Solved Numericals on Common Ion Effect | Physical Chemistry PDF Download
Q.1. 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?
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.
Q.2. 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.
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.
Q.3. 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?
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 10–2
According 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.
Q.4. What are concentration of [Na+], [Cl−], [Ca2+], and [H+] in a solution containing 0.10 M each of NaCl, CaCl2, and HCl?
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.
Q.5. What is the importance of the common ion effect?
Common ion effect plays a critical role in physical chemistry.
- It helps in controlling the pH of the reaction.
- It helps to estimate the solubility of a slightly soluble salt.
Q.6. What is the solubility of AgCl (s) if the solubility product of AgCl is 1.6 × 10−10 in 0.1 M NaCl solution?
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.
Q.7. If the pH of a saturated solution of Ca(OH)2 is 9. What is the solubility product (Ksp) of Ca(OH)2?
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 10-5
According to the law of conservation of ions, the concentration of calcium would be half of hydroxide ions.
[Ca2+] = 0.5 X 10-5
Solubility 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
Q.8. 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?
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
Q.9. What is Le Chatlier’s principle?
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.
Q.10. What is the common ion effect?
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.
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