JEE Mains Chemistry · Equilibrium

Buffer Solutions

Finding the pH of a weak acid or base mixed with its salt through the Henderson equation, including buffers made by part-neutralising a weak acid or base and buffers after strong acid is added.

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Why this matters

Fourteen PYQs, seven of them numerical, and four from 2026. Seven apply the Henderson equation to a stated mixture; seven make the buffer by part-neutralising a weak acid or base, or add strong acid to one.

Concept 1 of 2: The Henderson equation

A buffer holds a weak acid and its conjugate base in comparable amounts. Added acid is taken up by the base, and added base by the acid, so the pH barely moves. Taking the log of Ka gives the pH directly from the ratio of the two.

Definition

Acidic buffer (weak acid + its salt): $\mathrm{pH}=\mathrm{p}K_a+\log\frac{[\text{salt}]}{[\text{acid}]}$ .
Basic buffer (weak base + its salt): $\mathrm{pOH}=\mathrm{p}K_b+\log\frac{[\text{salt}]}{[\text{base}]}$ .
For the base $\mathrm{NH_3}$ , $\mathrm{p}K_a(\mathrm{NH_4^+})=14-\mathrm{p}K_b$ .
Both parts share one volume, so a mole ratio works as well as a concentration ratio.
A buffer needs a weak acid or base with its conjugate, in comparable amounts. HCl with NaCl is not a buffer. Blood is buffered by $\mathrm{H_2CO_3/HCO_3^-}$ .
If a weak acid is ionised to a fraction x, then $\mathrm{pH}-\mathrm{p}K_a=\log\frac{x}{1-x}$ .

Henderson equation

\mathrm{pH}=\mathrm{p}K_a+\log\frac{[\text{salt}]}{[\text{acid}]}

Worked example

A solution is 0.2 M in a weak acid HA (

\mathrm{p}K_a=4.7

) and 0.4 M in its sodium salt. Find the pH. (

\log2=0.30

)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2022 · 28 June 2022 · Q124Moderate

Example 1 · Equilibrium · Buffer Solutions

A student needs to prepare a buffer solution of propanoic acid and its sodium salt with

pH4

. The ratio of

\frac{\left\lbrack CH_{3}CH_{2}COO^{-} \right\rbrack}{\left\lbrack CH_{3}CH_{2}COOH \right\rbrack}

required to make buffer is Given:

K_{a}\left( CH_{3}CH_{2}COOH \right)= 1.3 \times10^{- 5}

The ratio flips for a basic buffer

pOH uses salt over base:

\mathrm{pOH}=\mathrm{p}K_b+\log\frac{[\text{salt}]}{[\text{base}]}

. Written for pH it becomes

\mathrm{pH}=\mathrm{p}K_a+\log\frac{[\text{base}]}{[\text{salt}]}

. Mixing the two forms gives a pH on the wrong side of

\mathrm{p}K_a

Salt over acid, not acid over salt

For a weak acid ionised to a fraction x, the salt form is x and the acid form is

1-x

, so

\mathrm{pH}-\mathrm{p}K_a=\log\frac{x}{1-x}

. The inverted fraction is always offered.

Drill 6 more on the henderson equation

Concept 2 of 2: Buffers made by part-neutralisation

Mix a weak acid with less than an equal amount of strong base. The strong base is used up completely, and each mole of it turns one mole of weak acid into its salt. What is left is a weak acid with its salt, which is a buffer. Adding strong acid to a buffer works the same way in reverse.

Definition

Weak acid $n_a$ + strong base $n_b$ with $n_b<n_a$ : salt $=n_b$ , acid left $=n_a-n_b$ .
$\mathrm{pH}=\mathrm{p}K_a+\log\frac{n_b}{n_a-n_b}$ .
Half-neutralised ( $n_b=\tfrac12n_a$ ): $\mathrm{pH}=\mathrm{p}K_a$ .
Strong reagent equal to or more than the weak one: no buffer. It is a salt solution, or the excess strong reagent sets the pH.
Strong acid added to a basic buffer turns base into salt: base falls, salt rises by the same amount.
With equal molarities, the volumes can stand in for moles.

Weak acid part-neutralised by strong base

\mathrm{pH}=\mathrm{p}K_a+\log\frac{n_b}{n_a-n_b}

Worked example

15 mL of 0.2 M NaOH is added to 40 mL of 0.1 M weak acid HA (

\mathrm{p}K_a=4.8

). Find the pH. (

\log3=0.477

)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 13 April 2023 · Q146Moderate

Example 2 · Equilibrium · Buffer Solutions

20\text{ }mL

0.1MNaOH

is added to

50\text{ }mL

0.1M

acetic acid solution. The

pH

of the resulting solution is____

\times10^{- 2}

(Nearest integer) Given:

pKa\left( CH_{3}COOH \right)= 4.76

\log2 = 0.30

\log3 = 0.48

Base left is y minus x, not y

Mixing x mL of HCl with y mL of a weak base of the same molarity leaves salt x and base

y-x

. Putting

x/y

into the Henderson equation instead of

x/(y-x)

gives a wrong pair of volumes that is always among the options.

Excess strong acid is not a buffer

\mathrm{NH_4OH}

with an equal or larger amount of HCl leaves no free base. For a pH of

\mathrm{p}K_a(\mathrm{NH_4^+})=9.25

the mixture must hold equal amounts of

\mathrm{NH_3}

and

\mathrm{NH_4^+}

, so the base must be exactly twice the acid.

Drill 6 more on buffers made by part-neutralisation

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 (2)

The Henderson equation
Henderson equation
$\mathrm{pH}=\mathrm{p}K_a+\log\frac{[\text{salt}]}{[\text{acid}]}$
Buffers made by part-neutralisation
Weak acid part-neutralised by strong base
$\mathrm{pH}=\mathrm{p}K_a+\log\frac{n_b}{n_a-n_b}$

Watch out for (4)

The ratio flips for a basic buffer
→ The Henderson equation
Salt over acid, not acid over salt
→ The Henderson equation
Base left is y minus x, not y
→ Buffers made by part-neutralisation
Excess strong acid is not a buffer
→ Buffers made by part-neutralisation

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