Struggling with solubility and solubility product in chemistry? Learn the easy-to-understand guide with examples, formulas, and tips that make solving Ksp problems simple and exam-ready. Perfect for Class XI & XII students!
Welcome to Learn Chemistry by Inam Jazbi! Today, we’re diving into one of the most important topics in chemistry – Solubility and Solubility Product (Ksp). Don’t worry if it sounds tricky – I’ll break it down step by step, with simple explanations, examples, and tips that make it super easy to understand. By the end of this post, you’ll be able to solve any solubility problem like a pro, boost your exam confidence, and actually enjoy chemistry!
Solubility and Factors Affecting Solubility
Definition of
Saturated Solution
A solution that contains the maximum equilibrium amount of solute at a specific temperature is called saturated solution. A saturated solution cannot dissolve any more amount of solute at a given temperature.
Definition and Unit of Solubility
The solubility (dissolution) of a substance in a given solvent is a physical property. Solubility is a measure of the amount of solute that will dissolve in a solvent at a specific temperature. It is the extent (tendency) of a solvent to dissolve solute.
Solubility is the maximum equilibrium amount of solute (expressed either in grams or mole) needed to prepare a
saturated solution in a given quantity of solvent (usually 100 g or 100 ml or
in one litre or dm3 or 1000 mL of a solvent) at specified
temperature.
[In fact, the solubility of a solute in a solvent is the concentration of the saturated solution at a specified temperature. Thus it denotes maximum amount of solute which be held in a solution at a particular temperature.]
Miscible and Immiscible Liquids
Pairs of liquids that are mutually soluble in all proportions are
said to be miscible e.g. ethyl alcohol and water are miscible.
Pairs of liquids that do not mix are said to be immiscible e.g. water and methane are immiscible.
Unit of Solubility
Solubility is usually expressed as:
g/litre or g/dm3→
grams of solute per litre of solvent
mol/L (Molarity) → moles of solute per litre of solution i.e. Molar solubility (s) (practical and S.I unit)
g/100 g water or g/100 ml water → commonly used in data tables and solubility curves
Formula to calculate Solubility
Solubility of Some Solutes at 20°C
1. KNO₃ = 37 g (31.6 g)
2. NaCl = 36.0 g
3. KCl = 34.0 g
4. NaBr = 94.32 g
5. AgNO₃ = 222.0 g
6. NaNO₃ = 91.2 g
Types of Compounds according to Solubility
1. Soluble compounds; With solubility 10 or more gram per litre or
Solubility > 0.1 M
2. Slightly soluble compounds; With solubility 0.1 to 10 gram per litre or
Solubility = 0.01-0.1
M
3. Insoluble compounds; With solubility less than 0.1 gram per litre or Solubility < 0.01 M
Factors Affecting Solubility
Solubility depends upon following factors:
1. Temperature
2. Pressure
3. Nature of solute and solvent
1. Effect of Nature of Solute and Solvent
1. Substances with similar types of polarity or intermolecular
attractive forces tend to be soluble in one another
(i.e. substances may mix and dissolve in each other if they have approximately
the same type of polarity).
This generalization is a guiding rule of solubility and is often simply stated
as “Like dissolves like” which means
non-polar substances are soluble in non-polar solvent (e.g. benzene is soluble in CCl4 as both are
non-polar. Similarly, naphthalene or methane being a non-polar solute is more soluble in non-polar solvents like
benzene or carbon tetrachloride than in water) while polar or ionic solutes are soluble in polar
solvents. (e.g common salt (NaCl) being an ionic compound dissolves more readily in polar
solvent like water but it is insoluble in non-polar (organic) solvents like benzene or petrol).
The polar (or ionic) nature of polar solutes and polar
solvent like water produces particularly strong solute-solvent interactions in
the form of ion-dipole forces or dipole-dipole forces thereby favouring
solubility. On the other hand, non-polar solutes and polar solvents like water
have weak solute-solvent interactions in the form of dispersion forces which is
not sufficient enough to overcome the stronger hydrogen bonding initially
present in water.]
2. Hydrogen bonding interactions between solute and solvent greatly increase the solubility of non-polar compounds in water. e.g. ethanol is completely miscible with water due to hydrogen bonding(and also both have same type of structure having polarized –OH group). Similarly sugar, glucose, glycerine although non-polar in nature are very soluble in water due to hydrogen bonding as both have similar type of structure containing –OH groups].
3. The
solubility of the compound increases with the increase of –OH group along the
carbon chains of organic compounds.
e.g.
(i) glucose with
five –OH groups on a six-carbon framework is very soluble in water (83 g/100 mL of
water at 17.5°C).
(ii) Similarly, sugar with 10 –OH groups on a twelve-carbon framework is highly soluble in water (179 g/100 mL of water at 0°C).
4. The solubility of gases increases with increasing molecular mass
e.g.
i). Solubility of N2 gas (molecular mass 28 amu) is 0.63 × 10‒3 M
ii) Solubility of CO gas (molecular mass 28 amu) is 1.04 × 10‒3 M
iii) Solubility of O2 gas (molecular mass 32 amu) is 1.38 × 10‒3 M
iv) Solubility of Ar gas (molecular mass 40 amu) is 1.50 × 10‒3 M
v) Solubility of Kr gas (molecular mass 83.7amu) is 2.79 × 10‒3 M
5. The solubility of gas increases if it reacts with solvent. e.g. the solubility of Cl2 in water is 0.102M (much higher than predicted) as it reacts with water on dissolution. Similarly, solubility of CO2 and NH3 are also much higher than expected because of their reaction with solvent water.
Cl2 + H2O → HOCl + HCl
NH3 + H2O ⇌ NH4+ + OH‒
CO2 + H2O ⇌ H2CO3
6.The
solubility of the alcohols decreases with increasing molecular masses (as the –OH group is more tightly bound and molecules become more like a hydrocarbon).
e.g.
(i) solubility of butanol is 0.11 mol/100 g
H2O at 20°C
(ii) solubility
of pentanol is 0.03 mol/100 g H2O at 20°C.
Summary of Effect of Nature of Solute and
Solvent on Solubility
The rule “Like dissolves like” applies:
Polar solutes dissolve in polar solvents (e.g., NaCl in water).
Non-polar solutes dissolve in non-polar solvents (e.g., wax in
benzene).
Polar substances dissolve well due to ion–dipole or hydrogen bonding interactions.
2. Effect of
Temperature
Solubilities are temperature-dependent. The solubilities of most ionic and molecular solids in liquids (or solubility of partially miscible liquids) usually increase with increasing temperature though the solubilities of some (NaCl, NaBr) are almost unchanged and the solubilities of others [CaSO4, Na2SO4, Ce2(SO4)3] decrease i.e. solubility of solids in liquids is directly proportional to temperature. The dissolution for ionic solids is endothermic (heat is absorbed), so by Le-Chatelier’s principle the solubility increases with increasing temperature.
When a solid substance is dissolved in water, either heat is
evolved (exothermic) or heat is absorbed (endothermic).
(i) For endothermic
solubility process, solubility increases with increase in temperature.
(ii) For exothermic solubility process, solubility decreases with
increase in temperature.
For example;
i) Solubility
of sugar in water at 0°C is 179 g/100 ml of water whereas 100°C it is 487 g/100
ml of water.
ii) Similarly the solubility of KNO3 at 0°C is 13.5 g/100 g of water but at 100°C it is 247 g/100 g of water.
However, solubility of some solutes in liquids decreases with the
increase in temperature. e.g. CaSO4, Na2SO4,
Ce2(SO4)3. The dissolution for some solids is
exothermic (heat is released), so by Le-Chatelier’s principle the solubility
decreases with rising temperature.
e.g. solubility of sodium sulphate decahydrate (Na2SO4.10H2O) increases with the rise of temperature till 32.4°C reached and then its solubility decreases. This is due to the decomposition of hydrated salt to the anhydrous Na2SO4.
The solubility of gases in a liquid (unlike solids) decreases with
increasing temperature i.e. gases are more soluble in cold water than in hot
water. Thus when a solution comprising of gas in a liquid is heated, gases are
evolved
e.g.
(i) the solubility of oxygen in water at
0°C is 4.8 cm3/100 cm3 water while at 100°C it is only
1.72 cm3.
(ii) at 0°C, the solubility of CO2 in water is 171.3 cm3 but at 60°C its solubility is reduced to 35.9 cm3.
consequences of rising
temperatures on gas solubility
Some consequences of rising temperatures on gas solubility are
discussed below:
i). When a glass of cold tape water is warmed, the bubbles of dissolved air are seen on the inside of the glass. [The boiled water has a characteristic flat taste because dissolved air (and also the salts) has been expelled during boiling. That is why it is advised to fill boiled water in bottles after cooling].
ii). The effect of rise in temperature on gas solubility is obvious when carbonated drinks bubble continuously as they warm up to room temperature after being refrigerated. Soon carbonated beverages go ‘flat’ due to escape of dissolved CO2 gas. At 0°C, the solubility of CO2 in water is 171.3 cm3 but at 60°C its solubility is reduced to 35.9 cm3. Thus it is advised by soft drink companies that carbonated beverages give their best taste when chilled.]
iii). A much more important consequence is the damage to aquatic life that can result from the decrease in dissolved oxygen when hot water is discharged from power stations into lakes and rivers, an effect known as Thermal Pollution. [The effect is particularly serious in deep lakes because warm water is less dense than cold water. It therefore, tends to remain on top of cold water, at the surface. This situation impedes the dissolving of oxygen into the deeper layers, thus stifling the respiration of all aquatic life needing oxygen. Fish may suffocate and die in these circumstances].
iv). On a hot summer day, an experienced fisherman knowing rules of gas solubility usually picks a deep spot in the river or lake to cast the bait because the oxygen content is greater in the deeper, cooler region, most fish will be found there.
Summary of Effect of Temperature on Solubility
Effect on
Solids in Liquids:
Solubility usually
increases with temperature because heat provides more kinetic energy to
particles, helping them separate and mix.
Example:
Solubility of KNO₃ increases as temperature rises.
Effect on Gases
in Liquids:
Solubility decreases
with temperature because gas molecules gain energy and escape from the liquid.
Example: Warm water holds less CO₂ (why cold drinks go flat when warm).
3. Effect of Pressure
The solubilities of solids and liquids in liquid solvent are not affected by pressure (i.e. pressure has practically no effect on the solubility of solids and liquids). The solubility of gases in liquids considerably increases with increasing pressure.
The quantitative relationship between gas solubility and pressure
is given by Henry’s Law which states that the solubility of gas (i.e. the
amount of gas dissolved in a given amount of liquid solvent) at constant
temperature is directly proportional to the partial pressure of the gas over
the solution.
m ∝ Pg And m = k Pg Where;
m = amount of gas
dissolved.
P = Partial pressure of the gas over the solution.
k = Henry’s Law Constant (with unit mol/liter-atm),
which is temperature dependent and characteristic
of a specific gas.
e.g.
at S.T.P. (standard temperature, 0°C and standard pressure, 1 atmosphere), 0.335 g of CO2 dissolves per 100 cm3 of water but if pressure is doubled 0.67g of CO2 will dissolve.]
This effect is used in the manufacture of soft drinks bottled such as Coca-Cola, 7-up etc. These are bottled in which CO2 is filled under high pressure slightly greater than 1 atm. [CO2 gas is slightly soluble in water at S.T.P., so CO2 gas is filled in soda water bottles under high pressure].
The most common example of Henry’s law behaviour occurs when a can or bottle of soda or other carbonated drinks (in which CO2 is filled under pressure slightly greater than 1 atm) are opened to air, bubbles of CO2 gas comes fizzing out of the solution with effervescence because the pressure of CO2 above the solution drops and CO2 suddenly becomes less soluble
A more serious example of Henry’s law behaviour occurs when a deep-sea diver surfaces too quickly and develops a painful and life-threatening condition called the “bend”. Deep-sea divers rely on compressed air for their oxygen supply. If a diver is suddenly exposed to atmospheric pressure (where the solubility of gases is less), large amounts of nitrogen dissolved in the blood at high underwater pressure, form bubbles in the blood-stream blocking capillaries and inhibiting blood flow, and affect nerve impulses giving rise to the condition known as the bends or decompression sickness. [The bends can be prevented by using an oxygen/helium mixture (98% He and 2% O2) for breathing rather than air (O/N2), because helium has a much lower solubility in blood than nitrogen].
Summary of Effect of Pressure on Solubility
Pressure affects only gases dissolved in liquids.
According to Henry’s Law:
“The solubility of a gas in a liquid is directly proportional to
the pressure of that gas above the liquid.”
Example:
⇒ Soft drinks are bottled under high pressure to keep CO₂
dissolved.
⇒ When opened, pressure drops → gas escapes → fizz appears.
4. Common Ion
Effect
Solubility of a
sparingly soluble salt decreases when a common ion is added from another
source.
This is
explained by Le-Chatelier’s Principle — the system shifts toward the solid
phase to reduce ion concentration.
Example:
AgCl(s)⇌Ag⁺ + Cl⁻
If NaCl is
added → [Cl⁻] increases → equilibrium shifts left → less AgCl dissolves.
5. pH of the
Solution
The solubility
of salts of weak acids (like carbonates, sulfides, and hydroxides) increases in
acidic medium because H⁺ reacts with the anion.
Examples:
CaCO₃ (s)+2H⁺ →Ca2⁺ + CO₂ + H₂O
FeS(s)+2H⁺→Fe2⁺ + H₂S↑
6. Formation of
Complex Ions
Solubility of
some sparingly soluble salts increases in the presence of a complexing agent,
because new soluble complexes form.
Example:
AgCl(s)+2NH₃→[Ag(NH₃)₂]⁺ + Cl−
Here, AgCl
dissolves due to the formation of the soluble diammine silver complex.
7. Lattice Energy and Hydration Energy
Lattice Energy
(U): Energy needed to separate ions in a crystal lattice.
Hydration
Energy (H): Energy released when ions are surrounded by water molecules.
A salt
dissolves if hydration energy > lattice energy.
Example:
NaCl dissolves easily (hydration > lattice), but AgCl does not (lattice >
hydration).
Saturated Solution and Solubility Product (Ksp)
Solubility
Solubility is the extent (tendency) of a solvent to dissolve
solute. It is defined as the amount of solute in gram, which is required to
prepare a saturated solution of 1 liter (1 dm3). It is expressed in
g/liter or g/dm3. The
practical unit is mol/liter.
Types of Compounds according to Solubility
Salts are
classified into three types on the basis of their solubility in the following
table.
Saturated Solution
A solution that contains the maximum equilibrium amount of solute and cannot dissolve any more amount of solute at a specific temperature is called saturated solution. A saturated solution is in a state of dynamic equilibrium between the dissolved, dissociated, ionic compound and the undissolved solid.
Definition of
Solubility Product
The value of Ksp is a measure of the solubility of an ionic salt (ionic compound). The larger the value of Ksp, the greater is the concentration of ions in the solution and hence greater is the solubility of the salt and vice versa. The smaller the value of Ksp, the more insoluble is the salt.
The product of
molar ionic concentration of its positive and negative ions of dissolved
sparingly soluble salt, each raised to an appropriate power equal to its
co-efficient in balanced ionized equilibrium equation in a saturated solution
at a specific temperature is called Solubility Product represented by Ksp. It is
expressed in chemical unit of concentration i.e. molar concentration (mol/dm3).
OR
It is the product of the concentration of the ions of a
dissolved salt produced in a saturated solution; each concentration is raised
to power equal to its co-efficient given in the balanced equation at a given
temperature.
Ksp is the equilibrium constant for the equilibrium between a solid ionic solute and its ions in a saturated solution.
It represents the equilibrium between solid phase and ions in a saturated solution of ionic compounds of relatively low solubility. Solubility products of a salt in a saturated solution remains constant. Like other equilibrium constants it changes with temperature.
General Representation of Solubility Product
For any sparingly soluble salt like AmBn, the Ksp expression is given as:
Explanation
Ksp represents the
maximum value of the ionic product (i.e. Ksp represents the
saturated state of the solution). Thus a salt cannot be dissolved in a solvent
after the achievement of its Ksp. The precipitation of a salt can
only occur when its ionic product is greater than its Ksp.
The solubility product characterizes the solubility of a salt at a given temperature. If there are two similar salts (e.g. CaSO4, BaSO4), the solubility is greater for that salt whose solubility product is greater.
Unit of Solubility Product
Ksp is represented in chemical unit of concentration i.e. molar
concentration (moles/dm3). It is expressed in mol2/dm6
(or M2) or mol3/dm9 (or M3) etc.
Factor affecting
Solubility Product
Ksp varies directly
with temperature
(as increase in temperature increases solubility, which in turn raises
ionic concentrations).
Derivation for a General Ksp Expression
Various ionic compounds such as AgCl, BaSO4, Ca3(PO4)2 etc. are practically very slightly soluble in water and commonly known as sparingly soluble salts. When a sparingly or slightly soluble salt like AmBn is dissolved in water, its very small part becomes ionized and it forms a saturated after certain time. At this stage an equilibrium is established between dissolved ions of salt and its solid phase (undissolved salt). By applying law of mass action, the equilibrium constant (Kc) can be given as:
Since for a saturated solution, the concentration of undissolved
solid salt (AmBn) remains constants, therefore; [AmBn(s)]
= K, it is not included in the equilibrium expression and thus Kc is
replaced by Ksp which is known as solubility product constant or
simply solubility product. By applying law of mass action, the equilibrium
constant (Kc) and the equilibrium expression of this ionic dissolution
equilibrium can be given as:
Derivation for Ksp Expression
Various ionic compounds such as AgCl, BaSO4, Ca3(PO4)2 etc. are practically very slightly soluble in water and commonly known as sparingly soluble salts. CaSO4 is slightly soluble salt, when it is dissolved in water, its very small part becomes ionized and it forms a saturated after certain time. At this stage an equilibrium is established between dissolved ions of salt and its solid phase (undissolved salt).
Since for a saturated solution, the concentration of undissolved
solid salt (CaSO4) remains constants, therefore; [CaSO4(s)]
= K, it is not included in the equilibrium expression and thus Kc is
replaced by Ksp which is known as solubility product constant or
simply solubility product. By applying law of mass action, the equilibrium
constant (Kc) and the equilibrium expression of this ionic
dissolution equilibrium can be given as:
When a sparingly or slightly soluble salt like AgCl is dissolved in
water, then it forms saturated solution after certain time. The amount of salt
in a saturated solution represents the maximum limit of dissolved salt at given
temperature. At this stage an equilibrium is established between dissolved ions
of salt and undissolved salt. By applying law of mass action, the equilibrium
constant (Kc) can be given as:
Applications of Solubility Product
Precipitation is a kind of double displacement reaction in which two solutions of different salts mixed together to form two products, one of these product is insoluble in solution and called precipitate (ppt).
In order to predict precipitation, we have to calculate Ionic Product or reaction quotient (Qsp or QI) denoted by Qsp or QI which is the product of initial ionic molar concentrations of a dissolved salt (not necessarily correspond to those at equilibrium) each raised to the power by its coefficient mentioned in net ionic equation of ionization. Ionic product is relevant to both saturated and unsaturated solution while Ksp is only applicable to saturated solution (which involves a dynamic equilibrium between an insoluble salt and its aqueous ions).
The solubility product (Ksp) of salts helps to predict
whether the precipitation will occur or not from the solution of known ionic
concentrations. Ksp value represents the saturated state of the
solution. The following three
possible relations between Qsp (ionic product) and Ksp
are used to predict the nature of solution even if the salt forms precipitate
by mixing two different salt solutions.
Common Ion Effect
Definition
The process of decreasing the solubility of a sparingly soluble salt in solution by the addition of a highly soluble salt with one common ion refers as common ion effect.
The suppression the degree of ionization of a weak
electrolyte (or sparingly soluble salts) in a saturated solution at a given
temperature by the addition of common ion (+ive or –ive) to decrease its
solubility and for its precipitation is known as Common Ion Effect.
OR
It is the phenomenon of lowering the degree of ionization
of a weak electrolyte (or sparingly soluble salts) by adding a common ion
(cation or anion) to that salt to decrease its solubility and for its
precipitation.
OR
The decrease in the solubility of the salt in a solution that already contains an ion common to that salt is called Common Ion Effect.
Due to Common Ion Effect, when a salt is dissolved in a solution that already contains one of its ions, its solubility is less than in pure water. The solubility of a slightly soluble salt is lowered by adding a soluble salt having same cation or anion to the suspension of the slightly soluble salt in water. e.g.
⇒ With the addition of CaCl2 to CaC2O4,
solubility of CaC2O4 decreases.
⇒ With the addition of KCl to KClO4, solubility
ofKClO4 decreases
⇒ With the addition of HCl to H2S, solubility of the concentration of S2- decreases
⇒ With the addition of NH4Cl to NH4OH, solubility of the concentration of OH- decreases
⇒ With the addition of HCl to NaCl, solubility of the equilibrium of NaCl shifts to left
⇒ With the addition of NaCl to AgCl, solubility of the equilibrium of AgCl shifts to left
Example
The solubility of AgCl is suppressed by
the addition of KCl to its saturated solution.
AgCl(s) ⇌ Ag+(aq) + Cl– (aq)
KCl(s) ⇌ K+(aq) + Cl– (aq)
When saturated solution of sliver chloride is prepared in water, equilibrium is established. If we add some amount of a soluble salt like sodium chloride (NaCl) having Cl− ion common to silver chloride, then according to Le-Chatelier’s principle, the increased concentration of Cl– ion (which is a common ion) in a solution produce a stress on AgCl equilibrium. To reduce this effect effectively and to decrease the added amount of Cl− ions, the excess Cl− ions reacts with some Ag+ ions and shift the equilibrium to the left producing more and more AgCl in the form of its precipitates. Conclusively, solubility of silver chloride in the solution decreases by the addition of NaCl. This effect to decrease the solubility of a sparingly soluble salt by the addition of highly soluble salt having one common ion is called Common ion Effect.
Explanation with Example
The solubility of AgCl is suppressed by the addition of KCl to its saturated solution. According to Le-Chatelier’s Principle, the increased concentration of Cl– ion (which is a common ion) in a solution shifts the equilibrium towards left thereby producing more of AgCl by consuming Cl- ions and thus decrease the amount of Cl–. Thus solubility of AgCl is decreased and AgCl is precipitated. This effect to decrease the solubility of a sparingly soluble salt is called Common ion Effect.
AgCl(s) ⇌ Ag+(aq) + Cl– (aq)
KCl(s) ⇌ K+(aq) + Cl–
(aq)
Common
ion
Importance
Since common ion effect is related to the lowering in the solubility of slightly soluble salt in the precipitation formation, it plays versatile roles in many areas of analytical chemistry such as buffering of solutions, purification of salts, soap formation and other qualitative and quantitative analysis.
Relation between Ksp
and Common Ion Effect
The product of the molar ionic concentration of an electrolyte in a saturated solution at specific temperature is called its Solubility Product (Ksp) which is constant at given temperature.
Ksp of salt (AmBn) = [An+]m [Bm–]n
As long as the ionic product is equal to the Ksp, the solution is saturated. Thus a salt cannot be dissolved in a solvent after the achievement of its solubility product. [If by any means the ionic product is made to exceed to the Ksp, the solution becomes super-saturated and precipitation of the electrolyte takes place and ionization of electrolyte is suppressed]. Thus an electrolyte is precipitated when the concentration of its ions exceed the Ksp. The precipitation of an electrolyte can be occurred by increasing the concentration of any one of its ion. Thus by adding common ion, the solubility of electrolyte is suppressed and its precipitation occurs. For example; the addition of NaCl in a solution of AgCl produces common Cl– ion which exceeds the ionic product from the Ksp of AgCl. As a result of which ionization of AgCl is decreased and AgCl is precipitated out. This is known as Common Ion Effect.
AgCl(s)
⇌ Ag+(aq)
+ Cl– (aq)
NaCl(s) ⇌ Na+(aq) + Cl– (aq)
Thus an electrolyte is precipitated only when the concentrations of
its ions is greater than the Ksp i.e.
[Ag+] [Cl–] or Ionic Product (Q) > Ksp (Precipitation occurs)
Relation between Ksp and Molar solubility (s)
Applications Common Ion Effect in Qualitative Salt Analysis & Purification of Common Salt
(a) Application in
Qualitative Salt Analysis
In Qualitative Analysis, the cations of Group II and Group IV are precipitated as their sulphides but under different conditions due to their different solubility products. In fact the Ksp of Group IV sulphides is higher than that of Group II sulphides.
(i) Precipitation of Basic
Radicals of Group II as their respective Sulphides
Ksp of sulphides of group II cations is very low(less
than 10–28. The cations of Group II basic radicals (Cu2+,
Hg2+, Pb2+, Cd2+, Bi3+, As3+,
Sb3+, Sn2+) are precipitated as sulphides (whose Ksp are
low) by passing H2S through their solution in presence of dilute
HCl. The HCl suppresses the ionization of H2S through common ion
effect by providing common H+ ions thereby decreasing S2–
ions concentrations, which is sufficient to precipitate cations of group II but
not enough to precipitate cations of higher groups due to their high Ksp. [The
sulphide (S2–) ions combine with H+ ions to form
undissociated H2S there-by lowering the concentration of S2–
ions. This reduced concentration of S2– ions is still enough to
exceed the solubility product of Group II sulphides because Ksp of
sulphides of Group II cations is very low. In this low concentration of S2–
ions, the sulphides of all other cations are not precipitated due to their
higher solubility products.]
(ii) Precipitation
of Basic Radicals of IV as their respective Sulphides
Ksp of sulphides of group IV cations is higher than
those of group II sulphides, hence higher concentrations of S2- ions
(from H2S) is required for their precipitation. The cations of group
IV of basic radicals (Zn2+, Mn2+, Co2+, Ni2+)
are precipitated as sulphides by passing H2S gas through their
solution in presence of NH4OH. In order to increase the ionization of H2S,
NH4OH is added which ionized to give OH- ions that combines with H+
ions of H2S to form H2O, thereby shifting the equilibrium
to right giving excess of sulphides ions which is enough to exceed the Ksp
for the precipitation of sulphides of Group IV as CoS, NiS, ZnS etc. This high
concentration of S2- ions may be high enough to exceed Ksp of
sulphides of higher groups. Thus, NH4Cl is added which controls the
ionization of NH4OH, which in turns controls the ionization of H2S.
(iii) Precipitation of
Basic Radicals of III as their respective Hydroxides
The cations of Group III of basis radicals (Cr3+, Fe2+,
Fe3+, Al3+) are precipitated as hydroxides by NH4OH
in presence of NH4Cl. The Ksp of hydroxides of Group III
is very low.
NH4OH which is a precipitating agent is a weak
electrolyte and dissociates partially to give OH– ions and NH4+
ions. In spite of partial dissociation, the concentration of OH–
ions is high enough to exceed the Ksp of Group III as well as
cations of higher group too. To encounter this effect, NH4Cl is
added before adding NH4OH because NH4Cl is a good
Ionizer, produces Common Ion Effect by providing NH4+ ion
which combines with OH– ion to form undissociated NH4OH
thereby lowering the concentration of OH– ions. NH4Cl depresses
the dissociation of NH4OH to such an extent at which OH–
ions concentration is just sufficient to precipitate the hydroxides of Group
III only. The hydroxides of other higher group cations especially group IV
cations (Zn, Ni, Co) are not precipitated due to their higher Ksp.
🧪 MDCAT Chemistry MCQs on Solubility, Solubility Product (Ksp), and Common Ion Effect
Learn Chemistry by Inam Jazbi
Solubility and Solubility Product (Ksp) are essential topics in physical chemistry, especially for MDCAT, ECAT, and FSc exams. Understanding how ionic equilibrium and the common ion effect influence solubility helps students solve conceptual and numerical problems easily.
Let’s test your preparation with some important conceptual MCQs with brief explanations below!
MCQ 1
When a saturated solution of NaCl is prepared, the solution is said to be:
A) Unsaturated
B) Supersaturated
C) Saturated
D) Dilute
Answer: ✅ C) Saturated
Explanation: A saturated solution contains the maximum amount of solute that can dissolve at a given temperature. Any additional solute remains undissolved.
MCQ 2
Solubility product (Ksp) applies to:
A) All types of solutions
B) Only soluble salts
C) Only slightly soluble salts
D) Only gaseous solutions
Answer: ✅ C) Only slightly soluble salts
Explanation: Ksp is used for sparingly soluble ionic compounds that dissociate partially in water to establish equilibrium between solid and ions.
MCQ 3
For the salt ⁺⁻, the solubility product expression is:
A) Ksp = [AgCl]
B) Ksp = [AgCl][H2O]
C) Ksp = [Ag⁺][Cl⁻]
D) Ksp = [Ag⁺]/[Cl⁻]
Answer: ✅ C) Ksp = [Ag⁺][Cl⁻]
Explanation: Solubility product is the product of the molar concentrations of ions each raised to the power of their stoichiometric coefficients.
MCQ 4
The value of Ksp depends on:
A) Pressure
B) Concentration
C) Temperature
D) Volume
Answer: ✅ C) Temperature
Explanation: Ksp changes with temperature because solubility usually increases with temperature for most salts.
MCQ 5
Which of the following decreases the solubility of AgCl in water?
A) Addition of NaNO₃
B) Addition of NaCl
C) Increasing temperature
D) Decreasing pressure
Answer: ✅ B) Addition of NaCl
Explanation: NaCl provides common Cl⁻ ions, reducing the dissociation of AgCl due to the common ion effect, thus decreasing solubility.
MCQ 6
When Na₂SO₄ is added to a saturated solution of BaSO₄, the concentration of Ba²⁺ ions:
A) Increases
B) Decreases
C) Remains same
D) Becomes zero
Answer: ✅ B) Decreases
Explanation: Na₂SO₄ adds SO₄²⁻ ions, which shift equilibrium toward solid BaSO₄, reducing Ba²⁺ concentration — a classic common ion effect example.
MCQ 7
If solubility of PbCl₂ is ‘s’, then its Ksp =
A) s²
B) 2s²
C) 4s³
D) 27s³
Answer: ✅ C) 4s³
Explanation: PbCl₂ ⇌ Pb²⁺ + 2Cl⁻
So, Ksp = [Pb²⁺][Cl⁻]² = (s)(2s)² = 4s³
MCQ 8
Which one has the lowest solubility if all have same Ksp value?
A) AgCl
B) Ag₂CrO₄
C) Ag₂S
D) Ca₃(PO₄)₂
Answer: ✅ D) Ca₃(PO₄)₂
Explanation: The greater the number of ions produced per formula unit, the lower the molar solubility for the same Ksp value.
MCQ 9
The common ion effect is a consequence of:
A) Henry’s law
B) Le-Chatelier’s principle
C) Dalton’s law
D) Raoult’s law
Answer: ✅ B) Le-Chatelier’s principle
Explanation: Addition of a common ion shifts the equilibrium to oppose the change, decreasing solubility — consistent with Le-Chatelier’s principle.
MCQ 10
Which of the following salts has the highest solubility in water at room temperature?
A) AgCl
B) BaSO₄
C) NaCl
D) CaCO₃
Answer: ✅ C) NaCl
Explanation: NaCl is highly soluble in water and does not have a measurable Ksp like sparingly soluble salts such as AgCl or BaSO₄.
MCQ 11
When HCl is added to a solution of CH₃COOH, the degree of ionization of acetic acid:
A) Increases
B) Decreases
C) Remains same
D) Becomes zero
Answer: ✅ B) Decreases
Explanation: HCl provides common H⁺ ions, suppressing the ionization of acetic acid (CH₃COOH ⇌ H⁺ + CH₃COO⁻). This is the common ion effect.
MCQ 12
The solubility of CaF₂ decreases on addition of NaF because:
A) Temperature decreases
B) Ionic product increases
C) Common ion (F⁻) is added
D) Ionic strength decreases
Answer: ✅ C) Common ion (F⁻) is added
Explanation: NaF provides F⁻ ions, shifting equilibrium of CaF₂ ⇌ Ca²⁺ + 2F⁻ to the left, thus decreasing solubility.
MCQ 13
Which of the following pairs will show a common ion effect?
A) H₂SO₄ and NaOH
B) Na₂CO₃ and NaHCO₃
C) NaCl and NaNO₃
D) HCl and NaNO₃
Answer: ✅ B) Na₂CO₃ and NaHCO₃
Explanation: Both contain the carbonate ion (CO₃²⁻) as a common ion, leading to the common ion effect.
MCQ 14
Addition of NH₄Cl to NH₄OH solution:
A) Increases ionization
B) Decreases ionization
C) Does not affect ionization
D) Increases solubility
Answer: ✅ B) Decreases ionization
Explanation: NH₄Cl provides NH₄⁺ ions, a common ion that reduces the dissociation of NH₄OH — this is a common ion effect example.
MCQ 15
Which of the following will not show a common ion effect with AgCl?
A) NaCl
B) HCl
C) KNO₃
D) NaBr
Answer: ✅ C) KNO₃
Explanation: KNO₃ has no Cl⁻ ions, hence it does not share any common ion with AgCl.
MCQ 16
The solubility of BaSO₄ in water decreases when:
A) NaCl is added
B) Na₂SO₄ is added
C) HCl is added
D) Temperature increases
Answer: ✅ B) Na₂SO₄ is added
Explanation: Na₂SO₄ gives SO₄²⁻ ions that shift equilibrium to the left, reducing solubility — a common ion effect.
MCQ 17
Common ion effect is used in:
A) Buffer preparation
B) Electrolysis
C) Osmosis
D) Neutralization
Answer: ✅ A) Buffer preparation
Explanation: Buffers work on the common ion effect principle — weak acid and its salt share a common ion that resists pH change.
MCQ 18
If NH₄Cl is added to a saturated solution of Mg(OH)₂, the solubility of Mg(OH)₂:
A) Increases
B) Decreases
C) Remains constant
D) First increases then decreases
Answer: ✅ B) Decreases
Explanation: NH₄Cl increases H⁺ and NH₄⁺ ions, which suppress ionization of the weak base Mg(OH)₂ — an application of the common ion effect.
MCQ 19
Which statement about common ion effect is incorrect?
A) It reduces the ionization of weak acids or bases
B) It decreases solubility of sparingly soluble salts
C) It increases conductivity of the solution
D) It follows Le-Chatelier’s principle
Answer: ✅ C) It increases conductivity of the solution
Explanation: The common ion effect reduces ionization, so the number of ions and conductivity both decrease, not increase.
MCQ 20
The pH of a solution containing acetic acid and sodium acetate remains nearly constant because of:
A) Hydrolysis
B) Common ion effect
C) Dilution
D) Neutralization
Answer: ✅ B) Common ion effect
Explanation: Sodium acetate provides CH₃COO⁻ ions, suppressing acetic acid ionization. This stabilizes pH — the common ion effect principle.
📚 Summary
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Solubility: Maximum amount of solute dissolved at a given temperature.
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Solubility Product (Ksp): Product of ion concentrations at equilibrium for sparingly soluble salts.
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Common Ion Effect: Suppression of ionization or Decrease in solubility due to the presence of a common ion.
Basis: Common Ion Effect is a direct application of Le-Chatelier’s principle.
Uses of Common Ion Effect: Used in buffer solutions, salt analysis, and precipitation reactions.