The common ion effect explains the decrease in the solubility of an ionic precipitate by adding a solution of a soluble compound with an ion common with the deposit. It is under Le Chatlier’s principle of ionic association or dissociation.
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Definition: The common ion effect explains the effect on equilibrium that occurs when a common ion is added to a solution. |
Common Ion Effect Chemistry Questions with Solutions
Q1. In a saturated solution of an electrolyte, the ionic product of their concentration is constant at a constant temperature, and this constant for electrolyte is known as
(a) Ionic product
(b) Solubility product
(c ) Ionization constant
(d) Dissociation constant
Answer: (b) In a saturated solution of an electrolyte, the ionic product of their concentration is constant at a constant temperature, and this constant for electrolyte is known as solubility product.
Q2. On passing a current of hydrochloric acid gas in a saturated solution of sodium chloride, the solubility of sodium chloride
(a) Decreases
(b) Increases
(c ) Remains unaffected
(d) Sodium chloride decomposes
Answer: (a) On passing a current of hydrochloric acid gas in a saturated solution of sodium chloride, the solubility of sodium chloride decreases.
Q3. The dissociation and ionisation are practically the same as both give
(a) Free anions only
(b) Free cations only
(c ) Both free cations and anions
(d) None of the above
Answer: (c ) The dissociation and ionisation are practically the same as both give both free cations and anions.
Q4. The solubility product is a kind of equilibrium constant, and its value depends on
(a) Volume
(b) Energy
(c ) Temperature
(d) None of the above
Answer: (c ) The solubility product is a kind of equilibrium constant, and its value depends on the temperature.
Q5. The solubility product increases with an increase in
(a) Energy
(b) Temperature
(c ) Pressure
(d) None of the above
Answer: (b) The solubility product increases with an increase in temperature.
Q6. What is the common ion effect?
Answer: The common ion effect describes the decrease in solubility of an ionic precipitate by adding a solution of a soluble compound with an ion common with the deposit. It is under Le Chatlier’s principle of ionic association or dissociation.
Q7. What is the importance of the common ion effect?
Answer: Common ion effect plays a critical role in physical chemistry.
1. It helps in controlling the pH of the reaction.
2. It helps to estimate the solubility of a slightly soluble salt.
Q8. What is Le Chatlier’s principle?
Answer: Le Chatlier’s principle states that the change in pressure, temperature and volume leads to a resisting change in the system to reach a new equilibrium state. It can either be in the direction of the reactant or the product.
Q9. What are concentration of [Na+], [Cl−], [Ca2+], and [H+] in a solution containing 0.10 M each of NaCl, CaCl2, and HCl?
Answer: By the law of conservation of ions, the concentration of sodium ions, calcium ions, and hydrogen ions will be equivalent, i.e. 0.10M.
[Na+] = [Ca2+] = [H+] = 0.10 M.but the concentration of [Cl−] will be 0.10 (Due to NaCl), 0.20 (Due to CaCl2) and 0.10 (Due to HCl).
Thus the total concentration of [Cl−] will be = 0.10 + 0.20 + 0.10 = 0.40 M.
Q10. John poured 10.0 mL of 0.10 M NaCl, 10.0 mL of 0.10 M KOH, and 5.0 mL of 0.20 M HCl solutions together and then he made the total volume 100.0 mL. What is the concentration of [Cl−] in the final solution?
Answer: Here,
M1= 0.10
M2= 0.20
V1= 10.0 mL
V2= 5.0 mL
V = 100.0 mL
Concentration of [Cl−] in the final solution = (M1V1 + M2V2) / V
Concentration of [Cl−] in the final solution = (0.10 X 10.0 + 0.20 X 5.0) / 100
Concentration of [Cl−] in the final solution = 2 / 100
Concentration of [Cl−] in the final solution = 0.02 M
Q11. If the pH of a saturated solution of Ba(OH)2 is 12. What is the value of solubility product (Ksp) of Ba(OH)2?
Answer: Reaction: Ba(OH)2 ⇄ Ba2+ + 2 OH−
The pH of a saturated solution of Ba(OH)2 = 12.
The pOH of a saturated solution of Ba(OH)2 = 14 – pH.
The pOH of a saturated solution of Ba(OH)2 = 14 – 12.
The pOH of a saturated solution of Ba(OH)2 = 2.
We will now calculate the concentration of OH– ions.
[OH–] = 10 – pOH [OH–] = 10 – 2 [OH–] = 1 X 10 – 2According to the law of conservation of ions, the concentration of barium would be half of hydroxide ions.
[Ba2+] = 0.5 X 10 – 2Solubility product Ksp = [Ba2+] [OH–]2
Solubility product Ksp = 0.5 X 10 – 2 X (1 X 10 – 2)2
Solubility product Ksp = 0.5 X 10 – 6
Solubility product Ksp = 5 X 10 – 7
Q12. If the pH of a saturated solution of Ca(OH)2 is 9. What is the solubility product (Ksp) of Ca(OH)2?
Answer: Reaction: Ca(OH)2 ⇌ Ca2+ + 2 OH−
The pH of a saturated solution of Ca(OH)2 = 9.
The pOH of a saturated solution of Ca(OH)2 = 14 – pH.
The pOH of a saturated solution of Ca(OH)2 = 14 – 9.
The pOH of a saturated solution of Ca(OH)2 = 5.
We will now calculate the concentration of OH– ions.
[OH–] = 10 – pOH [OH–] = 10 – 5 [OH–] = 1 X 10 – 5According to the law of conservation of ions, the concentration of calcium would be half of hydroxide ions.
[Ca2+] = 0.5 X 10 – 5Solubility product Ksp = [Ca2+] [OH–]2
Solubility product Ksp = 0.5 X 10 – 5 X (1 X 10 – 5)2
Solubility product Ksp = 0.5 X 10 – 15
Solubility product Ksp = 5 X 10 – 16
Q13. The solubility product (Ksp) of BaSO4 is 1.5 X 10−9. Calculate the solubility of barium sulphate in pure water and 0.1 M BaCl2.
Answer: Reaction: BaSO4 (s) ⟶ Ba2+ (aq) + SO42− (aq)
Hence, Ksp = [Ba2+] [SO42−] = x
Then, 1.5 X 10−9 = x X x
x2 = 15 X 10−10
x = 3.87 X 10−5
Then, the solubility of BaSO4 in pure water is 3.87 X 10−5.
Let the solubility of BaSO4 in 0.1 M BaCl2 be ′s′
Reaction: BaSO4 (s) ⟶ Ba2+ (aq) + SO42− (aq)
Initial (From BaCl2) 0
At equilibrium (0.1 M + s) s
Hence, 1.5 X 10−9 = (s + 0.1) X s = s X 0.1 (As s<<1)
s = 1.5 X 10−8
Thus, the solubility of BaSO4 in the presence of 0.1 M BaCl2 is 1.5×10−8.
Q14. What is the solubility of AgCl (s) if the solubility product of AgCl is 1.6×10−10 in 0.1 M NaCl solution?
Answer: Equation:
AgCl ⇌ Ag+ + Cl−
a 0 0
a – S S S + 0.1
The solubility product of AgCl Ksp = 1.6 X 10−10
The solubility product of AgCl Ksp = [Ag+] [Cl−]
The solubility product of AgCl Ksp = S (0.1 + S)
As the value of Ksp is very small.
We can ignore the value of S, with respect to 0.1 M.
1.6 X 10−10 = S X 0.1
S = 1.6 X 10−9 M
Hence, the solubility of AgCl (s) is 1.6 X 10−9 M.
Q15. If the Concentration of the Ag+ ions in a saturated solution of Ag2C2O4 is 2.2 X 10−4 molL−1. What is the solubility product of Ag2C2O4?
Answer: Given, Concentration of Ag+ = 2.2 X 10−4 molL−1
The concentration of C2O4 would be half of that of Ag.
Concentration of C2O4 = 0.5 X 2.2 X 10−4 molL−1
Concentration of C2O4 = 1.1 X 10−4 molL−1
Ksp = [Ag+]2 [C2O4]
Ksp = (2.2 X 10−4 molL−1)2 X 1.1 X 10−4 molL−1
Ksp = 5.3 X 10-12
Hence, the solubility product of Ag2C2O4 is 5.3 X 10-12.
Practise Questions on Common Ion Effect
Q1. If the solubility of BaSO4 in water is 2.42 X 10−3 g L−1 at 298 K., What will be the solubility product (Ksp) of BaSO4? Given the molar mass of BaSO4 is 233 gmol−1.
Q2. The Ksp of Ag2CrO4, AgCl, AgBr and AgI are respectively, 1.1 X 10−12, 1.8 X 10−10, 5.0 X 10−13, 8.3 X 10−17. Which one of the following salts will precipitate last if AgNO3 solution is added to the solution containing equal moles of NaCl, NaBr, NaI and Na2CrO4?
Q3. Find out the solubility of Ni(OH)2 in 0.1 M NaOH. Given that the ionic product of Ni(OH)2 is 2 X 10−15.
Q4. Let the solubilities of AgCl in H2O, 0.01 M CaCl2, 0.01 M NaCl and 0.05 M AgNO3 be s1,s2,s3 and s4 respectively. What will be the correct relationship between these quantities?
Q5. If the pH of a saturated solution of Mg(OH)2 is 9. What is the solubility product (Ksp) of Mg(OH)2?
Click the PDF to check the answers for Practice Questions.
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