The Common Ion Effect was first discovered by the chemist William Hyde Wollaston in the starting of 19th century. However, this phenomenon was explained by the other scientist Gilbert N. Lewis in year of 1923. He provides very good and easy explanation of the in the 1923. The common ion effect was observed When the solubility od salt is change in the solution where the command ion were present. William observe that the solubility of the particularly is started decrease when the same ion are already present in the solution.
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So the common idea effect idea is refer to as the change in the solubility of ionic compound in the solution which alredy ha the same ion. And this process occur fur to the change in the position of the equilibrium position according to Le Chatelier's principle explanation which is in a solution of a weak acid or base, adding a common ion shifts the equilibrium towards the formation of the undissociated compound, by which the concentration of the is reduced in that solution. For example, adding a salt that contains the common ion to a solution as that of weak acid decreases the dissociation of the acid and by which the concentration of the hydride ion is decreased.
The iso-hydric effect and the common ion effect are both related to each in such a way that the presence of ions affects the equilibrium of chemical reactions, but they describe different phenomena. the iso-hydric effect is a specific application of the common ion effect, particularly use for the system of acids and bases . Both of these effects depict that how the addition of a common ion affects equilibrium, but the iso-hydric effect often focuses on pH-related changes in equilibrium involving weak acids and bases.
The Common Ion Effect is defined as the phenomenon where the solubility of an ionic compound decreases in a solution that already contains one of the ions present in the compound.In other words, the presence of a common ion in a solution reduces the dissociation of a weak acid or base and shifts the equilibrium of the solubility reaction, according to Le Chatelier's principle. This occurs because the increased concentration of one of the ions suppresses the dissociation of the ionic compound.
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The value of the degree of dissociation for a weak electrolyte is decreased by the addition of a strong electrolyte having a common ion. As a result of this effect, the concentration of the uncommon ion of the weak electrolyte decreases.
For example:
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These are the solutions having the same concentration of common ions. An isohydric solution refers to solutions that maintain the same pH level, indicating an identical concentration of hydrogen ions. This can occur when solutions of different acids or bases are mixed in such a way that the resulting solutions have the same hydrogen ion concentration. When mixing solutions of different acids or bases, if the total concentration of hydrogen ions remains constant, the solutions can be considered isohydric. This principle is used in studying the buffering capacity of solutions.
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The pH of a mixture of a weak acid and a strong acid can be understood using the following example.
Weak acid: H2S (0.1M)
Strong acid: HCl (0.3M)
The Ka value for this mixture = 1.2 x 10-20
The chemical equation for H2S is given as follows:
$\mathrm{H}_2 \mathrm{~S} \rightleftharpoons 2 \mathrm{H}^{+}+\mathrm{S}^{2-}$
Initial: c 0 0
Equil: c - c? c? c?
Thus, the equilibrium constant Ka is given as follows:
$\begin{aligned} & \mathrm{K}_{\mathrm{a}}=\frac{[\mathrm{c} \alpha]^2 \cdot \mathrm{c} \alpha}{\mathrm{c}(1-\alpha)}=\frac{\left[\mathrm{H}^{+}\right]^2 \cdot \mathrm{c} \alpha}{\mathrm{c}(1-\alpha)} \\ & 1.2 \times 10^{-20}=\frac{[\mathrm{c} \alpha+0.3] \cdot \mathrm{c} \alpha}{\mathrm{c}(1-\alpha)}=(0.3)^2=0.09 \alpha \\ & \text { Thus, } \alpha=\frac{1.2 \times 10^{-20}}{0.09}=13.3 \times 10^{-20}\end{aligned}$
Again, the chemical equation for HCl is given as follows:
$\mathrm{HCl} \rightarrow \mathrm{H}^{+}+\mathrm{Cl}^{-}$
Initial: 0.3 0 0
Equil: 0 0.3 0.3
Thus, total [H+] for the mixture = c? + 0.3
[H+] = 3 x 10-1 (? is very small for weak acids)
Now, the pH for the mixture is given as follows:
$\begin{aligned} & \mathrm{pH}=-\log \left[\mathrm{H}^{+}\right]=-\log \left[3 \times 10^{-1}\right]=-\log 3+1=-0.47+1 \\ & \mathrm{pH}=0.53\end{aligned}$
Example.1
1.The solubility product of Mg(OH)2 is 1.2 x 10-11. What minimum OH- concentration must be attained (for example, by adding NaOH) to decrease the Mg2+ concentration in a solution of Mg(NO3)2 to less than 1.1 x 10-10 M?
1) (correct)0.33
2)0.33-11
3)0
4)None of the above
Solution
Ksp expression:
Ksp = [Mg2+] [OH]2
We set [Mg2+] = 1.1 x 10-10 and [OH] = S. Substituting into the Ksp expression:
1.2 x 10-11 = (1.1 x 10-10) (S)2
S = 0.33 M
Any sodium hydroxide solution greater than 0.33 M will reduce the [Mg2+] to less than 1.1 x 10-10 M.
Hence, the answer is the option (1).
Example.2
2.The expression for solubility product of Al2(SO4)3 is:
1)$\mathrm{K}_{\mathrm{sp}}=\left[\mathrm{A}^{3+}\right]\left[\mathrm{SO}_4^{2-}\right]$
2) (correct)$\mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Al}^{3+}\right]^2\left[\mathrm{SO}_4^{2-}\right]^3$
3)$\mathrm{Ksp}=\left[\mathrm{Al}^{3+}\right]^3\left[\mathrm{SO}_4^{2-}\right]^2$
4)None of above
Solution
The solubility of Al2(SO4)3.
Al2(SO4)3 ⇌ 2Al3+ +3SO42-
Hence,Ksp = [Al3+]2 [SO42-]3
Hence, the answer is the option (2).
Example.3
3.On addition of ammonium chloride to a solution of ammonium hydroxide:
1)Dissociation of NH4OH increases
2) Concentration of OH− increases
3) (correct) Concentration of OH− decreases
4)Concentration of NH4+ and OH− increases
Solution
Common ion effect - The value of the degree of dissociation for a weak electrolyte is decreased by the addition of a strong electrolyte having a common ion. As a result of this effect, the concentration of the uncommon ion of the weak electrolyte decreases.
For example:
Hence, the answer is the option (2)
Example.4
4.Find the pH of $0.004 \mathrm{M} \mathrm{NH}_4 \mathrm{OH}$ having $3.2 \%$ dissociation.
1)-3.8894
2)38.894
3)4.55
4) (correct)3.8894
Solution
Given,
$\begin{array}{lll}\mathrm{NH}_4 \mathrm{OH} \rightarrow \mathrm{NH}_4^{+}+\mathrm{OH}^{-}(\text {weak base }) & \\ \mathrm{c} \quad 0 \quad 0 & \text { conc. before ionization } \\ \mathrm{c}(1-\mathrm{a}) \quad 0 \quad \text { c.a } \quad \text { c.a } & \text { conc. after ionization } \\ {\left[\mathrm{H}^{+}\right]=\mathrm{c} \cdot a=4 \times 10^{-3} \times \frac{3.2}{100}} & \\ & \\ \mathrm{pH}=-\log \left[\mathrm{H}^{+}\right]=-\log \left[1.28 \times 10^{-2}\right] & \\ \mathrm{pH}=1.8927+2 \approx 3.8894 & \end{array}$
Hence, the answer is the option (4).
Example.5
5.How many grams of NaOH must be dissolved in one litre of solution to give it a pH value of 11?
1) (correct)0.04
2)0.08
3)0.4
4)4
Solution
$\begin{aligned} & \text { Given } \mathrm{pH} \text { of solution }=11 \\ & \mathrm{pH}=-\log \left[\mathrm{H}^{+}\right] \\ & 11=-\log \left[\mathrm{H}^{+}\right] \\ & {\left[\mathrm{H}^{+}\right]=10^{-11} \mathrm{M}} \\ & \mathrm{Kw}=\left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right]=1 \times 10^{-14} \\ & {\left[\mathrm{OH}^{-}\right]=\frac{1 \times 10^{-14}}{10^{-11}}=1 \times 10^{-3} \mathrm{M}} \\ & \text { Strength }=\text { Molarity } \times \mathrm{mol} . \text { weight } \\ & =10^{-3} \times 40 \\ & =0.04 \mathrm{~g} / \text { litre. }\end{aligned}$
Hence, the answer is the option (1).
Example.6
6.What will happen to the degree of dissociation of $\mathrm{CH}_3 \mathrm{COOH}$ if HCl is added to the solution?
1)It will increase because of the common ion
2) (correct)It will decrease because of the common ion
3)It will remain the same as HCl is a strong acid
4)It will first increase and then decrease
Solution
$\begin{aligned} & \mathrm{CH}_3 \mathrm{COOH} \rightleftharpoons \mathrm{CH}_3 \mathrm{COO}^{-}+\mathrm{H}^{+} \\ & \mathrm{HCl} \longrightarrow \mathrm{H}^{+}+\mathrm{Cl}^{-}\end{aligned}$As HCl is a strong acid, it will produce ions which will cause the dissociation equilibrium to shift in the backward direction.
Hence, the degree of dissociation will decrease.
Hence, the answer is the option (2).
The common ion effect and the isohydric effect are important topics in chemistry, particularly in the relation of acid and base equilibrium and also in the buffer solution. The common ion Effect is very important in Knowing and for making rhe buffer Solutions. Adding that ion in the the solution Which is already present in the solution will change the position of the equilibrium. Which is helpfuk for maintaining the constant ph of Solution.
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NCERT Chemistry Notes:
When a common ion (an ion that is already present in the solution) is added to a solution, the common ion effect defines the effect on equilibrium that happens. The solubility of a solution is generally reduced by the common ion effect.
The common ion effect is a phenomena in which adding a common ion to two solutes produces precipitation or decreases ionisation. When sodium chloride (NaCl) is introduced to a solution of HCl and water, the common ion effect occurs.
For gravimetric measurement, the common ion effect is employed to completely precipitate one of the ions as a sparingly soluble salt with a very low solubility product value. Silver ions precipitates to form silver chloride and barium ions precipitates to form barium sulphate.
If a reaction at equilibrium is subjected to a change in parameters such as temperature, pressure, or the concentration of reactants and products, Le Chatelier's principle asserts that the reaction equilibrium shifts in a direction that eliminates the change.
Common ion effect is neutralised when any ion from other electrolyte absorbs any ion from interfering electrolyte. This effect is called the odd-ion effect.
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