JEE Mains Chemistry · Equilibrium

Solubility Product

Linking Ksp to the molar solubility of a sparingly soluble salt, lowering that solubility with a common ion, and deciding whether and in what order a precipitate forms.

All Equilibrium notes JEE Mains Chemistry strategy

Why this matters

Twenty-four PYQs, fifteen of them multiple choice, and three from 2026. Thirteen convert between Ksp and solubility; six add a common ion; five decide whether, or in what order, a precipitate forms. It is the biggest page in the chapter.

Concept 1 of 3: Ksp and molar solubility

If s mol of a salt dissolves per litre, each ion's concentration is s times its count in the formula. Put those into the Ksp expression and every salt type gives a number times a power of s. The number is each count raised to itself, and the power is the total number of ions.

Definition

$\mathrm{A_xB_y}\rightleftharpoons x\mathrm{A^{y+}}+y\mathrm{B^{x-}}$ : $K_{sp}=(xs)^x(ys)^y=x^xy^ys^{x+y}$ .
AB: $s^2$ . $\mathrm{AB_2}$ or $\mathrm{A_2B}$ : $4s^3$ . $\mathrm{AB_3}$ , such as $\mathrm{Cr(OH)_3}$ : $27s^4$ . $\mathrm{A_2B_3}$ or $\mathrm{A_3B_2}$ : $108s^5$ . $\mathrm{A_3B_4}$ : $6912s^7$ .
Mass solubility to molar: divide g/L by the molar mass. A value in g per 100 mL is first multiplied by 10.
Across different salt types, compare s, not Ksp.
Increasing Ksp: $\mathrm{HgS<PbS<AgBr<Ca(OH)_2}$ .

Solubility product

K_{sp}=x^x\,y^y\,s^{x+y}\qquad(\mathrm{A_xB_y})

Worked example

K_{sp}

\mathrm{CaF_2}

3.2\times10^{-11}

. Find its molar solubility.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 22 Jan 2025 · Q112Moderate

Example 1 · Equilibrium · Solubility Product

The molar solubility(s) of zirconium phosphate with molecular formula

\left( Zr^{4 +} \right)_{3}\left( PO_{4}^{3 -} \right)_{4}

is given by relation :

Each ion is its count times s

\mathrm{Ag_2CrO_4}

[\mathrm{Ag^+}]=2s

, so

K_{sp}=(2s)^2s=4s^3

. Writing

s^2\cdot s

drops the factor 4.

Divide first, then take the root

From

K_{sp}=27s^4

s=\left(\frac{K_{sp}}{27}\right)^{1/4}

. Taking the root of

K_{sp}

alone, or a square root out of habit, gives the offered wrong answers.

Drill 12 more on ksp and molar solubility

Concept 2 of 3: Common ion and solubility

If the solution already holds one of the salt's ions, the equilibrium is pushed back and much less salt dissolves. The added ion's concentration is fixed by the strong electrolyte, so it goes straight into Ksp and s is found by division. Solubility is highest in pure water.

Definition

With the common ion at concentration C (from a strong electrolyte), put C for that ion in $K_{sp}$ .
AB in C of $\mathrm{B^-}$ : $s=\frac{K_{sp}}{C}$ .
$\mathrm{AB_2}$ in C of $\mathrm{B^-}$ : $s=\frac{K_{sp}}{C^2}$ .
Count the common ion per formula: 0.02 M $\mathrm{CaCl_2}$ gives 0.04 M $\mathrm{Cl^-}$ .
Passing HCl gas into a saturated NaCl or $\mathrm{BaCl_2}$ solution raises $[\mathrm{Cl^-}]$ and the salt precipitates.
A salt of a weak acid dissolves more in acid: $\mathrm{H^+}$ removes its anion. For AgCN in acid, $s^2=\frac{K_{sp}}{K_a}[\mathrm{H^+}]$ .

Solubility with a common ion

s=\frac{K_{sp}}{C^{\,n}}\qquad(n=\text{count of the common ion in the formula})

Worked example

K_{sp}

\mathrm{Mg(OH)_2}

1.2\times10^{-11}

. Find its molar solubility in 0.01 M NaOH.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 8 Apr 2023 · Q141Moderate

Example 2 · Equilibrium · Solubility Product

The solubility product of

BaSO_{4}

1 \times10^{- 10}

at 298K. The solubility of

BaSO_{4}

0.1MK_{2}SO_{4}\left( aq \right)

solution is_____

\times10^{- 9}\text{ }g{\text{ }L}^{- 1}

(nearest integer). Given: Molar mass of

BaSO_{4}

233\text{ }g{\text{ }mol}^{- 1}

Square the common ion for AB₂

For

\mathrm{Zn(OH)_2}

in NaOH,

s=\frac{K_{sp}}{[\mathrm{OH^-}]^2}

. Dividing by

[\mathrm{OH^-}]

once gives an answer too large by a factor of

1/[\mathrm{OH^-}]

Convert to grams only at the end

Find s in mol/L first, then multiply by the molar mass if the answer is asked in g/L. Using a mass concentration inside

K_{sp}

is wrong.

Drill 5 more on common ion and solubility

Concept 3 of 3: Will it precipitate, and in what order

Write the ionic product Q in the same form as Ksp, using the concentrations actually present. If Q is more than Ksp, solid forms until Q falls back to Ksp. When a reagent is added slowly to a mixture, each salt starts to form when its own Q reaches its own Ksp, so the salt that needs the least reagent comes out first.

Definition

Ionic product Q uses the concentrations after mixing. Equal volumes halve each one.
$Q>K_{sp}$ : precipitate. $Q=K_{sp}$ : just saturated. $Q<K_{sp}$ : no precipitate.
Onset for $\mathrm{M(OH)_n}$ : $[\mathrm{OH^-}]=\left(\frac{K_{sp}}{[\mathrm{M^{n+}}]}\right)^{1/n}$ . The ion needing the smaller $[\mathrm{OH^-}]$ precipitates first.
The smaller Ksp does not decide the order when the salts are of different types.
Sulphides: $[\mathrm{S^{2-}}]=\frac{K_{a1}K_{a2}[\mathrm{H_2S}]}{[\mathrm{H^+}]^2}$ , so the pH controls which sulphide precipitates.

Precipitation condition

Q>K_{sp}\ \Rightarrow\ \text{precipitate}

Worked example

Equal volumes of

2\times10^{-4}

\mathrm{AgNO_3}

and

2\times10^{-4}

M NaCl are mixed.

K_{sp}(\mathrm{AgCl})=1.8\times10^{-10}

. Does AgCl precipitate?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 23 January 2025 · Q43Moderate

Example 3 · Equilibrium · Solubility Product

Which of the following happens when

NH_{4}OH

is added gradually to the solution containing

1MA^{2 +}

and

1MB^{3 +}

ions ? Given :

K_{\text{sp~}}\left\lbrack A(OH)_{2} \right\rbrack= 9 \times10^{- 10}

and

K_{sp}\left\lbrack \text{ }B(OH)_{3} \right\rbrack= 27 \times10^{- 18}\text{~at~}298\text{ }K

Mixing halves both concentrations

When equal volumes of two solutions are mixed, each ion is at half its original concentration. Using the original values makes Q four times too large for an AB salt, and eight times for

\mathrm{AY_2}

The smaller Ksp is not always first

For salts of different types, compare the reagent concentration each one needs, not the Ksp values. A hydroxide with a larger Ksp can still start first.

Drill 4 more on will it precipitate, and in what order

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)

Ksp and molar solubility
Solubility product
$K_{sp}=x^x\,y^y\,s^{x+y}\qquad(\mathrm{A_xB_y})$
Common ion and solubility
Solubility with a common ion
$s=\frac{K_{sp}}{C^{\,n}}\qquad(n=\text{count of the common ion in the formula})$
Will it precipitate, and in what order
Precipitation condition
$Q>K_{sp}\ \Rightarrow\ \text{precipitate}$

Watch out for (6)

Each ion is its count times s
→ Ksp and molar solubility
Divide first, then take the root
→ Ksp and molar solubility
Square the common ion for AB₂
→ Common ion and solubility
Convert to grams only at the end
→ Common ion and solubility
Mixing halves both concentrations
→ Will it precipitate, and in what order
The smaller Ksp is not always first
→ Will it precipitate, and in what order

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