MHT-CET Chemistry · Ionic Equilibria

Solubility Product (Ksp)

For a sparingly soluble salt, the solubility product Ksp is the product of the molar concentrations of its ions, each raised to the power of its coefficient; it links directly to the salt's solubility S through a fixed stoichiometry factor.

All Ionic Equilibria notes MHT-CET Chemistry strategy

Why this matters

This is the single biggest subtopic in MHT-CET Ionic Equilibria (30 PYQs) and almost every question is one of two mechanical steps: get S from Ksp, or get Ksp from S. The whole battle is picking the right relation for the salt type — S squared for AB, 4S cubed for AB2 or A2B, and 108S to the fifth for A2B3. Learn the stoichiometry factors cold, remember to take the correct root (square, cube, or fifth), and this becomes a guaranteed-scoring block.

Concept 1 of 6

Solubility product expression

Intuition

When a sparingly soluble salt sits in its saturated solution, a tiny amount dissolves into ions and the rest stays solid. The solubility product is just the equilibrium constant for that dissolving — the product of the ion concentrations, each raised to how many of that ion appear in the formula. The solid itself never enters the expression, only the dissolved ions do.

Definition

Solubility product for a salt $A_x B_y$ :

The salt dissolves as $A_x B_y \rightleftharpoons x\,A^{y+} + y\,B^{x-}$ .
Its solubility product is $K_{sp} = [A^{y+}]^x\,[B^{x-}]^y$ — each ion concentration raised to its coefficient.
The undissolved solid does not appear (its activity is 1); only the ions count.
$K_{sp}$ is a constant at a given temperature — it does not change when you add a common ion, only the individual concentrations adjust.

General solubility product

K_{sp} = [A^{y+}]^x\,[B^{x-}]^y \qquad (A_x B_y \rightleftharpoons x\,A^{y+} + y\,B^{x-})

K_{sp}solubility product (constant at a given temperature)
[A^{y+}]molar concentration of the cation
[B^{x-}]molar concentration of the anion
x, ynumber of cations and anions in the formula (their exponents)

Worked example

Write the solubility product expression for aluminium hydroxide, Al(OH)3.

The salt dissolves as $\text{Al(OH)}_3 \rightleftharpoons \text{Al}^{3+} + 3\,\text{OH}^-$ .
Raise each ion concentration to its coefficient: $\text{Al}^{3+}$ to the power 1, $\text{OH}^-$ to the power 3.

Answer:

K_{sp} = [\text{Al}^{3+}][\text{OH}^-]^3

Practice this conceptself-check · 4 quick reps

Try it yourself

Write the solubility product expression for calcium phosphate,

\text{Ca}_3(\text{PO}_4)_2

Practice — Level 1 (4 reps)

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

1.
Write the Ksp expression for AgCl.
2.
Write the Ksp expression for $\text{PbI}_2$ .
3.
Does the undissolved solid appear in the Ksp expression?
4.
Write the Ksp expression for $\text{Ag}_2\text{CrO}_4$ .

From the bank · past-year question

Example 1Ionic EquilibriaEASY

What is the expression for solubility product of silver chromate if its solubility is expressed as

S\,\text{mol L}^{-1}

[Q55 · 12th May Shift 1 · 2024]

Raise each ion to its own coefficient

For

\text{Ag}_2\text{CrO}_4 \rightleftharpoons 2\text{Ag}^+ + \text{CrO}_4^{2-}

, the silver-ion concentration is squared:

K_{sp} = [\text{Ag}^+]^2[\text{CrO}_4^{2-}]

. Dropping the exponent (writing

[\text{Ag}^+][\text{CrO}_4^{2-}]

) is the most common slip.

The solid is left out

The solubility product involves only the aqueous ions. The concentration of the pure solid salt is taken as constant (activity 1) and never appears in

K_{sp}

Concept 2 of 6

Ksp in terms of solubility, by salt type

Intuition

If the molar solubility is S, then each ion's concentration is S times its coefficient. Feed those into the Ksp expression and you get a single tidy formula per salt type. Memorising the four common shapes — S squared, 4S cubed, 27S to the fourth, 108S to the fifth — lets you write down the answer without re-deriving it each time.

Definition

If the molar solubility of the salt is $S$ , the ion concentrations follow the stoichiometry and $K_{sp}$ reduces to a power of $S$ :

AB (like AgCl, AgBr, CaCO3): $K_{sp} = S^2$ .
AB2 or A2B (like PbI2, PbCl2, Ag2CrO4): $K_{sp} = 4S^3$ .
AB3 or A3B (like AlCl3-type, FeCl3-type): $K_{sp} = 27S^4$ .
A2B3 or A3B2 (like Ca3(PO4)2, Al2(SO4)3): $K_{sp} = 108\,S^5$ .

The numerical factor is just the product of each coefficient raised to itself: e.g. $2^2 = 4$ , $3^3 = 27$ , $3^3 \times 2^2 = 108$ .

Salt type	Dissociation	Ksp in terms of S	Example salt
AB (1:1)	$AB \rightleftharpoons A^+ + B^-$	$K_{sp} = S^2$	AgCl, AgBr, CaCO3, NiS Most common type in the bank. Recover S by a single square root: $S = \sqrt{K_{sp}}$ .
AB2 or A2B (1:2)	$AB_2 \rightleftharpoons A^{2+} + 2B^-$	$K_{sp} = 4S^3$	PbI2, PbCl2, Ag2CrO4, Ba(OH)2 Recover S by $S = \sqrt[3]{K_{sp}/4}$ — divide by 4 first, then take the cube root.
AB3 or A3B (1:3)	$AB_3 \rightleftharpoons A^{3+} + 3B^-$	$K_{sp} = 27S^4$	Fe(OH)3-type, AlCl3-type Recover S by $S = \left(\dfrac{K_{sp}}{27}\right)^{1/4}$ .
A2B3 or A3B2 (2:3)	$A_2B_3 \rightleftharpoons 2A^{3+} + 3B^{2-}$	$K_{sp} = 108\,S^5$	Ca3(PO4)2, Al2(SO4)3 Factor is $2^2 \times 3^3 = 4 \times 27 = 108$ . Recover S by $S = \left(\dfrac{K_{sp}}{108}\right)^{1/5}$ .

The numerical factor is the product of each coefficient raised to its own power; the exponent on S is the total number of ions produced.

Practice this conceptself-check · 4 quick reps

Try it yourself

Which equation represents the relation between solubility S and solubility product for the salt

\text{B}_3\text{A}_2

Practice — Level 1 (4 reps)

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

1.
Ksp in terms of S for an AB salt like AgCl?
2.
Ksp in terms of S for $\text{PbI}_2$ ?
3.
Ksp in terms of S for $\text{Ca}_3(\text{PO}_4)_2$ ?
4.
Ksp in terms of S for a $BA_3$ salt?

From the bank · past-year question

Example 2Ionic EquilibriaEASY

Solubility of

Ca_{3}\left( PO_{4} \right)_{2}

is ' S '

moldm^{- 3}

. Find solubility product.

[Q51 · 20 April Shift I · 2025]

AB2 is 4S cubed, not S squared

For

AB_2 \rightleftharpoons A^{2+} + 2B^-

[B^-] = 2S

, so

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

. Treating it as

S^2

(the AB formula) is the classic mistake — the coefficient 2 both squares AND multiplies in the factor 4.

Match the root to the exponent on S

From

K_{sp} = 4S^3

you take a cube root (after dividing by 4); from

K_{sp} = 108S^5

a fifth root. Taking a square root out of habit gives a wildly wrong S.

Drill 2 more on ksp in terms of solubility, by salt type

Concept 3 of 6

Solubility of a 1:1 (AB) salt: Ksp = S squared

Intuition

For a simple 1:1 salt each formula unit gives one cation and one anion, so both ion concentrations equal the solubility S. The solubility product is then just S times S. To go from Ksp to solubility you take a single square root; to go the other way you square the solubility.

Definition

For a binary (1:1) salt $AB \rightleftharpoons A^+ + B^-$ :

Both ions have concentration equal to the solubility: $[A^+] = [B^-] = S$ .
Therefore $K_{sp} = S \times S = S^2$ .
From Ksp to solubility: $S = \sqrt{K_{sp}}$ .
From solubility to Ksp: $K_{sp} = S^2$ .

Handy squares: $\sqrt{4.9\times 10^{-13}} = 7.0\times 10^{-7}$ , $\sqrt{6.4\times 10^{-5}} = 8.0\times 10^{-3}$ , $\sqrt{1.6\times 10^{-10}} = 1.26\times 10^{-5}$ .

AB salt: solubility and solubility product

K_{sp} = S^2 \qquad\Longleftrightarrow\qquad S = \sqrt{K_{sp}}

Smolar solubility (mol dm^-3), equals each ion concentration
K_{sp}solubility product of the 1:1 salt

Worked example

The solubility product of AgCl is

1.21 \times 10^{-10}

. Calculate its solubility in mol dm

^{-3}

$\text{AgCl} \rightleftharpoons \text{Ag}^+ + \text{Cl}^-$ , a 1:1 salt, so $K_{sp} = S^2$ .
$S = \sqrt{K_{sp}} = \sqrt{1.21 \times 10^{-10}} = \sqrt{12.1 \times 10^{-11}}$ .
$\sqrt{1.21} = 1.1$ and $\sqrt{10^{-10}} = 10^{-5}$ .

Answer:

S = 1.1 \times 10^{-5}\ \text{mol dm}^{-3}

Practice this conceptself-check · 4 quick reps

Try it yourself

The solubility of a sparingly soluble 1:1 salt BA is

2.0 \times 10^{-5}\ \text{mol dm}^{-3}

. Calculate its solubility product.

Practice — Level 1 (4 reps)

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

1.
$K_{sp}$ of a 1:1 salt is $4.9 \times 10^{-9}$ . Its solubility?
2.
Solubility of an AB salt is $7 \times 10^{-5}$ . Its Ksp?
3.
$K_{sp}$ of NiS is $4.9 \times 10^{-5}$ . Solubility?
4.
$K_{sp}$ of AgBr is $6.4 \times 10^{-13}$ . Solubility?

From the bank · past-year question

Example 3Ionic EquilibriaEASY

The solubility product of a sparingly soluble salt AX is

4.9 \times10^{- 13}

. What is its solubility in mol dm

\ ^{- 3}

[Q91 · 19 April Shift I · 2025]

Take the square root — do not report Ksp as the solubility

For AgBr with

K_{sp} = 4.9 \times 10^{-13}

, the solubility is

\sqrt{K_{sp}} = 7.0 \times 10^{-7}

, NOT

4.9 \times 10^{-13}

. The distractor equal to

K_{sp}

itself is always offered — remember to root it.

Handle the power correctly under the root

Rewrite the mantissa so the exponent is even:

\sqrt{4.9 \times 10^{-13}} = \sqrt{49 \times 10^{-14}} = 7 \times 10^{-7}

. Splitting

10^{-13}

49 \times 10^{-14}

keeps the arithmetic clean.

Drill 14 more on solubility of a 1:1 (ab) salt: ksp = s squared

Concept 4 of 6

Solubility of AB2, A2B and A2B3 salts

Intuition

When one formula unit releases more than one of an ion, that ion's concentration is a multiple of S, and the Ksp picks up a numerical factor plus a higher power of S. The recipe is always the same: substitute the ion concentrations (S, 2S, 3S ...) into the Ksp expression and simplify. Going backwards, divide out the factor first, then take the matching root.

Definition

Substitute the ion concentrations into $K_{sp}$ and simplify:

AB2 / A2B: $[A^{2+}] = S$ , $[B^-] = 2S$ , so $K_{sp} = (S)(2S)^2 = 4S^3$ , giving $S = \sqrt[3]{K_{sp}/4}$ .
A2B3: $[A^{3+}] = 2S$ , $[B^{2-}] = 3S$ , so $K_{sp} = (2S)^2(3S)^3 = 108\,S^5$ , giving $S = \left(K_{sp}/108\right)^{1/5}$ .
The cube-root trick: write $K_{sp}/4$ as (a number) $\times 10^{-(\text{multiple of }3)}$ so both parts have exact roots, e.g. $27 \times 10^{-9} \Rightarrow 3 \times 10^{-3}$ .

AB2 / A2B salt: solubility and solubility product

K_{sp} = 4S^3 \qquad\Longleftrightarrow\qquad S = \sqrt[3]{\dfrac{K_{sp}}{4}}

Smolar solubility (mol dm^-3)
K_{sp}solubility product of the AB2 / A2B salt
4the factor 2^2 from the doubly-produced ion (2S)^2

Worked example

The solubility product of

\text{PbCl}_2

3.2 \times 10^{-5}

. Calculate its solubility in mol dm

^{-3}

$\text{PbCl}_2 \rightleftharpoons \text{Pb}^{2+} + 2\text{Cl}^-$ , so $K_{sp} = (S)(2S)^2 = 4S^3$ .
$S^3 = \dfrac{K_{sp}}{4} = \dfrac{3.2 \times 10^{-5}}{4} = 8 \times 10^{-6}$ .
$S = \sqrt[3]{8 \times 10^{-6}} = \sqrt[3]{8}\times\sqrt[3]{10^{-6}} = 2 \times 10^{-2}$ .

Answer:

S = 2 \times 10^{-2}\ \text{mol dm}^{-3}

Practice this conceptself-check · 4 quick reps

Try it yourself

The solubility of the salt

B_2A

has

K_{sp} = 3.2 \times 10^{-11}

. Find its solubility in mol dm

^{-3}

Practice — Level 1 (4 reps)

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

1.
Solubility of $AX_2$ is $1 \times 10^{-4}$ . Its Ksp?
2.
Solubility of $BA_2$ is $4 \times 10^{-4}$ . Its Ksp?
3.
$K_{sp}$ of $AB_2$ is $2.56 \times 10^{-10}$ . Solubility?
4.
Solubility of $A_2B_3$ is $1 \times 10^{-3}$ . Its Ksp?

From the bank · past-year question

Example 4Ionic EquilibriaMODERATE

The solubility product of

PbI_{2}

1.08 \times10^{- 7}

. Calculate its solubility in

moldm^{- 3}

at 298 K .

[Q65 · 21 April Shift I · 2025]

Divide by the factor BEFORE taking the root

For

K_{sp} = 4S^3

, first isolate

S^3 = K_{sp}/4

, THEN cube-root. Cube-rooting

K_{sp}

directly (forgetting the 4) gives an answer too large by

\sqrt[3]{4} \approx 1.59

Group the power of ten into a multiple of the root

To cube-root

1.08 \times 10^{-7} / 4 = 2.7 \times 10^{-8}

, rewrite it as

27 \times 10^{-9}

; then

\sqrt[3]{27} = 3

and

\sqrt[3]{10^{-9}} = 10^{-3}

, giving

3 \times 10^{-3}

. Choosing a clean exponent makes the root exact.

Drill 7 more on solubility of ab2, a2b and a2b3 salts

Concept 5 of 6

Ksp from pH and from mass solubility

Intuition

Sometimes the solubility is handed to you indirectly — through the pH of a saturated hydroxide solution, or as grams per litre instead of moles per litre. The trick is to convert to the ion concentrations (or to molar solubility) first, then apply the ordinary Ksp relation. For a hydroxide, the pH gives you the hydroxide-ion concentration directly; for a mass solubility, divide by the molar mass.

Definition

Two common indirect routes:

From pH (for a hydroxide): $\text{pOH} = 14 - \text{pH}$ , then $[\text{OH}^-] = 10^{-\text{pOH}}$ . For $M(\text{OH})_2 \rightleftharpoons M^{2+} + 2\text{OH}^-$ , the metal ion is $[M^{2+}] = \tfrac{1}{2}[\text{OH}^-]$ , and $K_{sp} = [M^{2+}][\text{OH}^-]^2$ .
From mass solubility: molar solubility $S = \dfrac{\text{mass solubility (g dm}^{-3})}{\text{molar mass (g mol}^{-1})}$ ; then apply the usual $K_{sp}$ -vs- $S$ relation for the salt type.
Both routes end in the same Ksp formulas — only the way you obtain the concentrations changes.

Molar solubility from mass solubility

S = \dfrac{m}{M} \qquad\text{and}\qquad [\text{OH}^-] = 10^{-\text{pOH}},\ \ \text{pOH} = 14 - \text{pH}

Smolar solubility (mol dm^-3)
mmass solubility (g dm^-3)
Mmolar mass of the salt (g mol^-1)
[\text{OH}^-]hydroxide-ion concentration from the pH

Worked example

A saturated solution of

\text{Ba(OH)}_2

has pH 12. Calculate its

K_{sp}

$\text{pOH} = 14 - 12 = 2$ , so $[\text{OH}^-] = 10^{-2}\ \text{M}$ .
$\text{Ba(OH)}_2 \rightleftharpoons \text{Ba}^{2+} + 2\text{OH}^-$ , so $[\text{Ba}^{2+}] = \tfrac{1}{2}[\text{OH}^-] = 5 \times 10^{-3}\ \text{M}$ .
$K_{sp} = [\text{Ba}^{2+}][\text{OH}^-]^2 = (5 \times 10^{-3})(10^{-2})^2$ .
$= 5 \times 10^{-3} \times 10^{-4} = 5 \times 10^{-7}$ .

Answer:

K_{sp} = 5 \times 10^{-7}

Practice this conceptself-check · 3 quick reps

Try it yourself

A binary (1:1) sparingly soluble salt has a solubility of

1.12 \times 10^{-4}\ \text{g dm}^{-3}

. Its molar mass is 112 g mol

^{-1}

. Calculate its solubility product.

Practice — Level 1 (3 reps)

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

1.
A saturated hydroxide solution has pH 12. What is $[\text{OH}^-]$ ?
2.
For $\text{Ba(OH)}_2$ with $[\text{OH}^-] = 10^{-2}$ , what is $[\text{Ba}^{2+}]$ ?
3.
Mass solubility $2.24 \times 10^{-4}\ \text{g dm}^{-3}$ , molar mass 112. Molar solubility?

From the bank · past-year question

Example 5Ionic EquilibriaMODERATE

What is the value of

K_{sp}

for saturated solution of

\text{Ba(OH)}_2

having pH 12?

[Shift || · 2025]

Metal-ion concentration is HALF the hydroxide in M(OH)2

For

\text{Ba(OH)}_2 \rightleftharpoons \text{Ba}^{2+} + 2\text{OH}^-

, two hydroxides come from each barium, so

[\text{Ba}^{2+}] = \tfrac{1}{2}[\text{OH}^-]

. Using

[\text{Ba}^{2+}] = [\text{OH}^-]

doubles the answer.

Convert grams to moles before using Ksp

K_{sp}

relations use molar solubility. Plugging a mass solubility (g dm

^{-3}

) straight into

K_{sp} = S^2

is wrong — divide by the molar mass first to get S in mol dm

^{-3}

Drill 2 more on ksp from ph and from mass solubility

Concept 6 of 6

Common ion effect on solubility

Intuition

Adding an ion that is already part of a sparingly soluble salt pushes its dissolution equilibrium backward (Le Chatelier), so the salt becomes LESS soluble. The same suppression of ionization by a common ion is what makes a buffer work. Quantitatively, the common ion fixes one ion's concentration, so the solubility drops to Ksp divided by that concentration.

Definition

The common ion effect is the suppression of the dissociation (or dissolution) of a sparingly soluble salt or weak electrolyte on adding a strong electrolyte that shares a common ion:

For a salt $\text{AB} \rightleftharpoons \text{A}^+ + \text{B}^-$ , adding extra $\text{B}^-$ shifts the equilibrium to the left, lowering the solubility of AB.
If the common ion $\text{B}^-$ is present at concentration $C$ (from a strong electrolyte), then $[\text{A}^+] = S$ and $[\text{B}^-] \approx C$ , so $K_{sp} = S \times C$ and the new solubility is $S = K_{sp}/C$ .
The same effect suppresses the ionization of a weak acid by its salt — the basis of an acidic buffer.

Solubility in presence of a common ion

S = \dfrac{K_{sp}}{C}

Smolar solubility of the salt in the common-ion solution
K_{sp}solubility product of the salt
Cconcentration of the common ion (from the added strong electrolyte)

Worked example

The solubility product of AgCl is

1.8 \times 10^{-10}

. Find its solubility in 0.1 M NaCl solution.

NaCl supplies the common ion $\text{Cl}^-$ at $C = 0.1\ \text{M}$ , far more than AgCl alone would give.
$[\text{Ag}^+] = S$ and $[\text{Cl}^-] \approx 0.1$ , so $K_{sp} = S \times 0.1$ .
$S = K_{sp}/0.1 = (1.8 \times 10^{-10})/0.1 = 1.8 \times 10^{-9}\ \text{M}$ .

Answer:

S = 1.8 \times 10^{-9}\ \text{M}

— far less than in pure water (

\sqrt{K_{sp}} = 1.34 \times 10^{-5}\ \text{M}

Practice this conceptself-check · 3 quick reps

Try it yourself

The solubility product of AgBr is

5 \times 10^{-13}

. Find its solubility in 0.05 M KBr solution.

Practice — Level 1 (3 reps)

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

1.
Does adding NaCl increase or decrease the solubility of AgCl?
2.
Solubility of a 1:1 salt ( $K_{sp} = 10^{-12}$ ) in a 0.01 M common-ion solution?
3.
What does sodium acetate do to the ionization of acetic acid?

A common ion LOWERS solubility

The common ion effect always DECREASES the solubility of a sparingly soluble salt (and suppresses a weak acid's ionization). Expecting more dissolving is the classic error — Le Chatelier pushes the equilibrium the other way.

Use the common-ion concentration, not the square root

In pure water

S = \sqrt{K_{sp}}

for a 1:1 salt, but with a common ion at concentration

C

you use

S = K_{sp}/C

. The two answers differ by orders of magnitude — check whether a common ion is present before choosing the formula.

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 (5)

Solubility product expression
General solubility product
$K_{sp} = [A^{y+}]^x\,[B^{x-}]^y \qquad (A_x B_y \rightleftharpoons x\,A^{y+} + y\,B^{x-})$
Solubility of a 1:1 (AB) salt: Ksp = S squared
AB salt: solubility and solubility product
$K_{sp} = S^2 \qquad\Longleftrightarrow\qquad S = \sqrt{K_{sp}}$
Solubility of AB2, A2B and A2B3 salts
AB2 / A2B salt: solubility and solubility product
$K_{sp} = 4S^3 \qquad\Longleftrightarrow\qquad S = \sqrt[3]{\dfrac{K_{sp}}{4}}$
Ksp from pH and from mass solubility
Molar solubility from mass solubility
$S = \dfrac{m}{M} \qquad\text{and}\qquad [\text{OH}^-] = 10^{-\text{pOH}},\ \ \text{pOH} = 14 - \text{pH}$
Common ion effect on solubility
Solubility in presence of a common ion
$S = \dfrac{K_{sp}}{C}$

Reference tables (1)

Ksp in terms of solubility, by salt type4 rows

Salt type	Dissociation	Ksp in terms of S	Example salt
AB (1:1)	$AB \rightleftharpoons A^+ + B^-$	$K_{sp} = S^2$	AgCl, AgBr, CaCO3, NiS Most common type in the bank. Recover S by a single square root: $S = \sqrt{K_{sp}}$ .
AB2 or A2B (1:2)	$AB_2 \rightleftharpoons A^{2+} + 2B^-$	$K_{sp} = 4S^3$	PbI2, PbCl2, Ag2CrO4, Ba(OH)2 Recover S by $S = \sqrt[3]{K_{sp}/4}$ — divide by 4 first, then take the cube root.
AB3 or A3B (1:3)	$AB_3 \rightleftharpoons A^{3+} + 3B^-$	$K_{sp} = 27S^4$	Fe(OH)3-type, AlCl3-type Recover S by $S = \left(\dfrac{K_{sp}}{27}\right)^{1/4}$ .
A2B3 or A3B2 (2:3)	$A_2B_3 \rightleftharpoons 2A^{3+} + 3B^{2-}$	$K_{sp} = 108\,S^5$	Ca3(PO4)2, Al2(SO4)3 Factor is $2^2 \times 3^3 = 4 \times 27 = 108$ . Recover S by $S = \left(\dfrac{K_{sp}}{108}\right)^{1/5}$ .

The numerical factor is the product of each coefficient raised to its own power; the exponent on S is the total number of ions produced.

Watch out for (12)

Raise each ion to its own coefficient
→ Solubility product expression
The solid is left out
→ Solubility product expression
AB2 is 4S cubed, not S squared
→ Ksp in terms of solubility, by salt type
Match the root to the exponent on S
→ Ksp in terms of solubility, by salt type
Take the square root — do not report Ksp as the solubility
→ Solubility of a 1:1 (AB) salt: Ksp = S squared
Handle the power correctly under the root
→ Solubility of a 1:1 (AB) salt: Ksp = S squared
Divide by the factor BEFORE taking the root
→ Solubility of AB2, A2B and A2B3 salts
Group the power of ten into a multiple of the root
→ Solubility of AB2, A2B and A2B3 salts
Metal-ion concentration is HALF the hydroxide in M(OH)2
→ Ksp from pH and from mass solubility
Convert grams to moles before using Ksp
→ Ksp from pH and from mass solubility
A common ion LOWERS solubility
→ Common ion effect on solubility
Use the common-ion concentration, not the square root
→ Common ion effect on solubility

Mastery check — 5 interleaved questions

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

Example 1Ionic EquilibriaMODERATE

Which from the following equations represents the relation between solubility (mol L

^{-1}

) and solubility product for a salt

\text{B}_3\text{A}_2

[Q89 · 10th May Shift 2 · 2023]

Example 2Ionic EquilibriaEASY

Calculate the solubility in

\text{mol dm}^{-3}

of sparingly soluble salt BA if its solubility product

4.9 \times 10^{-13}

at same temperature.

[Q95 · 16th May Shift 1 · 2023]

Example 3Ionic EquilibriaHARD

Solubility of a salt

A_2B_3

1\times10^{-3}\,mol\,dm^{-3}

. What is the value of its solubility product?

[Q92 · 9th May Shift 1 · 2024]

Example 4Ionic EquilibriaMODERATE

Solubility of binary sparingly soluble salt is

1.12 \times 10^{-4}\,\text{g/dm}^3

. Calculate its solubility product (molar mass of salt = 112 g mol

^{-1}

)

[Q71 · 23 April Shift I · 2025]

Example 5Ionic EquilibriaMODERATE

Which of the following equations represents the relation between solubility and solubility product for salt

\text{BA}_3

[Q93 · 3rd May 2nd Shift · 2023]

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