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Learning Objectives
- Recognize common ions from various salts, acids, and bases.
- Calculate concentrations involving common ions.
- Calculate ion concentrations involving chemical equilibrium.
The common-ion effect is used to describe the effect on an equilibrium involving a substance that adds an ion that is a part of the equilibrium.
Introduction
The solubility products Ksp's are equilibrium constants in hetergeneous equilibria (i.e., between two different phases). If several salts are present in a system, they all ionize in the solution. If the salts contain a common cation or anion, these salts contribute to the concentration of the common ion. Contributions from all salts must be included in the calculation of concentration of the common ion. For example, a solution containing sodium chloride and potassium chloride will have the following relationship:
\[\mathrm{[Na^+] + [K^+] = [Cl^-]} \label{1}\]
Consideration of charge balance or mass balance or both leads to the same conclusion. The solubility product expression tells us that the equilibrium concentrations of the cation and the anion are inversely related. That is, as the concentration of the anion increases, the maximum concentration of the cation needed for precipitation to occur decreases—and vice versa—so that Ksp is constant. Consequently, the solubility of an ionic compound depends on the concentrations of other salts that contain the same ions. Adding a common cation or anion shifts a solubility equilibrium in the direction predicted by Le Chatelier’s principle. As a result, the solubility of any sparingly soluble salt is almost always decreased by the presence of a soluble salt that contains a common ion. The exceptions generally involve the formation of complex ions, which is discussed later.
Common Ions
When \(\ce{NaCl}\) and \(\ce{KCl}\) are dissolved in the same solution, the \(\mathrm{ {\color{Green} Cl^-}}\) ions are common to both salts. In a system containing \(\ce{NaCl}\) and \(\ce{KCl}\), the \(\mathrm{ {\color{Green} Cl^-}}\) ions are common ions.
\(\mathrm{NaCl \rightleftharpoons Na^+ + {\color{Green} Cl^-}}\)
\(\mathrm{KCl \rightleftharpoons K^+ + {\color{Green} Cl^-}}\)
\(\mathrm{CaCl_2 \rightleftharpoons Ca^{2+} + {\color{Green} 2 Cl^-}}\)
\(\mathrm{AlCl_3 \rightleftharpoons Al^{3+} + {\color{Green} 3 Cl^-}}\)
\(\mathrm{AgCl \rightleftharpoons Ag^+ + {\color{Green} Cl^-}}\)
For example, when \(\ce{AgCl}\) is dissolved into a solution already containing \(\ce{NaCl}\) (actually \(\ce{Na+}\) and \(\ce{Cl-}\) ions), the \(\ce{Cl-}\) ions come from the ionization of both \(\ce{AgCl}\) and \(\ce{NaCl}\). Thus, \(\ce{[Cl- ]}\) differs from \(\ce{[Ag+]}\). The following examples show how the concentration of the common ion is calculated.
Example \(\PageIndex{1}\)
What are \(\ce{[Na+]}\), \(\ce{[Cl- ]}\), \(\ce{[Ca^2+]}\), and \(\ce{[H+]}\) in a solution containing 0.10 M each of \(\ce{NaCl}\), \(\ce{CaCl2}\), and \(\ce{HCl}\)?
Solution
Due to the conservation of ions, we have
\[\mathrm{[Na^+] = [Ca^{2+}] = [H^+] = 0.10\: \ce M}\nonumber.\]
but
\[\begin{alignat}{3}
&\ce{[Cl- ]} &&= && && \:\textrm{0.10 (due to NaCl)}\nonumber \\
& && && + &&\mathrm{\:0.20\: (due\: to\: CaCl_2)}\nonumber\\
& && && + &&\mathrm{\:0.10\: (due\: to\: HCl)}\nonumber\\
& &&= && &&\mathrm{\:0.40\: M}\nonumber
\end{alignat}\]
Exercise \(\PageIndex{1}\)
John poured 10.0 mL of 0.10 M \(\ce{NaCl}\), 10.0 mL of 0.10 M \(\ce{KOH}\), and 5.0 mL of 0.20 M \(\ce{HCl}\) solutions together and then he made the total volume to be 100.0 mL. What is \(\ce{[Cl- ]}\) in the final solution?
\[\mathrm{[Cl^-] = \dfrac{0.1\: M\times 10\: mL+0.2\: M\times 5.0\: mL}{100.0\: mL} = 0.020\: M}\nonumber\]
Le Châtelier's Principle states that if an equilibrium becomes unbalanced, the reaction will shift to restore the balance. If a common ion is added to a weak acid or weak base equilibrium, then the equilibrium will shift towards the reactants, in this case the weak acid or base.
Example \(\PageIndex{2}\)
Consider the lead(II) ion concentration in this saturated solution of PbCl2. The balanced reaction is
\[ PbCl_{2 (s)} \rightleftharpoons Pb^{2+} _{(aq)} + 2Cl^-_{(aq)}\nonumber\]
Defining \(s\) as the concentration of dissolved lead(II) chloride, then:
\[[Pb^{2+}] = s\nonumber \]
\[[Cl^- ] = 2s\nonumber\]
These values can be substituted into the solubility product expression, which can be solved for \(s\):
\[\begin{align*} K_{sp} &= [Pb^{2+}] [Cl^-]^2 \\[4pt] &= s \times (2s)^2 \\[4pt] 1.7 \times 10^{-5} &= 4s^3 \\[4pt] s^3 &= \frac{1.7 \times 10^{-5}}{4} \\[4pt] &= 4.25 \times 10^{-6} \\[4pt] s &= \sqrt[3]{4.25 \times 10^{-6}} \\[4pt] &= 1.62 \times 10^{-2}\, mol\ dm^{-3} \end{align*}\]
The concentration of lead(II) ions in the solution is 1.62 x 10-2 M. Consider what happens if sodium chloride is added to this saturated solution. Sodium chloride shares an ion with lead(II) chloride. The chloride ion is common to both of them; this is the origin of the term "common ion effect".
Look at the original equilibrium expression again:
\[ PbCl_2 \; (s) \rightleftharpoons Pb^{2+} \; (aq) + 2Cl^- \; (aq)\nonumber \]
What happens to that equilibrium if extra chloride ions are added? According to Le Châtelier, the position of equilibrium will shift to counter the change, in this case, by removing the chloride ions by making extra solid lead(II) chloride.
Of course, the concentration of lead(II) ions in the solution is so small that only a tiny proportion of the extra chloride ions can be converted into solid lead(II) chloride. The lead(II) chloride becomes even less soluble, and the concentration of lead(II) ions in the solution decreases. This type of response occurs with any sparingly soluble substance: it is less soluble in a solution which contains any ion which it has in common. This is the common ion effect.
A Simple Example
If an attempt is made to dissolve some lead(II) chloride in some 0.100 M sodium chloride solution instead of in water, what is the equilibrium concentration of the lead(II) ions this time? As before, define s to be the concentration of the lead(II) ions.
\[[Pb^{2+}] = s \label{2}\]
The calculations are different from before. This time the concentration of the chloride ions is governed by the concentration of the sodium chloride solution. The number of ions coming from the lead(II) chloride is going to be tiny compared with the 0.100 M coming from the sodium chloride solution. In calculations like this, it can be assumed that the concentration of the common ion is entirely due to the other solution. This simplifies the calculation.
So we assume:
\[[Cl^- ] = 0.100\; M \label{3}\]
The rest of the mathematics looks like this:
\begin{equation} \begin{split} K_{sp}& = [Pb^{2+}][Cl^-]^2 \\ & = s \times (0.100)^2 \\ 1.7 \times 10^{-5} & = s \times 0.00100 \end{split} \end{equation}
therefore:
\begin{equation} \begin{split} s & = \dfrac{1.7 \times 10^{-5}}{0.0100} \\ & = 1.7 \times 10^{-3} \, \text{M} \end{split} \label{4} \end{equation}
Finally, compare that value with the simple saturated solution:
Original solution:
\[[Pb^{2+}] = 0.0162 \, M \label{5}\]
Solution in 0.100 M NaCl solution:
\[ [Pb^{2+}] = 0.0017 \, M \label{6}\]
The concentration of the lead(II) ions has decreased by a factor of about 10. If more concentrated solutions of sodium chloride are used, the solubility decreases further.
Adding a common ion to a system at equilibrium affects the equilibrium composition, but not the ionization constant.
Common Ion Effect with Weak Acids and Bases
Adding a common ion prevents the weak acid or weak base from ionizing as much as it would without the added common ion. The common ion effect suppresses the ionization of a weak acid by adding more of an ion that is a product of this equilibrium.
Example \(\PageIndex{3}\)
The common ion effect of H3O+ on the ionization of acetic acid
The common ion effect suppresses the ionization of a weak base by adding more of an ion that is a product of this equilibrium.
Example \(\PageIndex{4}\)
Consider the common ion effect of OH- on the ionization of ammonia
Adding the common ion of hydroxide shifts the reaction towards the left to decrease the stress (in accordance with Le Châtelier's Principle), forming more reactants. This decreases the reaction quotient, because the reaction is being pushed towards the left to reach equilibrium. The equilibrium constant, \(K_b=1.8 \times 10^{-5}\), does not change. The reaction is put out of balance, or equilibrium.
\[Q_a = \dfrac{[NH_4^+][OH^-]}{[NH_3]}\nonumber \]
At first, when more hydroxide is added, the quotient is greater than the equilibrium constant. The reaction then shifts right, causing the denominator to increase, decreasing the reaction quotient and pulling towards equilibrium and causing \(Q\) to decrease towards \(K\).
Common Ion Effect on Solubility
Consider, for example, the effect of adding a soluble salt, such as CaCl2, to a saturated solution of calcium phosphate [Ca3(PO4)2].
\[\ce{Ca3(PO4)2(s) <=> 3Ca^{2+}(aq) + 2PO^{3−}4(aq)} \label{Eq1}\]
We have seen that the solubility of Ca3(PO4)2 in water at 25°C is 1.14 × 10−7 M (Ksp = 2.07 × 10−33). Thus a saturated solution of Ca3(PO4)2 in water contains
\[3 × (1.14 × 10^{−7}\, M) = 3.42 × 10^{−7}\, M\, \ce{Ca^{2+}} \]
and
\[2 × (1.14 × 10^{−7}\, M) = 2.28 × 10^{−7}\, M\, \ce{PO4^{3−}}\]
according to the stoichiometry shown in Equation \(\ref{Eq1}\) (neglecting hydrolysis to form HPO42−). If CaCl2 is added to a saturated solution of Ca3(PO4)2, the Ca2+ ion concentration will increase such that [Ca2+] > 3.42 × 10−7 M, making Q > Ksp. The only way the system can return to equilibrium is for the reaction in Equation \(\ref{Eq1}\) to proceed to the left, resulting in precipitation of \(\ce{Ca3(PO4)2}\). This will decrease the concentration of both Ca2+ and PO43− until Q = Ksp.
Adding a common ion decreases solubility, as the reaction shifts toward the left to relieve the stress of the excess product. Adding a common ion to a dissociation reaction causes the equilibrium to shift left, toward the reactants, causing precipitation.
Example \(\PageIndex{5}\)
Consider the reaction:
\[ PbCl_2(s) \rightleftharpoons Pb^{2+}(aq) + 2Cl^-(aq)\nonumber \]
What happens to the solubility of PbCl2(s) when 0.1 M NaCl is added?
Solution
\[K_{sp}=1.7 \times 10^{-5}\nonumber\]
\[Q_{sp}= 1.8 \times 10^{-5}\nonumber\]
Identify the common ion: Cl-
Notice: Qsp > Ksp The addition of NaCl has caused the reaction to shift out of equilibrium because there are more dissociated ions. Typically, solving for the molarities requires the assumption that the solubility of PbCl2 is equivalent to the concentration of Pb2+ produced because they are in a 1:1 ratio.
Because Ksp for the reaction is 1.7×10-5, the overall reaction would be (s)(2s)2= 1.7×10-5. Solving the equation for s gives s= 1.62×10-2 M. The coefficient on Cl- is 2, so it is assumed that twice as much Cl- is produced as Pb2+, hence the '2s.' The solubility equilibrium constant can be used to solve for the molarities of the ions at equilibrium.
The molarity of Cl- added would be 0.1 M because Na+ and Cl- are in a 1:1 ration in the ionic salt, NaCl. Therefore, the overall molarity of Cl- would be 2s + 0.1, with 2s referring to the contribution of the chloride ion from the dissociation of lead chloride.
\[\begin{eqnarray} Q_{sp} &=& [Pb^{2+}][Cl^-]^2\nonumber \\ 1.8 \times 10^{-5} &=& (s)(2s + 0.1)^2 \\ s &=& [Pb^{2+}]\nonumber \\ &=& 1.8 \times 10^{-3} M\nonumber\\ 2s &=& [Cl^-]\nonumber\\ &\approx & 0.1 M \end{eqnarray} \]
Notice that the molarity of Pb2+ is lower when NaCl is added. The equilibrium constant remains the same because of the increased concentration of the chloride ion. To simplify the reaction, it can be assumed that [Cl-] is approximately 0.1M since the formation of the chloride ion from the dissociation of lead chloride is so small. The reaction quotient for PbCl2 is greater than the equilibrium constant because of the added Cl-. This therefore shift the reaction left towards equilibrium, causing precipitation and lowering the current solubility of the reaction. Overall, the solubility of the reaction decreases with the added sodium chloride.
The common ion effect usually decreases the solubility of a sparingly soluble salt.
Example \(\PageIndex{6}\)
Calculate the solubility of calcium phosphate [Ca3(PO4)2] in 0.20 M CaCl2.
Given: concentration of CaCl2 solution
Asked for: solubility of Ca3(PO4)2 in CaCl2 solution
Strategy:
- Write the balanced equilibrium equation for the dissolution of Ca3(PO4)2. Tabulate the concentrations of all species produced in solution.
- Substitute the appropriate values into the expression for the solubility product and calculate the solubility of Ca3(PO4)2.
Solution
A The balanced equilibrium equation is given in the following table. If we let x equal the solubility of Ca3(PO4)2 in moles per liter, then the change in [Ca2+] is once again +3x, and the change in [PO43−] is +2x. We can insert these values into the ICE table.
\[Ca_3(PO_4)_{2(s)} \rightleftharpoons 3Ca^{2+}_{(aq)} + 2PO^{3−}_{4(aq)}\]
| Ca3(PO4)2 | [Ca2+] | [PO43−] | |
| initial | pure solid | 0.20 | 0 |
| change | — | +3x | +2x |
| final | pure solid | 0.20 + 3x | 2x |
B The Ksp expression is as follows:
Ksp = [Ca2+]3[PO43−]2 = (0.20 + 3x)3(2x)2 = 2.07×10−33
Because Ca3(PO4)2 is a sparingly soluble salt, we can reasonably expect that x << 0.20. Thus (0.20 + 3x) M is approximately 0.20 M, which simplifies the Ksp expression as follows:
\[\begin{align*}K_{\textrm{sp}}=(0.20)^3(2x)^2&=2.07\times10^{-33}
\\[4pt] x^2&=6.5\times10^{-32}
\\[4pt] x&=2.5\times10^{-16}\textrm{ M}\end{align*}\]
This value is the solubility of Ca3(PO4)2 in 0.20 M CaCl2 at 25°C. It is approximately nine orders of magnitude less than its solubility in pure water, as we would expect based on Le Chatelier’s principle. With one exception, this example is identical to Example \(\PageIndex{2}\)—here the initial [Ca2+] was 0.20 M rather than 0.
Exercise \(\PageIndex{4}\)
Calculate the solubility of silver carbonate in a 0.25 M solution of sodium carbonate. The solubility of silver carbonate in pure water is 8.45 × 10−12 at 25°C.
- Answer
-
2.9 × 10−6 M (versus 1.3 × 10−4 M in pure water)
The Common Ion Effect in Solubility Products: https://youtu.be/_P3wozLs0Tc
Summary
Adding a common cation or common anion to a solution of a sparingly soluble salt shifts the solubility equilibrium in the direction predicted by Le Chatelier’s principle. The solubility of the salt is almost always decreased by the presence of a common ion.
Contributors and Attributions
- Emmellin Tung and Mahtab Danai (UCD)
Jim Clark (Chemguide.co.uk)
Chung (Peter) Chieh (Professor Emeritus, Chemistry @University of Waterloo)
FAQs
18.3: Common-Ion Effect in Solubility Equilibria? ›
The common ion effect suppresses the ionization of a weak base by adding more of an ion that is a product of this equilibrium. Adding the common ion of hydroxide shifts the reaction towards the left to decrease the stress (in accordance with Le Châtelier's Principle), forming more reactants.
How does the common ion effect influence solubility equilibria? ›Common Ion Effect on Solubility
Adding a common ion decreases solubility, as the reaction shifts toward the left to relieve the stress of the excess product. Adding a common ion to a dissociation reaction causes the equilibrium to shift left, toward the reactants, causing precipitation.
The common ion effect is an effect that suppresses the ionization of an electrolyte when another electrolyte (which contains an ion which is also present in the first electrolyte, i.e. a common ion) is added. It is considered to be a consequence of Le Chatlier's principle (or the Equilibrium Law).
Do common ions affect solubility? ›The solubility of an ionic compound is decreased by the presence of a common ion (an ion that is also present in the compound); this is known as the common-ion effect.
Does adding a common ion increase KSP? ›dealing with two of the same reactions the reaction with the ion will have a lesser Ksp and the reaction without the ion will have a greater Ksp. the common ion effect, uncommon ions increase the Ksp value.
What are the factors affecting solubility equilibrium? ›The solubility of most substances depends strongly on the temperature and, in the case of gases, on the pressure. The solubility of most solid or liquid solutes increases with increasing temperature.
Does a higher KSP mean more soluble? ›The solubility product constant (Ksp) describes the equilibrium between a solid and its constituent ions in a solution. The value of the constant identifies the degree to which the compound can dissociate in water. The higher the Ksp, the more soluble the compound is.
Does common ion effect increase concentration? ›Adding a strong acid (e.g. HCl) to a solution of acetic acid (a weak acid) shifts the equilibrium towards reactants, thus inhibiting the ionization of the weak acid. The common ion effect also suppresses the ionization of a weak base by again increasing the concentration of a product ion of the equilibrium.
Do common ions alter the equilibrium constant? ›Adding a common ion to a system at equilibrium affects the equilibrium composition, but not the ionization constant.
What is common ion effect with one example? ›When a salt of a weak acid is added to the acid itself, the dissociation of the weak acid is suppressed further. Acetic acid is a weak acid. It is not completely dissociated in an aqueous solution and hence the following equilibrium exists.
What are the solubility rules for the common ions? ›
- Salts containing Group I elements (Li+, Na+, K+, Cs+, Rb+) are soluble . ...
- Salts containing nitrate ion (NO3-) are generally soluble.
- Salts containing Cl -, Br -, or I - are generally soluble. ...
- Most silver salts are insoluble. ...
- Most sulfate salts are soluble. ...
- Most hydroxide salts are only slightly soluble.
Compounds with small ions are less soluble than compounds with large ions. Small ions are closer to each other, so they have strong attractive forces. It is more difficult for the water to break them apart, so they are less soluble.
What is the relation between KSP and ion product? ›The solubility product (Ksp) is used to calculate equilibrium concentrations of the ions in solution, whereas the ion product (Q) describes concentrations that are not necessarily at equilibrium.
Which pair of ions will have highest KSP? ›HC1, and due to high value of Ksp of group IV sulphides, group reagent is H2S in the presence of NH4OH and NH4C1.
Does KSP increase or decrease with solubility? ›It represents the level at which a solute dissolves in solution. The more soluble a substance is, the higher the Ksp value it has.
What are the 3 factors that affect the solubility of a solution? ›Solubility is affected by 4 factors – temperature, pressure, polarity, and molecular size. Solubility increases with temperature for most solids dissolved in liquid water. This is because higher temperatures increase the vibration or kinetic energy of the solute molecules.
What are the 4 factors affecting solubility? ›There are four major factors that affect the solubility of gases inside liquids. Pressure, temperature, size and chemical reactivity between the liquid and gas. By increasing the pressure, we usually increase the solubility of the gas inside the liquid as per Henry's law.
What are the 3 things conditions that affect solubility? ›The solubility of a substance depends on the physical and chemical properties of that substance. In addition to this, there are few conditions which can manipulate it. Temperature, pressure and the type of bond and forces between the particles are few among them.
What if Ksp is greater than 1? ›A Ksp value greater than 1 means that the dissociation of the solid compound into its constituent ions is highly favorable.
What does a low Ksp indicate? ›The more soluble a substance is, the higher its Ksp value. So if there is a relatively high concentration of dissolved particles in solution, Ksp will be higher. But if there's a relatively lower concentration of these dissolved particles, Ksp will be lower.
Which Ksp has highest solubility? ›
we know that S = √ ksp , according to the equation solubility is directly proportional to Ksp more the Ksp more will be the solubility So, the ZnS will have highest solubility product.
What is the conclusion of the common ion effect? ›Conclusion. The common ion effect is an effect that occurs when an electrolyte is mixed with another electrolyte that contains an ion that is also present in the first electrolyte, i.e. a common ion, and the ionisation of the first electrolyte is suppressed.
Does higher concentration mean higher percent ionization? ›The ionization constants increase as the strengths of the acids increase. Because the ratio includes the initial concentration, the percent ionization for a solution of a given weak acid varies depending on the original concentration of the acid, and actually decreases with increasing acid concentration.
What is the common ion effect experiment? ›When concentrated hydrochloric acid is added to a large test tube containing saturated sodium chloride solution, white sodium chloride precipitates out due to the common ion effect. This is a great demo to illustrate the common ion effect in a general chemistry course.
What causes an increase in equilibrium constant? ›Changing temperature
Where the forward reaction is endothermic, increasing the temperature increases the value of the equilibrium constant. The position of equilibrium also changes if you change the temperature.
The common ion effect suppresses the ionization of a weak acid by adding more of an ion that is a product of this equilibrium. Adding a common ion to a system at equilibrium affects the equilibrium composition, but not the ionization constant.
What ion causes the equilibrium to shift? ›The "stress" must be explained only in terms of the species present in the equilibrium reaction equation: sulfuric acid is a source of hydrogen ions and it is an increase in the hydrogen ion concentration that causes the equilibrium to shift.
What is the common ion effect in simple words? ›The common-ion effect refers to the decrease in solubility of an ionic precipitate by the addition to the solution of a soluble compound with an ion in common with the precipitate. This behaviour is a consequence of Le Chatelier's principle for the equilibrium reaction of the ionic association/dissociation.
What does the common ion effect promote? ›An unusual common ion effect promotes dissolution of metal salts in room-temperature ionic liquids: a strategy to obtain ionic liquids having organic–inorganic mixed cations.
What is the purpose of common ion effect? ›The common-ion effect plays important roles in controlling the pH of a solution, determining the solubility of a slightly soluble salt and thus can control the formation of a precipitate by either reversing the dissociation of the acid, if the acid had already dissociated or reducing the dissociation [10] if the salt ...
What is an example of common ion solubility? ›
An example of the common ion effect is when sodium chloride (NaCl) is added to a solution of HCl and water. The hydrochloric acid and water are in equilibrium, with the products being H3 O+ and Cl- . Then, some sodium chloride is added to the solution. The NaCl dissolves into the solution, forming Na+ and Cl-.
How do you determine the order of increasing solubility? ›Here to determine the increasing solubility of a solution we will use the concept of common ion effect which states that on the addition of a solution into a soluble compound, the solubility will decrease if common ions are present.
What is the golden rule of solubility? ›The golden rule of solubility is that like dissolves like. In other words, polar solvents dissolve polar materials, and non-polar solvents dissolve non-polar materials.
What is the relationship between ion size and solubility? ›That: solubility increases as the size of ions increases. To calculate solubility of substances temperature is kept constant throughout the system. Ksp $=$ solubility product constant. Solubility is calculated as the molarity of the material under excessive undissolved material in a solution.
Does higher charge mean less soluble? ›Solubility: Physical Principles
proportional to the magnitude of the charges of those ions. separated from each other and dissolved in water. If a salt is composed of highly charged ions, it is not very soluble. If a salt is composed of ions with lower charges, it is probably soluble.
The solubility increases because the ions dissolving in solution are stabilized by the presence of other ions (due to lower energy of ion-ion interactions as opposed to ion-dipole interactions).
What happens if ionic product is greater than KSP? ›A supersaturated solution is one in which the ion product exceeds the solubility product. A supersaturated solution is not at equilibrium, and no solid can ordinarily be present in such a solution. If some of the solid is added, the excess ions precipitate out and until solubility equilibrium is achieved.
What happens if the ion product is less than the KSP? ›The ionic product is the value that is used to determine whether a precipitate will form or not when two solutions containing ions that can form precipitate are mixed together. When the ionic product is less than the solubility product, the mixture is unsaturated which means that it will not form a precipitate.
What happens when ionic product is equal to solubility product? ›A precipitate is formed when ionic product becomes equal to the solubility product.
Which ion will precipitate first Ksp? ›Ag2C03 will precipitate first as its Ksp is less than that of CaC03.
Which has the lowest Ksp value? ›
▶"Cobalt(III) sulfide" has the lowest Ksp value.
Which of the following ions has the highest solubility in water? ›CsOH has the highest solubility of water because of low lattice energy due to large size.
What factors affect Ksp value? ›For a given chemical species and solvent system, the main factor which affects the value of Ksp is the temperature. Most often, an increase in the temperature causes an increase in the solubility and value.
How does Le Chatelier's principle apply to the common ion effect in acid base equilibria? ›Solution. Le Châtelier's Principle states that if an equilibrium becomes unbalanced, the reaction will shift to restore the balance. If a common ion is added to a weak acid or weak base equilibrium, then the equilibrium will shift towards the reactants, in this case the weak acid or base.
Is the common ion effect an application of Le Chatelier principle to solubility equilibria? ›The common-ion effect is an application of Le Chatelier's Principle to solubility equilibria. The solubility of a slightly soluble salt is decreased when a common ion (in the form of another, more soluble, salt) is added. For salts that contain an acidic or basic ion, pH can also affect solubility.
What three factors affect an equilibrium of a reaction according to Le Chatelier's principle? ›Temperature, pressure, and concentration all affect the state of equilibrium in a system.
What does Le Chatelier's principle have to do with equilibrium? ›Le Châtelier's principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium shifts to counteract the change to reestablish an equilibrium.
What is the effect of adding a common ion to the solubility of a slightly soluble salt? ›Adding a common cation or common anion to a solution of a sparingly soluble salt shifts the solubility equilibrium in the direction predicted by Le Chatelier's principle. The solubility of the salt is almost always decreased by the presence of a common ion.
What is an example of the common ion effect? ›When sodium chloride is added to the solution the concentration of Cl− ions will increases. The equilibrium shown above will be shifted to the left to form more of solid AgCl.
What is the principle of solubility equilibrium? ›Solubility equilibrium is a type of dynamic equilibrium that exists when a chemical compound in the solid state is in chemical equilibrium with a solution of that compound. The solid may dissolve unchanged, with dissociation, or with chemical reaction with another constituent of the solution, such as acid or alkali.
What is common ion effect and its application? ›
Common-ion effect is a shift in chemical equilibrium, which affects solubility of solutes in a reacting system. The phenomenon is an application of Le-Chatelier's principle for equilibrium reactions that has become a regular occurrence in chemistry analysis and industrial researches.
Which ion is always soluble according to solubility rules? ›1) Salts of ammonium and alkali metals (column 1A excluding hydrogen) are always soluble.