Solubility and Solubility Product

Solubility is a concept that was not discovered by any single person this concept was studied by multiple Scientists with time. Historically the understanding of solubility is the deeper study of solutions and mixtures. and after that, the Greeks and Romans observed that some substances can dissolve in the water. After some, the two scientists Robert Boyle and Isaac Newton performed some experiments for a better understanding of the solubility in the 1600s. In the 1750s, a Scottish scientist Joseph black made more contributions to heat and gases. Which directly influences the solubility on change temperature and pressure.

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This Story also Contains

1. Solubility And Solubility Product
2. Reaction Quotient
3. Some Solved Examples
4. Summary

Solubility And Solubility Product

Solubility

The maximum amount of a particular solute in grams, which can dissolve in 100 grams of solvent at a given temperature is called solubility. It is denoted by 's' and is expressed in g. The number of moles of solute in 1 L of saturated solution is known as molar solubility.

Solubility decreases with the increase in the concentration of common ion. It increases with temperature and increases in case the ions formed from the sparingly soluble salt undergo some sort of reaction like complexation.

For example, The solubility of AgCl in water in the presence of AgNO₃

Solubility increases due to complex ion formation. For example, AgCl has more solubility in ammonia due to complex formation$\mathrm{AgCl}+2 \mathrm{NH}_3 \rightleftharpoons \mathrm{Ag}\left(\mathrm{NH}_3\right)_2 \mathrm{Cl}$

Solubility Product:

It is the product of the molar concentrations of ions of an electrolyte in a saturated solution at a particular temperature. It is denoted by Ksp or S.

General Representation$\begin{aligned} & \mathrm{A}_{\mathrm{x}} \mathrm{B}_{\mathrm{y}} \rightleftharpoons \mathrm{xA}^{+y}+\mathrm{yB}^{-\mathrm{x}} \\ & \mathrm{Ksp}=\left[\mathrm{A}^{+\mathrm{y}}\right]^{\mathrm{x}} \times\left[\mathrm{B}^{-\mathrm{x}}\right]^{\mathrm{y}}\end{aligned}$

Relation between Solubility(s) and Solubility Product (Ksp)

$\mathrm{A}_{\mathrm{x}} \mathrm{B}_{\mathrm{y}} \rightleftharpoons \mathrm{xA}^{+y}+\mathrm{yB}^{-\mathrm{x}}$
s 0 0
- xs ysThus, Ksp $=\mathrm{x}^{\mathrm{x}} \mathrm{y}^{\mathrm{y}}(\mathrm{s})^{\mathrm{x}+\mathrm{y}}$

Solubility Product and Precipitation

• If K_sp≈ Ionic product
  The solution is saturated and for precipitation, more solute is to be added.
• If Ionic product > K_sp
  The solution is supersaturated so easily precipitated.
• If Ionic product < K_sp
  The solution is unsaturated so no precipitation takes place.

The solubility product (Ksp) is a constant that provides the equilibrium concentration of ions in a saturated solution of a sparingly soluble salt. It reflects how much of the salt can dissolve in water to form a solution. For a salt ABₓ, which dissociates into A^n⁺ and B^m⁻ ions

Precipitation occurs when the concentration of ions in a solution exceeds the solubility product, causing the ions to combine and form an insoluble solid, or precipitate. This happens when a solution is supersaturated, meaning the ion concentration surpasses the equilibrium limit defined by the Ksp.

Reaction Quotient

Reaction coefficient/quotient -

It is defined as the ratio of the concentration of products to the concentration of the reacting species raised to their stoichiometric coefficient at any point of time other than the equilibrium stage. It has the exact same expression as that of the Equilibrium constant except that the concentration values are at any instant. Mathematically, it can be determined as follows:

If we consider a reaction$\mathrm{mA}+\mathrm{nB} \rightleftharpoons \mathrm{pC}+\mathrm{qD}$

$\mathrm{Q}=\frac{[\mathrm{C}]^{\mathrm{p}}[\mathrm{D}]^{\mathrm{q}}}{[\mathrm{A}]^{\mathrm{m}}[\mathrm{B}]^{\mathrm{n}}}$
Q can be denoted as Q_c or Q_p if we use concentration in terms of mole per litre or partial pressure respectively.

The value of Q is useful to determine the direction in which the equilibrium will shift at any instant for a particular set of activities of the species involved.

• When Q = K, the reaction is at equilibrium and the rate of forward and backward reactions are equal.
• When Q > K, the reaction will proceed or favour a backward direction. This means products convert into reactants to attain equilibrium.
• When Q < K, the reaction will proceed or favour a forward direction. This means that the reactants convert into products to attain equilibrium.

Recommended topic video on ( Solubility and Solubility Product)

Some Solved Examples

Example.1 Solubility product of silver bromide is 5.0 × 10^-13 . The quantity of potassium bromide (molar mass taken as 120 g mol^-1) to be added to 1 litre of 0.05M solution of silver nitrate to start the precipitation of AgBr is

1)5.0 × 10^-8 g

2)1.2 x 10^-10 g

3) (correct)1.2 x 10^-9 g

4)6.2 x 10^-5 g

Solution

$\begin{aligned} & \mathrm{AgBr} \rightleftharpoons \mathrm{Ag}^{+}+\mathrm{Br}^{-} \\ & K_{\text {sp }} \text { of } \mathrm{AgBr}=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Br}^{-}\right] \\ & 5 \times 10^{-13}=0.05\left[B r^{-}\right] \\ & {\left[\mathrm{Br}^{-}\right]=\frac{5 \times 10^{-13}}{0.05}=1 \times 10^{-11} \mathrm{M}}\end{aligned}$

we know, b mole of KBr have b mole of K⁺ and b mole of Br^-

so, Moles of KBr $=1 \times 10^{-11}$

The weight of KBr = moles X molar mass $=1 \times 10^{-11} \times 120$ gram $=1.2 \times 10^{-9} \mathrm{~g}$

Hence, the answer is the option (3).

Example.2 The incorrect statement is :

< 1)$C u^{2+}$ salts give red coloured borax bead test in reducing flame.

2) (correct)$C u^{2+}$ and $N i^{2+}$ ions give black precipitate with H₂S in presence of HCl solution.

3)Ferric ion gives blood red colour with potassium thiocyanate.

4)$C u^{2+}$ ion gives chocolate coloured precipitate with potassium ferrocyanide solution.

Solution

$\mathrm{Cu}^{2+}$ and $\mathrm{Ni}^{2+}$ ions belong to Group II and Group IV, respectively,
$
\begin{aligned}
& \mathrm{Cu}^{2+}+\mathrm{H}_2 \mathrm{~S} \xrightarrow{\text { dil. } \mathrm{HCl}} \underset{(\text { Black })}{\mathrm{CuS}} \downarrow \\
& \mathrm{Ni}^{2+}+\mathrm{H}_2 \mathrm{~S} \xrightarrow{\mathrm{NH}_4 \mathrm{OH}} \underset{(\text { Black })}{\mathrm{NiS}} \downarrow \\
&
\end{aligned}
$

Group II sulphides precipitated out even under low concentrations of sulphides, while Group IV sulphides require higher concentrations of sulphides. Due to the common ion effect, sufficient concentration is not produced and a precipitate of NiS does not form. To fulfil this reaction condition, dil. HCl is chosen in Group II reagents and NH₄OH is chosen in Group IV reagents.

Hence, the answer is the option (2).

Example.3 An aqueous solution contains an unknown concentration of Ba²⁺. When 50 mL of a 1 M solution of Na₂SO₄ is added, BaSO₄
just begins to precipitate. The final volume is 500 mL. The solubility product of BaSO₄ is 1×10⁻¹⁰. What is the original
concentration of Ba²⁺ ?

1)1.0×10⁻¹⁰ M

2)5×10⁻⁹ M

3)2×10⁻⁹ M

4) (correct)1.1×10⁻⁹ M

Solution

Let the original concentration of Ba²⁺ = x M

The volume of this solution is 500-50=450ml

On adding Na₂SO₄ solution. Concentration of Ba⁺² =$\frac{450 \times x}{500}=\frac{9 x}{10} M$

The concentration of $\mathrm{SO}_4^{-2}$ in the final solution $=\frac{1 \times 50}{500}=\frac{1}{10} M$

Precipitation just started so $Q_{s p}=K_{s p}$

$\begin{aligned} & \Rightarrow\left[\mathrm{Ba}^{+2}\right]\left[\mathrm{SO}_4^{-2}\right]=1 \times 10^{-10} \\ & \Rightarrow \frac{9 x}{10} \times \frac{1}{10}=1 \times 10^{-10} \\ & \therefore x=\frac{1}{9} \times 10^{-8} \cong 1.1 \times 10^{-9} \mathrm{M}\end{aligned}$

Hence, the answer is an option (4).

EXAMPLE.4 If solubility product of $\mathrm{Zr}_3\left(\mathrm{PO}_4\right)_4$ is denoted by $K_{s p}$ and its molar solubility is denoted by S, then which of the following relation between S and $K_{s p}$ is correct ?

1)$S=\left(\frac{K_{s p}}{216}\right)^{1 / 7}$

2) (correct)$S=\left(\frac{K_{s p}}{6912}\right)^{1 / 7}$

3)$S=\left(\frac{K_{s p}}{144}\right)^{1 / 6}$

4)$S=\left(\frac{K_{s p}}{929}\right)^{1 / 9}$

Solution

$\begin{aligned} & \mathrm{Zr}_3\left(\mathrm{PO}_4\right)_4 \rightleftharpoons 3 \mathrm{Zr}^{4+}(\mathrm{aq})+4 \mathrm{PO}_4^{3-}(\mathrm{aq}) \\ & \mathrm{s} \quad 3 \mathrm{~S} \\ & \mathrm{~K}_{\mathrm{sp}}=\left[\mathrm{Zr}^{4+}\right]^3\left[\mathrm{PO}_4^{3-}\right]^4 \\ &=(3 \mathrm{~S})^3 \times(4 \mathrm{~S})^4=6912 \mathrm{~S}^7 \\ & \mathrm{~S}=\left(\frac{\mathrm{K}_{\text {sp }}}{6912}\right)^{1 / 7}\end{aligned}$

Hence, the answer is the option (2).

EXAMPLE.5 The molar solubility ( in mol L ^-1) of a sparingly soluble salt $\mathrm{MX}_4$ is s . The corresponding solubility product is $\mathrm{K}_{\mathrm{sp}}$.The relation between s and $\mathrm{K}_{\mathrm{sp}}$ is given by

1)$s=\left(K_{s p} / 128\right)^{1 / 4}$

2)$s=\left(128 K_{s p}\right)^{1 / 4}$

3)$s=\left(256 K_{s p}\right)^{1 / 5}$

4) (correct)$s=\left(K_{s p} / 256\right)^{1 / 5}$

Solution

$
\begin{aligned}
& \mathrm{MX}_4(\mathrm{~s}) \rightleftharpoons \mathrm{M}_{(\mathrm{aq})}^{4+}+4 \mathrm{X}_{(\mathrm{aq})}^{-} \\
& \text {s } \quad 4 \mathrm{~s} \\
&
\end{aligned}
$

Solubility product, $\mathrm{K}_{\mathrm{sp}}=\mathrm{s} \times(4 \mathrm{~s})^4=256 \mathrm{~s}^5$
$
\therefore \mathrm{s}=\sqrt[5]{\frac{\mathrm{K}_{\mathrm{sp}}}{256}}=\left(\frac{\mathrm{K}_{\mathrm{sp}}}{256}\right)^{1 / 5}
$

Hence, the answer is the option (4).

Summary

The concept of solubility is very important as it has various applications in the industrial field and various pharmaceutical company and also in the biological process. Solubility determines how substances dissolve in solvents, which is crucial for reactions that occur in solution. Drugs must dissolve in bodily fluids to be absorbed and effective. Solubility plays a role in the behavior of pollutants and nutrients in water. It is used to access the mobility of the environment. Many industrial processes, such as those in mining, water treatment, and manufacturing, rely on solubility for operations like extraction, purification, and formulation. Solubility affects the formulation of food products, including flavorings, colorings, and preservatives, ensuring they mix properly and remain stable. Solubility affects everyday activities, from dissolving salt in cooking to the effectiveness of cleaning products.