresponding to a ratio [A⁻]/[HA] between 0.1 and 10.

When does a buffer have maximum capacity?

When [A⁻] = [HA], i.e., when pH = pKₐ, because both forms are equally abundant to neutralize added H⁺ or OH⁻.

An acetate buffer has [CH₃COOH] = 0.20 M and [CH₃CO⁻] = 0.10 M. pKₐ = 4.76. What is the pH?

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

Why can't you use Henderson-Hasselbalch for strong acids?

Strong acids fully dissociate, leaving no appreciable [HA] at equilibrium, so the ratio [A⁻]/[HA] is undefined and no buffering occurs.

You add 0.01 mol HCl to 1 L of a buffer with 0.30 M HA and 0.30 M A⁻. What are the new concentrations?

[A⁻] = 0.29 M (decreased), [HA] = 0.31 M (increased), because H⁺ reacts with A⁻ to form HA.

What happens to buffer pH if you add a small amount of NaOH?

pH increases slightly because OH⁻ reacts with HA to form A⁻, increasing the [A⁻]/[HA] ratio and thus the log term.

For an NH₃/NH₄⁺ buffer, which species is the acid in the Henderson-Hasselbalch equation?

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

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Buffer solutionsek aisa mixture hai jo pH ko stable rakhta hai jab ap thoda acid ya base add karte ho. Socho agar aapke pas ek weak acid (jaise acetic acid, CH₃COOH) aur uska conjugate base (acetate ion, CH₃COO⁻) dono ek sath hain. Jab aap extra H⁺ (acid) daalte ho, to acetate ion use pakad leta hai aur acetic acid ban jata hai. Jab aap OH⁻ (base) daalte ho, to acetic acid H⁺ donate kar deta hai aur acetate ion ban jata hai. Is wajah se pH zyada change nahi hota — bufferek shock absorber ki tarah kaam karta hai.

Henderson-Hasselbalch equation yeh bata hai ki buffer ka pH kya hoga based on acid aur base ke ratio par: pH = pKₐ + log([base]/[acid]). Agar dono equal hain, to pH exactly pKₐ ke barabar hota hai.Agar zyada base hai, pH thoda upar jata hai; agar zyada acid hai, pH thoda neeche ata hai. Yeh equation biology aur medicine mein bohot important hai — jaise human blood ka pH 7.4 maintain hota hai bicarbonate buffer system se. Lab mein bhi experiments ke liye specific pH maintain karne ke liye bufers use hote hain. Formula derive karna simple hai: acid dissociation equilibrium Kₐ = [H⁺][A⁻]/[HA] se start karo, logarithm lo, aur rearrange kar lo.

Ek common mistake yeh hai ki log ammonia/ammonium buffer mein confused ho jate hain ki kaun acid hai aur kaun base. Yad rakho: NH₄⁺ (extra H⁺ wala) acid hai, NH₃ (neutral) base hai. Aur ek galti: strong acid (jaise HCl) ke saath buffer nahi banta kyunki woh completely dissociate ho jata hai — koi equilibrium nahi hota. Buffer ko zaroori hai ki weak acid/base ho aur dono forms appreciable amount mein present hon. Agar ratio10:1 ya 1:10 se zyada extreme ho gaya, to buffer ki capacity kam ho jati hai aur pH easily change hone lagta hai.

Buffer solutions — Henderson-Hasselbalch equation

What is the Henderson-Hasselbalch equation?

Derivation from first principles

When is this equation valid?

Buffer capacity and range

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Equilibrium

Connections