MHT-CET Chemistry · Ionic Equilibria

Buffer Solutions and the Henderson-Hasselbalch Equation

A buffer resists changes in pH; the Henderson-Hasselbalch equation lets you compute its pH from the salt-to-acid ratio and the pKa (or pOH from pKb for a basic buffer).

All Ionic Equilibria notes MHT-CET Chemistry strategy

Why this matters

This is one of the most reliably scored blocks in MHT-CET Ionic Equilibria — most PYQs are direct one-step plug-ins into pH = pKa + log([salt]/[acid]), and the numbers are picked so the log term is a clean log 2, log 5 or log 10. Two things earn the marks every year: identifying which mixture is a buffer (weak acid + its salt, or weak base + its salt) and keeping the ratio the right way up (salt over acid). When the salt and acid concentrations are equal, the log term vanishes and pH = pKa.

Concept 1 of 4

What a buffer is and how to recognise one

Intuition

A buffer is a solution that barely changes its pH when a little acid or base is added to it. It works because it holds a reservoir of a weak acid AND its conjugate base at the same time — the acid mops up added base, the conjugate base mops up added acid. So a buffer is always a weak acid with its salt, or a weak base with its salt.

Definition

Two kinds of buffer:

Acidic buffer (pH $< 7$ ): a weak acid together with its salt with a strong base. Example: $\text{CH}_3\text{COOH} + \text{CH}_3\text{COONa}$ (acetic acid + sodium acetate).
Basic buffer (pH $> 7$ ): a weak base together with its salt with a strong acid. Example: $\text{NH}_4\text{OH} + \text{NH}_4\text{Cl}$ (ammonium hydroxide + ammonium chloride).
How it resists change (Le Chatelier): the weak acid $\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-$ sits in equilibrium. Added $\text{H}^+$ is consumed by the large store of $\text{A}^-$ ; added $\text{OH}^-$ is neutralised by the large store of $\text{HA}$ . The ratio $[\text{salt}]/[\text{acid}]$ hardly moves, so the pH hardly moves.
Blood is buffered by the bicarbonate system $\text{H}_2\text{CO}_3 / \text{HCO}_3^-$ (carbonic acid and its salt), holding pH near $7.4$ .

Buffer type	Components	Example
Acidic buffer (pH < 7)	Weak acid + salt of that acid with a strong base	$\text{CH}_3\text{COOH} + \text{CH}_3\text{COONa}$ The salt supplies the conjugate base (acetate). A strong acid + salt is NOT a buffer.
Basic buffer (pH > 7)	Weak base + salt of that base with a strong acid	$\text{NH}_4\text{OH} + \text{NH}_4\text{Cl}$ The salt supplies the conjugate acid (ammonium). Note the components: weak base + its salt with a strong acid.
Blood buffer	Carbonic acid + its salt (bicarbonate)	$\text{H}_2\text{CO}_3 / \text{HCO}_3^-$ The bicarbonate buffer holds human blood pH near 7.4 — a frequently asked recall item.

A buffer always pairs a weak partner with its conjugate (from the salt).

Practice this conceptself-check · 5 quick reps

Try it yourself

Which of these is an acidic buffer: (i) HCl + NaCl, (ii) CH3COOH + CH3COONa, (iii) NH4OH + NH4Cl?

Practice — Level 1 (5 reps)

Quick reps to lock in the method. Try each, then check.

1.
Acidic buffer = weak acid + its salt with which kind of base?
2.
Basic buffer = weak base + its salt with which kind of acid?
3.
Which buffer maintains the pH of human blood?
4.
Is HCl + NaCl a buffer?
5.
Name an everyday acidic buffer.

From the bank · past-year question

Example 1Ionic EquilibriaEASY

Which from following mixtures in water acts as a buffer?

[Q93 · 16th May Shift 1 · 2023]

A strong acid + its salt is NOT a buffer

A buffer needs a weak acid (or weak base).

\text{HCl} + \text{NaCl}

has no weak partner to soak up added base, so it cannot resist pH change. Only weak-acid/salt or weak-base/salt pairs buffer.

Match the salt to the right partner

An acidic buffer's salt comes from the weak acid + strong base (giving the conjugate base). A basic buffer's salt comes from the weak base + strong acid (giving the conjugate acid). Mixing acetic acid with ammonium chloride is NOT a buffer — the salt is not the conjugate of the acid.

Drill 3 more on what a buffer is and how to recognise one

Concept 2 of 4

Henderson-Hasselbalch equation — pH of an acidic buffer

Intuition

For an acidic buffer, the pH sits at the pKa of the weak acid, nudged up or down by the log of the salt-to-acid ratio. More salt (conjugate base) than acid pushes the pH above pKa; more acid than salt pulls it below. The ratio always goes salt over acid — get that the wrong way up and the answer lands on the mirror-image option.

Definition

Henderson-Hasselbalch (acidic buffer):

$\text{pH} = pK_a + \log\dfrac{[\text{salt}]}{[\text{acid}]}$ , where $pK_a = -\log K_a$ .
[salt] over [acid] — salt (conjugate base) on top, weak acid on the bottom.
When the salt is more concentrated than the acid, $\log > 0$ and the pH rises above $pK_a$ .
Because both concentrations share the same volume, you may use moles or molarity directly — the ratio is what matters, so no volume conversion is needed.
Rearranged for hydrogen-ion concentration: $[\text{H}^+] = K_a\,\dfrac{[\text{acid}]}{[\text{salt}]}$ (note the ratio flips to acid over salt).

Henderson-Hasselbalch (acidic buffer)

\text{pH} = pK_a + \log\dfrac{[\text{salt}]}{[\text{acid}]}

pK_aacid dissociation exponent, = -\log K_a
[\text{salt}]concentration of the conjugate base (the salt)
[\text{acid}]concentration of the weak acid

Worked example

A buffer is made from 0.05 M weak acid and 0.5 M of its salt with a strong base. The pKa of the acid is 4.60. Find the pH.

Acidic buffer, so use $\text{pH} = pK_a + \log\dfrac{[\text{salt}]}{[\text{acid}]}$ .
Ratio $\dfrac{[\text{salt}]}{[\text{acid}]} = \dfrac{0.5}{0.05} = 10$ .
$\log 10 = 1$ , so $\text{pH} = 4.60 + 1 = 5.60$ .

Answer:

\text{pH} = 5.60

Practice this conceptself-check · 4 quick reps

Try it yourself

A buffer contains 0.02 M weak acid and 0.08 M of its salt with a strong base, with pKa = 4.30. What is the pH? (log 4 = 0.602)

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Buffer: 0.1 M acid, 0.2 M salt, pKa = 4.7. pH? (log 2 = 0.301)
2.
Buffer: 0.1 M salt, 0.01 M acid, pKa = 4.5. pH?
3.
Which concentration goes on top of the log ratio, salt or acid?
4.
Buffer: Ka = 6.6e-10, acid 0.01 M, salt 0.02 M. Find [H+].

From the bank · past-year question

Example 2Ionic EquilibriaEASY

What is the pH of buffer solution prepared by mixing 0.01 M weak acid and 0.02 M salt of weak acid with strong base? (

pK_a = 4.680

)

[Shift || · 2025]

Ratio is salt over acid — don't invert it

The most common wrong answer inverts the ratio to

\log([\text{acid}]/[\text{salt}])

, which flips the sign of the log term and lands on the decoy option. Keep it

\log\dfrac{[\text{salt}]}{[\text{acid}]}

: more salt raises the pH. (Only in the

[\text{H}^+] = K_a\,[\text{acid}]/[\text{salt}]

form does acid go on top.)

Use concentrations directly — no volume conversion

When equal volumes are mixed, both the salt and acid are diluted by the same factor, so the ratio is unchanged. Plug the given molarities straight in; converting to moles first is extra work that changes nothing.

Drill 11 more on henderson-hasselbalch equation — ph of an acidic buffer

Concept 3 of 4

Equal salt and acid — pH equals pKa

Intuition

When a buffer holds the salt and the weak acid at the SAME concentration, the ratio is 1, and log 1 = 0. The whole log term disappears and the pH is exactly the pKa of the weak acid. This is the maximum-buffering point, and it turns a calculation into a one-liner: just convert Ka to pKa.

Definition

Equal-concentration buffer:

If $[\text{salt}] = [\text{acid}]$ , then $\dfrac{[\text{salt}]}{[\text{acid}]} = 1$ and $\log 1 = 0$ .
So $\text{pH} = pK_a = -\log K_a$ .
This is the point of maximum buffer capacity — the buffer resists pH change best when salt and acid are equal.
Watch for the phrase "equal concentrations" or "equal moles" in the stem — it signals you skip the log term entirely.

Buffer with equal salt and acid

[\text{salt}] = [\text{acid}] \;\Rightarrow\; \text{pH} = pK_a = -\log K_a

K_aacid dissociation constant of the weak acid
pK_a= -\log K_a

Worked example

A buffer contains equal concentrations of a weak acid and its salt. The acid's Ka is 1.0 x 10^-5. Find the pH.

Equal concentrations, so $\dfrac{[\text{salt}]}{[\text{acid}]} = 1$ and $\log 1 = 0$ .
Then $\text{pH} = pK_a = -\log(1.0 \times 10^{-5})$ .
$-\log(10^{-5}) = 5$ .

Answer:

\text{pH} = 5

Practice this conceptself-check · 4 quick reps

Try it yourself

A weak acid with Ka = 4.0 x 10^-6 is mixed with an equal concentration of its sodium salt. What is the pH? (log 4 = 0.602)

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Equal salt and acid, Ka = 1.0 x 10^-4. pH?
2.
What is log 1?
3.
At which ratio does a buffer resist pH change best?
4.
Equal salt and acid, Ka = 1.8 x 10^-5. pH? (log 1.8 = 0.255)

From the bank · past-year question

Example 3Ionic EquilibriaEASY

A buffer solution contains equal concentrations of weak acid and its salt with strong base. Calculate pH of buffer solution if dissociation constant of weak acid is

1.8 \times 10^{-5}

[Q67 · 10th May Shift 2 · 2023]

Equal concentrations means the log term is zero

When

[\text{salt}] = [\text{acid}]

\log\dfrac{[\text{salt}]}{[\text{acid}]} = \log 1 = 0

, so

\text{pH} = pK_a

. Don't waste time on the log — just compute

-\log K_a

. If the stem gives

K_a

rather than

pK_a

, converting it is the whole job.

Concept 4 of 4

Basic buffers — the pOH form and converting to pH

Intuition

A basic buffer (weak base + its salt) is handled the same way, but with the base version of the equation: pOH = pKb + log([salt]/[base]). You find the pOH first, then get pH by subtracting from 14. The single most common slip is forgetting that last step and quoting the pOH as the pH.

Definition

Henderson-Hasselbalch (basic buffer):

$\text{pOH} = pK_b + \log\dfrac{[\text{salt}]}{[\text{base}]}$ , where $pK_b = -\log K_b$ .
Here [salt] is the concentration of the salt (conjugate acid, e.g. $\text{NH}_4\text{Cl}$ ) and [base] is the weak base (e.g. $\text{NH}_4\text{OH}$ ).
Convert to pH with ** $\text{pH} = 14 - \text{pOH}$ ** at 25 °C.
As with acidic buffers, equal volumes mixed keep the ratio unchanged — use the given concentrations directly.

Henderson-Hasselbalch (basic buffer)

\text{pOH} = pK_b + \log\dfrac{[\text{salt}]}{[\text{base}]} \qquad \text{pH} = 14 - \text{pOH}

pK_bbase dissociation exponent, = -\log K_b
[\text{salt}]concentration of the salt (conjugate acid)
[\text{base}]concentration of the weak base

Worked example

A basic buffer is made by mixing equal volumes of 0.1 M NH4OH and 1.0 M NH4Cl. The pKb of NH4OH is 4.75. Find the pOH and the pH.

Basic buffer, so $\text{pOH} = pK_b + \log\dfrac{[\text{salt}]}{[\text{base}]}$ .
$\dfrac{[\text{salt}]}{[\text{base}]} = \dfrac{1.0}{0.1} = 10$ , and $\log 10 = 1$ .
$\text{pOH} = 4.75 + 1 = 5.75$ .
$\text{pH} = 14 - 5.75 = 8.25$ .

Answer:

\text{pOH} = 5.75

\text{pH} = 8.25

Practice this conceptself-check · 4 quick reps

Try it yourself

A buffer of 0.5 M weak base and 0.5 M of its salt has pKb = 4.60. Find the pH. (equal concentrations)

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Basic buffer: pOH = 5.6. What is the pH?
2.
Which equation gives pOH of a basic buffer?
3.
Basic buffer, equal base and salt, pKb = 4.74. pH?
4.
In pOH = pKb + log([salt]/[base]), what is [salt] for NH4OH/NH4Cl?

From the bank · past-year question

Example 4Ionic EquilibriaMODERATE

What is the value of pOH if a buffer solution is prepared by mixing equal volumes of 0.4 M NH

_4

OH and 0.5 M NH

_4

Cl solutions. pK

_b

= 4.730

[Q75 · 2nd May Shift 1 · 2023]

Find pOH first, then subtract from 14

pK_b + \log([\text{salt}]/[\text{base}])

gives the pOH, not the pH. A basic buffer has pH

> 7

, so quoting the pOH (a number below 7) as the pH is the classic decoy. Always finish with

\text{pH} = 14 - \text{pOH}

Use pKb for a base, pKa for an acid

Don't plug a weak base's data into the acidic form. A basic buffer (weak base + its salt) uses

pK_b

and the pOH equation; an acidic buffer uses

pK_a

and the pH equation directly.

Drill 1 more on basic buffers — the poh form and converting to ph

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Henderson-Hasselbalch equation — pH of an acidic buffer
Henderson-Hasselbalch (acidic buffer)
$\text{pH} = pK_a + \log\dfrac{[\text{salt}]}{[\text{acid}]}$
Equal salt and acid — pH equals pKa
Buffer with equal salt and acid
$[\text{salt}] = [\text{acid}] \;\Rightarrow\; \text{pH} = pK_a = -\log K_a$
Basic buffers — the pOH form and converting to pH
Henderson-Hasselbalch (basic buffer)
$\text{pOH} = pK_b + \log\dfrac{[\text{salt}]}{[\text{base}]} \qquad \text{pH} = 14 - \text{pOH}$

Reference tables (1)

What a buffer is and how to recognise one3 rows

Buffer type	Components	Example
Acidic buffer (pH < 7)	Weak acid + salt of that acid with a strong base	$\text{CH}_3\text{COOH} + \text{CH}_3\text{COONa}$ The salt supplies the conjugate base (acetate). A strong acid + salt is NOT a buffer.
Basic buffer (pH > 7)	Weak base + salt of that base with a strong acid	$\text{NH}_4\text{OH} + \text{NH}_4\text{Cl}$ The salt supplies the conjugate acid (ammonium). Note the components: weak base + its salt with a strong acid.
Blood buffer	Carbonic acid + its salt (bicarbonate)	$\text{H}_2\text{CO}_3 / \text{HCO}_3^-$ The bicarbonate buffer holds human blood pH near 7.4 — a frequently asked recall item.

A buffer always pairs a weak partner with its conjugate (from the salt).

Watch out for (7)

A strong acid + its salt is NOT a buffer
→ What a buffer is and how to recognise one
Match the salt to the right partner
→ What a buffer is and how to recognise one
Ratio is salt over acid — don't invert it
→ Henderson-Hasselbalch equation — pH of an acidic buffer
Use concentrations directly — no volume conversion
→ Henderson-Hasselbalch equation — pH of an acidic buffer
Equal concentrations means the log term is zero
→ Equal salt and acid — pH equals pKa
Find pOH first, then subtract from 14
→ Basic buffers — the pOH form and converting to pH
Use pKb for a base, pKa for an acid
→ Basic buffers — the pOH form and converting to pH

Mastery check — 5 interleaved questions

Try each one before clicking. Questions are interleaved across the concepts above, not grouped — interleaving sharpens transfer.

Example 1Ionic EquilibriaEASY

Which from following mixtures in water acts as a basic buffer?

[Q64 · 4th May Shift 1 · 2023]

Example 2Ionic EquilibriaEASY

What is pH of solution containing 50 mL each of 0.1 M sodium acetate and 0.01 M acetic acid? (

pK_a(\text{CH}_3\text{COOH})=4.50

)

[Q68 · 12th May Shift 1 · 2024]

Example 3Ionic EquilibriaMODERATE

A buffer solution is prepared by mixing 0.2 M NH

_4

OH and 1 M NH

_4

Cl. What is the pH value of buffer solution? (Give

\text{p}K_{b} = 7.744

)

[Q87 · 15th May Shift 1 · 2023]

Example 4Ionic EquilibriaEASY

Acidic buffer solution is prepared by mixing proportionate quantity of

[Q62 · 15th May Shift 2 · 2023]

Example 5Ionic EquilibriaEASY

Calculate the pH of buffer solution containing 0.04 M NaF and 0.02 M HF (pK

_a

= 3.142).

[Q86 · 9th May Shift 2 · 2023]

Drill every past-year question on this subtopic

19 questions from the bank — paginated, with cart and Word-export support.

Drill the 19 questions

Related notes

The pH scale and common substances →