Solubility Equilibria Of Sparingly Soluble Salts

ChemistryEquilibriumFor JEE aspirants

The solubility product $K_{s p}$ is the equilibrium constant for the dissolution of a sparingly soluble salt into its constituent ions. For $A_{x} B_{y} (s) ⇌ x A^{y +} + y B^{x -}$ , $K_{s p} = [A^{y +}]^{x} [B^{x -}]^{y} = x^{x} y^{y} (s)^{x + y}$ , where $s$ is the molar solubility. The ionic product $Q$ compared with $K_{s p}$ predicts whether a precipitate will form: $Q > K_{s p}$ means precipitation, $Q = K_{s p}$ means saturation, $Q < K_{s p}$ means undersaturation. Solubility is suppressed by common ions, altered by pH (for salts of weak acids), and dramatically increased by complex formation.

Key Formulas - Quick Reference

General $K_{s p}$ : For $A_{x} B_{y} ⇌ x A^{y +} + y B^{x -}$ , $K_{s p} = x^{x} \cdot y^{y} \cdot s^{x + y}$ .
1:1 salt (AgCl): $K_{s p} = s^{2}$ ; $s = K_{s p}$ .
1:2 salt (CaF $_{2}$ ): $K_{s p} = 4 s^{3}$ ; $s = (K_{s p} /4)^{1/3}$ .
2:1 salt (Ag $_{2}$ CrO $_{4}$ ): $K_{s p} = 4 s^{3}$ ; $s = (K_{s p} /4)^{1/3}$ .
1:3 salt (Fe(OH) $_{3}$ ): $K_{s p} = 27 s^{4}$ ; $s = (K_{s p} /27)^{1/4}$ .
Precipitation condition: $Q > K_{s p}$ .
Common ion effect: solubility in presence of common ion $c$ is $s \approx K_{s p} / c^{n}$ where $n$ is the stoichiometry.
Solubility in buffered acid: $s = K_{s p} (1 + \frac{[ H ^{+} ]}{K _{a 2}} + \frac{[ H ^{+} ] ^{2}}{K _{a 1} K _{a 2}})$ for MS in $H_{2} S$ .

1. Solubility and Molar Solubility

Solubility is the maximum amount of a salt that dissolves in a given amount of solvent at a given temperature to form a saturated solution. When expressed in moles of dissolved salt per litre of solution, it is called molar solubility and denoted $s$ .

Sparingly soluble salts have $s$ typically below $1 0^{- 2}$ M. Examples: $A g C l$ , $B a S O_{4}$ , $C a F_{2}$ , $F e (O H)_{3}$ , $A g_{2} C r O_{4}$ , most metal sulphides. These salts form dynamic equilibria in solution:

A_{x} B_{y} (s) ⇌ x A^{y +} (a q) + y B^{x -} (a q)

Figure 1: Dynamic equilibrium in a saturated solution. Ions dissolve and crystallise at equal rates, and

K_{s p} = [A g^{+}] [C l^{-}]

is fixed at any given temperature.

2. Solubility Product Ksp - Definition

Because the activity of a pure solid is 1, the equilibrium constant simplifies to:

K_{s p} = [A^{y +}]^{x} \cdot [B^{x -}]^{y}

The solubility product $K_{s p}$ is the product of ionic concentrations (each raised to its stoichiometric coefficient) in a saturated solution at a given temperature. Like any equilibrium constant, $K_{s p}$ depends only on temperature.

Smaller $K_{s p}$ means less soluble. For $A g C l$ , $K_{s p} \approx 1.8 \times 1 0^{- 10}$ ; for $B a S O_{4}$ , $K_{s p} \approx 1.1 \times 1 0^{- 10}$ ; for $F e (O H)_{3}$ , $K_{s p} \approx 4 \times 1 0^{- 38}$ (extremely insoluble).

3. General Expression: Ksp in Terms of Solubility

If molar solubility is $s$ , then $[A^{y +}] = x s$ and $[B^{x -}] = y s$ . Substituting:

K_{s p} = (x s)^{x} \cdot (y s)^{y} = x^{x} \cdot y^{y} \cdot s^{x + y}

Example: $L i_{3} N a_{3} [A l F_{6}]_{2}$ produces 3 $L i^{+}$ , 3 $N a^{+}$ , 2 $A l F_{6}^{3 -}$ per formula unit, so $K_{s p} = 3^{3} \cdot 3^{3} \cdot 2^{2} \cdot s^{8} = 2916 s^{8}$ .

4. Ksp for Different Salt Types

Salt type	Example	Ionisation	Ksp in terms of s
AB (1:1)	AgCl, BaSO $_{4}$	$s + s$	$s^{2}$
AB $_{2}$ (1:2)	CaF $_{2}$ , PbCl $_{2}$	$s + 2 s$	$4 s^{3}$
A $_{2}$ B (2:1)	Ag $_{2}$ CrO $_{4}$ , Ag $_{2}$ S	$2 s + s$	$4 s^{3}$
AB $_{3}$ (1:3)	Al(OH) $_{3}$ , Fe(OH) $_{3}$	$s + 3 s$	$27 s^{4}$
A $_{3}$ B (3:1)	Ag $_{3}$ PO $_{4}$	$3 s + s$	$27 s^{4}$
A $_{2}$ B $_{3}$ (2:3)	Ca $_{3}$ (PO $_{4}$ ) $_{2}$	$3 s + 2 s$	$108 s^{5}$

5. Ionic Product Q vs Ksp - Condition for Precipitation

For any actual mixture of ions, the ionic product $Q$ has the same algebraic form as $K_{s p}$ :

Q = [A^{y +}]^{x} \cdot [B^{x -}]^{y}

(using current concentrations)

Condition	State of solution	What happens
$Q < K_{s p}$	Undersaturated	More solid can dissolve
$Q = K_{s p}$	Saturated	Dynamic equilibrium
$Q > K_{s p}$	Supersaturated	Precipitation occurs

This gives a clean quantitative test for whether mixing two solutions will produce a precipitate.

6. Common Ion Effect on Solubility

Adding a source of one of the salt's ions (a common ion) suppresses its dissolution:

For

A g C l

in a solution already 0.1 M in

C l^{-}

K_{s p} = [A g^{+}] \cdot 0.1 \approx 1.8 \times 1 0^{- 10}

, so

[A g^{+}] \approx 1.8 \times 1 0^{- 9}

In pure water, $[A g^{+}] = [C l^{-}] = K_{s p} \approx 1.3 \times 1 0^{- 5}$ M. Adding 0.1 M $C l^{-}$ drops silver solubility by a factor of about $1 0^{4}$ .

7. Salt Effect (Increased Solubility from Inert Electrolytes)

Adding a non-common-ion electrolyte (like $N a N O_{3}$ to an $A g C l$ suspension) slightly increases solubility. The added ions reduce the activity coefficients of $A g^{+}$ and $C l^{-}$ , so higher molar concentrations are needed to maintain the same activity product. In JEE problems this effect is usually neglected unless activities are explicitly discussed.

8. Simultaneous Solubility of Two Salts with Common Ion

If two sparingly soluble salts sharing a common ion are dissolved together, each equilibrium expression uses the total concentration of the shared ion. For $A g S C N$ (solubility $x$ ) and $A g B r$ (solubility $y$ ) in the same solution:

K_{s p} (A g S C N) = (x + y) \cdot x

;

K_{s p} (A g B r) = (x + y) \cdot y

Given the two $K_{s p}$ values, we can solve the two equations for $x$ and $y$ . Both solubilities are lower than they would be if each salt were dissolved separately.

9. Fractional / Selective Precipitation JEE Advanced

When a mixture contains two ions that can form insoluble salts with the same reagent, we can precipitate one before the other by carefully controlling the reagent concentration.

Worked Example: Separating Ba²⁺ from Sr²⁺

Setup: both ions at 0.1 M, chromate ( $C r O_{4}^{2 -}$ ) added slowly as the precipitating reagent.

$B a C r O_{4}$	$K_{s p} = 1.2 \times 1 0^{- 10}$	less soluble
$S r C r O_{4}$	$K_{s p} = 3.5 \times 1 0^{- 5}$	more soluble

Threshold [CrO₄²⁻] to begin precipitation:

$B a C r O_{4}$ : $[C r O_{4}^{2 -}] = \frac{1.2 \times 1 0 ^{- 10}}{0.1} = 1.2 \times 1 0^{- 9}$ M

$S r C r O_{4}$ : $[C r O_{4}^{2 -}] = \frac{3.5 \times 1 0 ^{- 5}}{0.1} = 3.5 \times 1 0^{- 4}$ M

Because $B a C r O_{4}$ 's threshold is ~10⁵ times lower, it precipitates almost completely before $S r C r O_{4}$ starts to form.

Result: When

[C r O_{4}^{2 -}]

finally reaches

3.5 \times 1 0^{- 4}

M (the moment

S r C r O_{4}

begins to precipitate), the remaining

[B a^{2 +}]

in solution is:

$[B a^{2 +}] = \frac{1.2 \times 1 0 ^{- 10}}{3.5 \times 1 0 ^{- 4}} \approx 3.4 \times 1 0^{- 7}$ M

That is 0.00034% of the original — near-complete separation of the two ions.

Figure 2: Fractional precipitation of

C d^{2 +}

and

Z n^{2 +}

using

H_{2} S

at controlled

[H^{+}]

. Because

K_{s p} (C d S) ≪ K_{s p} (Z n S)

, cadmium precipitates first at high

[H^{+}]

, and zinc is left in solution to be recovered separately by lowering

[H^{+}]

10. Solubility of Salts of Weak Acids (Hydrolysis Effect)

Salts like $C a C O_{3}$ , $Z n S$ , and $A g_{2} S$ have anions that are weak conjugate bases and hydrolyse in water:

C O_{3}^{2 -} + H_{2} O ⇌ H C O_{3}^{-} + O H^{-}

Hydrolysis removes free $C O_{3}^{2 -}$ from solution, lowering $Q$ below $K_{s p}$ , so more solid dissolves to compensate. The measured solubility is therefore higher than a naive $s = K_{s p}$ calculation would predict, especially in acidic media where the effect is amplified.

11. Solubility of Salts of Weak Acids in Acidic Medium

Salts whose anion is a weak conjugate base (like $C a C O_{3}$ , $Z n S$ , $A g_{2} S$ , $M g (O H)_{2}$ ) dissolve more readily in acidic solution than in neutral water. The mechanism:

In acidic medium, added $H^{+}$ reacts with the anion: $C O_{3}^{2 -} + 2 H^{+} \to H_{2} C O_{3}$ (or $S^{2 -} + 2 H^{+} \to H_{2} S$ ).
Free anion concentration falls, so the ionic product $Q$ drops below $K_{s p}$ .
To re-establish equilibrium, more solid dissolves, raising the metal ion concentration.

Salts of strong acids (like $B a S O_{4}$ , $A g C l$ ) show little pH sensitivity because their anions do not accept protons.

Application in qualitative analysis: Group II sulphides (

C u S

C d S

H g S

P b S

) have very small

K_{s p}

and precipitate even in dilute HCl where

[S^{2 -}]

is tiny. Group III B sulphides (

Z n S

M n S

C o S

N i S

) have larger

K_{s p}

and only precipitate when the medium is made basic (using

N H_{4} C l / N H_{4} O H

) to raise

[S^{2 -}]

. This is exactly how the classical scheme separates cations by controlling pH.

12. Solubility with Complex Formation (AgCl in NH₃) JEE Advanced

Some metal ions form stable complexes with ligands like ammonia, cyanide, or thiosulphate. Complex formation removes free metal ion from solution, driving more solid to dissolve:

A g C l (s) ⇌ A g^{+} + C l^{-}

;

K_{s p} = 1.8 \times 1 0^{- 10}

A g^{+} + 2 N H_{3} ⇌ [A g (N H_{3})_{2}]^{+}

;

K_{f} = \frac{[ A g ( N H _{3} ) _{2}^{+} ]}{[ A g ^{+} ] [ N H _{3} ] ^{2}}

Combining these, the effective dissolution reaction is:

A g C l (s) + 2 N H_{3} ⇌ [A g (N H_{3})_{2}]^{+} + C l^{-}

;

K = K_{s p} \cdot K_{f}

Even sparingly soluble $A g C l$ dissolves readily in concentrated ammonia because the very large $K_{f}$ ( $\sim 1 0^{7}$ ) more than compensates for the small $K_{s p}$ . This is the basis of Tollens' reagent and of separating $A g C l$ from $A g B r$ or $A g I$ (which have smaller $K_{s p}$ and do not dissolve as easily).

13. Application: Qualitative Analysis - Group Separation

In classical inorganic qualitative analysis, cations are separated into groups by controlled sulphide precipitation:

Group II (Cu, Cd, Hg, Pb, Bi, Sn, As, Sb): precipitated as sulphides in dilute HCl. High $[H^{+}]$ keeps $[S^{2 -}]$ very low, so only very-low- $K_{s p}$ sulphides precipitate.
Group III B (Zn, Mn, Co, Ni): precipitated as sulphides in neutral/basic $N H_{4} C l$ / $N H_{4} O H$ buffer. Lower $[H^{+}]$ raises $[S^{2 -}]$ enough to precipitate the moderately soluble sulphides.

Controlling $[H^{+}]$ (and hence $[S^{2 -}]$ ) via a buffer is what makes this classical scheme work.

14. Application: Gravimetric Analysis (BaSO₄)

Sulphate ion is determined gravimetrically by precipitation as $B a S O_{4}$ using excess $B a C l_{2}$ :

B a^{2 +} + S O_{4}^{2 -} \to B a S O_{4} (s)

;

K_{s p} \approx 1.1 \times 1 0^{- 10}

The very small $K_{s p}$ means precipitation is essentially complete, and the common ion effect from excess $B a^{2 +}$ further reduces residual sulphate to negligible levels. The precipitate is filtered, dried, and weighed to determine the original sulphate content with high accuracy.

Solved Example 1

Calculate

K_{s p}

L i_{3} N a_{3} [A l F_{6}]_{2}

if its molar solubility is

s

Solution:

$L i_{3} N a_{3} [A l F_{6}]_{2} \to 3 L i^{+} + 3 N a^{+} + 2 [A l F_{6}]^{3 -}$ .

$K_{s p} = [L i^{+}]^{3} [N a^{+}]^{3} [A l F_{6}^{3 -}]^{2} = (3 s)^{3} (3 s)^{3} (2 s)^{2} = 27 \cdot 27 \cdot 4 \cdot s^{8} = 2916 s^{8}$ .

Solved Example 2

Calculate the solubility of silver oxalate (

A g_{2} C_{2} O_{4}

) in

1 0^{- 2}

M potassium oxalate solution.

K_{s p} (A g_{2} C_{2} O_{4}) = 1 0^{- 10}

Solution:

Let solubility of $A g_{2} C_{2} O_{4}$ be $x$ . Then $[A g^{+}] = 2 x$ and $[C_{2} O_{4}^{2 -}] = 1 0^{- 2} + x \approx 1 0^{- 2}$ (common ion dominates).

$K_{s p} = (2 x)^{2} (1 0^{- 2}) = 1 0^{- 10}$ ; $4 x^{2} = 1 0^{- 8}$ ; $x^{2} = 2.5 \times 1 0^{- 9}$ ; $x = 5 \times 1 0^{- 5}$ M.

Solved Example 3

Calculate the simultaneous solubility of

A g S C N

(

K_{s p} = 1 0^{- 12}

) and

A g B r

(

K_{s p} = 5 \times 1 0^{- 13}

) in the same aqueous solution.

Solution:

Let $x$ = solubility of $A g S C N$ , $y$ = solubility of $A g B r$ . Total $[A g^{+}] = x + y$ ; $[S C N^{-}] = x$ ; $[B r^{-}] = y$ .

Two equations: $(x + y) x = 1 0^{- 12}$ and $(x + y) y = 5 \times 1 0^{- 13}$ .

Dividing: $x / y = 2$ , so $x = 2 y$ . Substitute into (2): $(3 y) (y) = 5 \times 1 0^{- 13} \Rightarrow y^{2} = 1.67 \times 1 0^{- 13} \Rightarrow y = 4.08 \times 1 0^{- 7}$ M; $x = 8.16 \times 1 0^{- 7}$ M.

Solved Example 4

1 0^{- 5}

N a C l

and

1 0^{- 8}

A g N O_{3}

are mixed in 1:1 volume ratio. Will

A g C l

precipitate?

K_{s p} (A g C l) = 1 0^{- 10}

Solution:

After 1:1 mixing, $[C l^{-}] = 5 \times 1 0^{- 6}$ M, $[A g^{+}] = 5 \times 1 0^{- 9}$ M.

$Q = [A g^{+}] [C l^{-}] = 5 \times 1 0^{- 9} \times 5 \times 1 0^{- 6} = 2.5 \times 1 0^{- 14}$ .

$Q < K_{s p} = 1 0^{- 10}$ , so no precipitation.

Solved Example 5

Solubility product of

S r F_{2}

in water is

8 \times 1 0^{- 10}

. Find its solubility in 0.1 M NaF.

Solution:

$S r F_{2} \to S r^{2 +} + 2 F^{-}$ ; $K_{s p} = [S r^{2 +}] [F^{-}]^{2}$ .

Let $[S r^{2 +}] = s$ ; $[F^{-}] = 2 s + 0.1 \approx 0.1$ (common ion dominates).

$s \times (0.1)^{2} = 8 \times 1 0^{- 10} \Rightarrow s = 8 \times 1 0^{- 8}$ M.

Solved Example 6

What

[H^{+}]

must be maintained in a saturated

H_{2} S

solution (0.1 M) to precipitate

C d S

but not

Z n S

, if

[C d^{2 +}] = [Z n^{2 +}] = 0.1

K_{s p} (C d S) = 8 \times 1 0^{- 27}

;

K_{s p} (Z n S) = 1 0^{- 21}

;

K_{a} (H_{2} S)

overall =

1.1 \times 1 0^{- 21}

Solution:

To prevent $Z n S$ precipitation: $[Z n^{2 +}] [S^{2 -}] < K_{s p} (Z n S)$ ; $0.1 \cdot [S^{2 -}] < 1 0^{- 21}$ ; $[S^{2 -}] < 1 0^{- 20}$ M.

Using $K_{a}$ for $H_{2} S$ : $K_{a} = \frac{[ H ^{+} ] ^{2} [ S ^{2 -} ]}{[ H _{2} S ]} = \frac{x ^{2} \cdot 1 0 ^{- 20}}{0.1} = 1.1 \times 1 0^{- 21}$ .

$x^{2} = 1.1 \times 1 0^{- 2}$ ; $x = [H^{+}] = 0.105$ M.

Any $[H^{+}] > 0.105$ M keeps $[S^{2 -}]$ low enough that $Z n S$ stays in solution while $C d S$ still precipitates (because $K_{s p} (C d S)$ is much smaller).

Solved Example 7

What concentration of

N H_{3}

must be added to a solution of

4 \times 1 0^{- 3}

A g^{+}

containing 0.001 M

N a C l

to prevent

A g C l

precipitation?

K_{s p} (A g C l) = 1.8 \times 1 0^{- 10}

; formation constant

K_{f} = 1 0^{8} /6

for

[A g (N H_{3})_{2}]^{+}

Solution:

Maximum $[A g^{+}]$ tolerated: $[A g^{+}] = K_{s p} / [C l^{-}] = 1.8 \times 1 0^{- 10} /1 0^{- 3} = 1.8 \times 1 0^{- 7}$ M.

Nearly all silver must be complexed: $[A g (N H_{3})_{2}^{+}] \approx 4 \times 1 0^{- 3}$ M; free $[A g^{+}] = 1.8 \times 1 0^{- 7}$ M.

$K_{f} = \frac{[ A g ( N H _{3} ) _{2}^{+} ]}{[ A g ^{+} ] [ N H _{3} ] ^{2}} = \frac{4 \times 1 0 ^{- 3}}{1.8 \times 1 0 ^{- 7} \cdot [ N H _{3} ] ^{2}} = \frac{1 0 ^{8}}{6}$

$[N H_{3}]^{2} = \frac{4 \times 1 0 ^{- 3} \cdot 6}{1.8 \times 1 0 ^{- 7} \cdot 1 0 ^{8}} = 1.33 \times 1 0^{- 3}$ ; $[N H_{3}] \approx 0.037$ M.

Add slightly more (to also cover the $N H_{3}$ tied up in the complex): total $\approx 0.037 + 2 \times 4 \times 1 0^{- 3} \approx 0.045$ M.

Solved Example 8

Calculate the solubility of

B a S O_{4}

when

B a S O_{4}

and

C a S O_{4}

dissolve simultaneously.

K_{s p} (B a S O_{4}) = q

K_{s p} (C a S O_{4}) = p

; solubility of

C a S O_{4}

b

mol/L.

Solution:

Total $[S O_{4}^{2 -}]$ comes mainly from $C a S O_{4}$ dissolution (much more soluble): $[S O_{4}^{2 -}] \approx b$ .

For $B a S O_{4}$ : $[B a^{2 +}] [S O_{4}^{2 -}] = q$ ; $[B a^{2 +}] = q / b$ .

Solubility of $B a S O_{4}$ = $[B a^{2 +}] = q/b$ mol/L.

(A cross-check: dividing the two $K_{s p}$ expressions, $[B a^{2 +}] / [C a^{2 +}] = q / p$ , so $[B a^{2 +}] = b \cdot q / p$ ; both forms are equivalent given $b = p$ .)

Common Mistakes to Avoid

Watch out

Forgetting the stoichiometric coefficient inside the concentration bracket. For $A g_{2} C r O_{4} \to 2 A g^{+} + C r O_{4}^{2 -}$ , $[A g^{+}] = 2 s$ and $K_{s p} = (2 s)^{2} (s) = 4 s^{3}$ , not $s^{2} \cdot s$ .
Comparing $K_{s p}$ values directly to rank solubility across different salt types. A 1:1 salt with $K_{s p} = 1 0^{- 8}$ has $s \approx 1 0^{- 4}$ , but a 1:2 salt with the same $K_{s p}$ has $s = (1 0^{- 8} /4)^{1/3} \approx 1 0^{- 3}$ - actually more soluble.
Neglecting the common ion when the added ion concentration is comparable to salt solubility. Only ignore the salt's own contribution when the added ion is much larger.
Treating the ionic product $Q$ as always equal to $K_{s p}$ . Comparing them is the whole point of testing for precipitation.
Assuming the presence of a strong acid always increases solubility. It only does so for salts of weak acids (where the anion hydrolyses); for salts of strong acids like $B a S O_{4}$ , changing pH has little effect.
Forgetting activity effects. At high ionic strength, $K_{s p}$ formulas using concentrations become approximate. For JEE-level problems this is usually ignored.
Missing that complex formation dramatically increases solubility. $A g C l$ dissolves easily in excess $N H_{3}$ because $[A g (N H_{3})_{2}]^{+}$ ties up free $A g^{+}$ .
Using solubility from pure water when the salt is being dissolved in a solution already containing common ions - always start over with the appropriate ion concentrations.

Frequently Asked Questions

Q1. What is the difference between solubility and solubility product?

Solubility ( $s$ ) is the number of moles of salt that dissolves per litre to give a saturated solution - it has units of mol/L. Solubility product ( $K_{s p}$ ) is the equilibrium constant for the dissolution reaction, equal to the product of the ionic concentrations in a saturated solution, each raised to its stoichiometric coefficient. The two are related by expressions like $K_{s p} = s^{2}$ for AB salts or $K_{s p} = 4 s^{3}$ for AB $_{2}$ salts.

Q2. How do I decide whether a precipitate will form when two solutions are mixed?

Compute the ionic product $Q$ using the concentrations right after mixing (accounting for dilution). Compare $Q$ with $K_{s p}$ : if $Q > K_{s p}$ , precipitation occurs until $Q$ falls back to $K_{s p}$ ; if $Q = K_{s p}$ , the solution is exactly saturated; if $Q < K_{s p}$ , no precipitation and more solid could dissolve.

Q3. Why does adding a common ion decrease solubility?

The dissolution equilibrium $A B (s) ⇌ A^{+} + B^{-}$ has $K_{s p}$ fixed at constant temperature. If we increase one ion (say $B^{-}$ ) by adding a source of it, then $[A^{+}]$ must decrease to keep the product constant. So less salt can dissolve. Quantitatively, if a common ion is at concentration $c$ , solubility drops to about $K_{s p} / c$ .

Q4. Does temperature affect Ksp?

Yes. Like every equilibrium constant, $K_{s p}$ depends on temperature via the Van't Hoff equation. For most salts, dissolution is endothermic, so $K_{s p}$ (and hence solubility) increases with temperature - this is why more sugar or salt dissolves in hot water. A few salts have endothermic dissolution reversed at high $T$ and become less soluble.

Q5. Why does AgCl dissolve in ammonia solution?

$A g^{+}$ forms a very stable complex $[A g (N H_{3})_{2}]^{+}$ with ammonia (formation constant $K_{f} \sim 1 0^{7}$ ). This complexation removes free $A g^{+}$ from solution, so the equilibrium $A g C l (s) ⇌ A g^{+} + C l^{-}$ shifts to the right and more $A g C l$ dissolves. The net reaction $A g C l + 2 N H_{3} ⇌ [A g (N H_{3})_{2}]^{+} + C l^{-}$ has $K = K_{s p} \cdot K_{f} \sim 1 0^{- 3}$ , large enough to make $A g C l$ readily soluble in concentrated ammonia.

Q6. How is fractional precipitation used to separate mixtures?

When two ions in solution can be precipitated by the same reagent but their $K_{s p}$ values differ significantly, the less-soluble salt precipitates first as the reagent is added slowly. Continuing until the second salt just begins to precipitate leaves the first ion almost completely removed. This is the basis of many separations in qualitative analysis and hydrometallurgy.

Q7. Why is the solubility of sulphides pH-dependent?

Sulphide ion is the conjugate base of the weak acid $H S^{-}$ , so its concentration in solution depends strongly on $[H^{+}]$ : raising $[H^{+}]$ converts $S^{2 -}$ to $H S^{-}$ and then to $H_{2} S$ , lowering free $[S^{2 -}]$ . A lower $[S^{2 -}]$ means metal sulphides can hold more $[M^{2 +}]$ in solution before their $K_{s p}$ is exceeded, so solubility rises. This is used in classical qualitative analysis to control which sulphides precipitate.

Q8. Why can gravimetric determination of sulphate be done accurately?

$B a S O_{4}$ has a very small $K_{s p}$ ( $\sim 1 0^{- 10}$ ), so precipitation from solution is nearly quantitative. Adding a slight excess of $B a C l_{2}$ pushes any remaining $S O_{4}^{2 -}$ into precipitate via the common ion effect. The insoluble, dense, well-defined $B a S O_{4}$ crystals filter cleanly and weigh accurately, giving reliable determination of the original sulphate amount.

Q9. What is the salt effect and why is it usually ignored?

The salt effect is the small increase in solubility when an inert electrolyte (with no common ion) is added, because activity coefficients of the dissolving ions decrease at higher ionic strength. Ionic strength effects change effective concentrations by only a few percent in typical dilute solutions, and the effect is usually ignored in JEE-level problems that treat concentrations as activities.

Previous year questions on Solubility Equilibria Of Sparingly Soluble Salts

9 questions from past papers, each with a step-by-step solution.

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