Intermediate

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Buffer Solutions & Solubility Equilibrium

Master Buffers and Ksp for JEE & NEET. Covers Henderson-Hasselbalch equation, buffer capacity, Ksp expressions, solubility from Ksp, common ion effect on solubility.

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A buffer solution resists changes in pH upon addition of small amounts of acid or base. This remarkable property — essential in biological systems, industrial processes, and analytical chemistry — arises from the equilibrium between a weak acid (or base) and its conjugate salt. The Henderson-Hasselbalch equation is the key tool for buffer pH calculations. Alongside buffers, solubility product ( $K_{s p}$ ) governs when a sparingly soluble salt precipitates from solution — a concept directly tested in JEE and NEET through solubility calculations, common ion effect problems, and precipitation conditions. Together these two ideas complete the ionic equilibrium framework built in the previous topic.

1. Buffer Solutions

A buffer is a solution that resists significant changes in pH when small amounts of acid or base are added.

Types of Buffers

Type	Components	pH range	Example
Acidic buffer	Weak acid + its conjugate base (salt)	$p K_{a} \pm 1$	${CH}_{3} COOH + {CH}_{3} COONa$ (pH ≈ 3.74–5.74)
Basic buffer	Weak base + its conjugate acid (salt)	$14 - p K_{b} \pm 1$	${NH}_{3} + {NH}_{4} Cl$ (pH ≈ 8.26–10.26)

2. Henderson-Hasselbalch Equation

For an acidic buffer (weak acid HA + salt NaA):

$pH = p K_{a} + \log \frac{[A^{-}]}{[HA]} = p K_{a} + \log \frac{[salt]}{[acid]}$

For a basic buffer:

$pOH = p K_{b} + \log \frac{[{BH}^{+}]}{[B]} = p K_{b} + \log \frac{[salt]}{[base]}$

Key Properties from Henderson-Hasselbalch

When $[salt] = [acid]$ : $pH = p K_{a}$ (buffer pH equals pKa at equal concentrations — most effective buffer point).
Buffer is most effective in the range $p K_{a} \pm 1$ (ratio of salt/acid between 0.1 and 10).
Diluting a buffer does NOT change its pH significantly (both salt and acid are diluted equally).

Worked Examples

Example 1: Buffer with $0.1 M$ acetic acid + $0.1 M$ sodium acetate ( $K_{a} = 1.8 \times 10^{- 5}$ ; $p K_{a} = 4.74$ ):

$pH = 4.74 + \log \frac{0.1}{0.1} = 4.74 + 0 = 4.74$

Example 2: Buffer with $0.2 M$ acetic acid + $0.1 M$ sodium acetate:

$pH = 4.74 + \log \frac{0.1}{0.2} = 4.74 + \log (0.5) = 4.74 - 0.30 = 4.44$

3. Buffer Capacity

Buffer capacity = number of moles of acid or base that can be added to 1 L of buffer without changing pH by more than 1 unit. It is maximum when $[salt] = [acid]$ (i.e., $pH = p K_{a}$ ). Buffer capacity increases with increasing concentration of buffer components.

4. Solubility Product ( $K_{s p}$ )

For a sparingly soluble salt $M_{x} A_{y}$ dissolving in water:

$M_{x} A_{y} (s) ⇌ x M^{y +} (aq) + y A^{x -} (aq)$

$K_{s p} = [M^{y +}]^{x} [A^{x -}]^{y}$

The solid is excluded (pure solid — constant "concentration").

Solubility and $K_{s p}$ Relationships

Salt type	Dissolution	$K_{s p}$ in terms of $s$	Example
1:1 (e.g., AgCl)	$AgCl ⇌ {Ag}^{+} + {Cl}^{-}$	$K_{s p} = s^{2}$ ; $s = \sqrt{K_{s p}}$	$s = \sqrt{1.8 \times 10^{- 10}} = 1.34 \times 10^{- 5} M$
2:1 (e.g., ${Ag}_{2} {CrO}_{4}$ )	${Ag}_{2} {CrO}_{4} ⇌ 2 {Ag}^{+} + {CrO}_{4}^{2 -}$	$K_{s p} = (2 s)^{2} \cdot s = 4 s^{3}$ ; $s = (K_{s p} / 4)^{1 / 3}$	$K_{s p} = 1.12 \times 10^{- 12}$ ; $s = 6.54 \times 10^{- 5} M$
1:2 (e.g., ${PbI}_{2}$ )	${PbI}_{2} ⇌ {Pb}^{2 +} + 2 I^{-}$	$K_{s p} = s \cdot (2 s)^{2} = 4 s^{3}$ ; $s = (K_{s p} / 4)^{1 / 3}$
1:3 (e.g., ${Fe(OH)}_{3}$ )	${Fe(OH)}_{3} ⇌ {Fe}^{3 +} + 3 {OH}^{-}$	$K_{s p} = s \cdot (3 s)^{3} = 27 s^{4}$ ; $s = (K_{s p} / 27)^{1 / 4}$

5. Precipitation Condition and Ionic Product

The ionic product (IP) at any moment has the same form as $K_{s p}$ but uses current (not equilibrium) concentrations.

Condition	Interpretation
$IP < K_{s p}$	Unsaturated — no precipitation; more salt can dissolve
$IP = K_{s p}$	Saturated — at equilibrium
$IP > K_{s p}$	Supersaturated — precipitation occurs

6. Common Ion Effect on Solubility

Adding a common ion (one already present from the dissolution) decreases the solubility of a sparingly soluble salt.

Example: Solubility of AgCl in $0.1 M NaCl$ :

$K_{s p} = [{Ag}^{+}] [{Cl}^{-}] = s \cdot (0.1 + s) \approx s \times 0.1 = 1.8 \times 10^{- 10}$

$s = \frac{1.8 \times 10^{- 10}}{0.1} = 1.8 \times 10^{- 9} M$ (vs $1.34 \times 10^{- 5} M$ in pure water)

The presence of ${Cl}^{-}$ from NaCl reduces AgCl solubility by a factor of ~7400.

$K_{s p}$ and Solubility — Key Formulae

Salt	$K_{s p}$ expression	Solubility $s$
AB type (AgCl)	$K_{s p} = s^{2}$	$s = \sqrt{K_{s p}}$
AB₂ or A₂B type	$K_{s p} = 4 s^{3}$	$s = (K_{s p} / 4)^{1 / 3}$
AB₃ or A₃B type	$K_{s p} = 27 s^{4}$	$s = (K_{s p} / 27)^{1 / 4}$
Common ion present	$K_{s p} = s \times [common ion]$	$s = K_{s p} / [common ion]$

JEE / NEET Power Tips — Buffers & Solubility

When $[salt] = [acid]$ , pH = pKa exactly. This is both the formula result and the condition of maximum buffer capacity. It's the most tested buffer fact.
Diluting a buffer does not change pH. Both acid and salt concentrations decrease by the same factor, so the ratio $[salt] / [acid]$ stays constant, and $pH = p K_{a} + \log (constant)$ stays constant.
Higher $K_{s p}$ does NOT always mean higher solubility between two salts of different stoichiometry. For example, $AgCl$ ( $K_{s p} = 1.8 \times 10^{- 10}$ , 1:1) vs ${Ag}_{2} {CrO}_{4}$ ( $K_{s p} = 1.1 \times 10^{- 12}$ , 2:1): $s (AgCl) = 1.34 \times 10^{- 5} M$ ; $s ({Ag}_{2} {CrO}_{4}) = 6.5 \times 10^{- 5} M$ . Despite lower $K_{s p}$ , Ag₂CrO₄ is MORE soluble. Always calculate $s$ from the $K_{s p}$ expression.
Common ion effect suppresses solubility and can be used for selective precipitation. Adding NaCl to a solution containing both ${Ag}^{+}$ and ${Pb}^{2 +}$ will preferentially precipitate AgCl first (lower $K_{s p}$ ), allowing separation of the two ions.

Practice Questions

Q1 (JEE Main / NEET): Calculate the pH of a buffer made by mixing $0.3 mol$ acetic acid and $0.1 mol$ sodium acetate in 1 L. ( $p K_{a} = 4.74$ )

Explanation:

For an acidic buffer, use the Henderson-Hasselbalch equation:
$pH = p K_{a} + \log (\frac{[salt]}{[acid]})$

Substitute the given values (since the volume is 1 L, moles = molarity):
$pH = 4.74 + \log (\frac{0.1}{0.3})$
$pH = 4.74 + \log (\frac{1}{3})$
$pH = 4.74 - \log (3)$
$pH = 4.74 - 0.477 = 4.26$

Q2 (JEE Main): The $K_{s p}$ of ${Ag}_{2} {CrO}_{4}$ is $1.12 \times 10^{- 12}$ . Calculate its molar solubility.

Explanation:

Write the dissociation equation for silver chromate:
${Ag}_{2} {CrO}_{4} (s) ⇌ 2 {Ag}^{+} (aq) + {CrO}_{4}^{2 -} (aq)$

If the molar solubility is $s$ , then:
$[{Ag}^{+}] = 2 s$
$[{CrO}_{4}^{2 -}] = s$

Set up the solubility product expression:
$K_{s p} = [{Ag}^{+}]^{2} [{CrO}_{4}^{2 -}]$
$K_{s p} = (2 s)^{2} (s) = 4 s^{3}$

Solve for $s$ :
$4 s^{3} = 1.12 \times 10^{- 12}$
$s^{3} = 0.28 \times 10^{- 12} = 280 \times 10^{- 15}$
$s = (280 \times 10^{- 15})^{1 / 3} \approx 6.54 \times 10^{- 5} mol/L$

Q3 (NEET): Will $AgCl$ precipitate if $0.001 M {AgNO}_{3}$ is mixed with $0.001 M NaCl$ ? ( $K_{s p} (AgCl) = 1.8 \times 10^{- 10}$ )

Explanation:

When mixing two equal volumes of solutions, the total volume doubles, causing the concentration of each ion to halve.
$[{Ag}^{+}] = \frac{0.001}{2} = 0.0005 M = 5 \times 10^{- 4} M$
$[{Cl}^{-}] = \frac{0.001}{2} = 0.0005 M = 5 \times 10^{- 4} M$

Calculate the Ionic Product (IP or $Q$ ):
$IP = [{Ag}^{+}] [{Cl}^{-}] = (5 \times 10^{- 4}) (5 \times 10^{- 4}) = 2.5 \times 10^{- 7}$

Compare IP with $K_{s p}$ :
Since $IP = 2.5 \times 10^{- 7} > K_{s p} = 1.8 \times 10^{- 10}$ , the solution is supersaturated.
Yes, $AgCl$ will precipitate.

Q4 (JEE Main): What is the solubility of $AgCl$ in a $0.1 M NaCl$ solution? ( $K_{s p} = 1.8 \times 10^{- 10}$ ). Compare it with its solubility in pure water.

Explanation:

Part 1: Solubility in pure water ( $s_{0}$ )
$s_{0} = \sqrt{K_{s p}} = \sqrt{1.8 \times 10^{- 10}} = 1.34 \times 10^{- 5} M$

Part 2: Solubility in $0.1 M NaCl$ ( $s$ )
$NaCl$ is a strong electrolyte and fully dissociates, providing a common ion $[{Cl}^{-}] = 0.1 M$ .
Let the new solubility be $s$ . The total chloride concentration is $(s + 0.1)$ . Because $s$ is extremely small compared to $0.1$ , we approximate $[{Cl}^{-}] \approx 0.1 M$ .

$K_{s p} = [{Ag}^{+}] [{Cl}^{-}]$
$1.8 \times 10^{- 10} = s \times 0.1$
$s = \frac{1.8 \times 10^{- 10}}{0.1} = 1.8 \times 10^{- 9} M$

Comparison: The solubility drops from $\sim 10^{- 5}$ to $\sim 10^{- 9}$ . The presence of the common ion reduces the solubility of $AgCl$ by a factor of roughly $7400$ .

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Buffer Solutions & Solubility Equilibrium

1. Buffer Solutions

Types of Buffers

2. Henderson-Hasselbalch Equation

Key Properties from Henderson-Hasselbalch

Worked Examples

3. Buffer Capacity

4. Solubility Product ( $K_{s p}$ )

Solubility and $K_{s p}$ Relationships

5. Precipitation Condition and Ionic Product

6. Common Ion Effect on Solubility

$K_{s p}$ and Solubility — Key Formulae

Practice Questions

Related Study Material

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Buffer Solutions & Solubility Equilibrium

1. Buffer Solutions

Types of Buffers

2. Henderson-Hasselbalch Equation

Key Properties from Henderson-Hasselbalch

Worked Examples

3. Buffer Capacity

4. Solubility Product (Ksp)

Solubility and Ksp Relationships

5. Precipitation Condition and Ionic Product

6. Common Ion Effect on Solubility

Ksp and Solubility — Key Formulae

Practice Questions

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4. Solubility Product ( $K_{s p}$ )

Solubility and $K_{s p}$ Relationships

$K_{s p}$ and Solubility — Key Formulae