All Exams >
JEE >
35 Years Chapter wise Previous Year Solved Papers for JEE >
All Questions
All questions of Equilibrium for JEE Exam
The conjugate base of 
- a)H3PO4
- b)P2O5
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
The conjugate base of
a)
H3PO4
b)
P2O5
c)
d)
|
|
Mr. Sameer answered |
The conjugate base of H2 PO4 - is HPO4 -2. becoz The removal of a proton (H+1) from a conjugate acid gives us its conjugate base.
Aqueous solutions of HNO3 , KOH, CH3COOH and CH3COONa of identical concentrations are provided. The pair(s) of solutions which form a buffer upon mixing is(are)- a)HNO3 and CH3COOH
- b)KOH and CH3COONa
- c)HNO3 and CH3COONa
- d)CH3COOH and CH3COONa
Correct answer is option 'C,D'. Can you explain this answer?
Aqueous solutions of HNO3 , KOH, CH3COOH and CH3COONa of identical concentrations are provided. The pair(s) of solutions which form a buffer upon mixing is(are)
a)
HNO3 and CH3COOH
b)
KOH and CH3COONa
c)
HNO3 and CH3COONa
d)
CH3COOH and CH3COONa
|
Gauri Rane answered |
Any solution of a weak acid and its salt with strong base acts as an acidic buffer solution.
If volume of HNO3 solution added is less as compared to that of CH3COONa solution, it results in the formation of an acidic buffer solution.
If volume of HNO3 solution added is less as compared to that of CH3COONa solution, it results in the formation of an acidic buffer solution.
The pH of 0.1 M solution of the following salts increases in the order.- a)NaCl < NH4Cl < NaCN < HCl
- b)HCl < NH4Cl < NaCl < NaCN
- c)NaCN < NH4Cl < NaCl < HCl
- d)HCl < NaCl < NaCN < NH4Cl
Correct answer is option 'B'. Can you explain this answer?
The pH of 0.1 M solution of the following salts increases in the order.
a)
NaCl < NH4Cl < NaCN < HCl
b)
HCl < NH4Cl < NaCl < NaCN
c)
NaCN < NH4Cl < NaCl < HCl
d)
HCl < NaCl < NaCN < NH4Cl
|
Geetika Shah answered |
The characteristics of the given solutions are:
NaCl – neutral solution
NH4Cl – slightly acidic due to the following reaction
NaCl – neutral solution
NH4Cl – slightly acidic due to the following reaction
NaCN – slightly alkaline due to the following reaction
HCl – highly acidic
The pH of the solution will follow the order highly acidic < slightly acidic < neutral < slightly alkaline i.e.
HCl < NH4Cl < NaCl < NaCN
The pH of the solution will follow the order highly acidic < slightly acidic < neutral < slightly alkaline i.e.
HCl < NH4Cl < NaCl < NaCN
PARAGRAPH 2Thermal decomposition of gaseous X2 to gaseous X at 298 K takes place according to the following equation:
The standard reaction Gibbs energy, ΔrG°, of this reaction is positive. At the start of the reaction, there is one mole of X2 and no X. As the reaction proceeds, the number of moles of X formed is given by β. Thus, βequilibrium is the number of moles of X formed at equilibrium. The reaction is carried out at a constant total pressure of 2 bar. Consider the gases to behave ideally.(Given R = 0.083 L bar K–1 mol–1)Q. The equilibrium constant Kp for this reaction at 298 K, in terms of βequilibrium, is - a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
PARAGRAPH 2
Thermal decomposition of gaseous X2 to gaseous X at 298 K takes place according to the following equation:
The standard reaction Gibbs energy, ΔrG°, of this reaction is positive. At the start of the reaction, there is one mole of X2 and no X. As the reaction proceeds, the number of moles of X formed is given by β. Thus, βequilibrium is the number of moles of X formed at equilibrium. The reaction is carried out at a constant total pressure of 2 bar. Consider the gases to behave ideally.
(Given R = 0.083 L bar K–1 mol–1)
Q. The equilibrium constant Kp for this reaction at 298 K, in terms of βequilibrium, is
a)
b)
c)
d)
|
Rohit Jain answered |
if KP = KC (RT)x where the symbols have usual meaning then the value of x is (assuming ideality):
A liquid is in equilibrium with its vapour at its boiling point. On the average, the molecules in the two phases have equal :- a)inter-molecular forces
- b)potential energy
- c)total energy
- d)kinetic energy
Correct answer is option 'C'. Can you explain this answer?
A liquid is in equilibrium with its vapour at its boiling point. On the average, the molecules in the two phases have equal :
a)
inter-molecular forces
b)
potential energy
c)
total energy
d)
kinetic energy
|
Sameer Saha answered |
Equilibrium between Liquid and Vapor
When a liquid is heated, its temperature gradually increases until it reaches a certain point called the boiling point. At this temperature, the liquid changes from its liquid phase to a gaseous phase, resulting in the formation of vapor. The process of liquid turning into vapor is known as boiling.
Equilibrium at Boiling Point
At the boiling point, the liquid and its vapor are in equilibrium with each other. This means that the rate at which molecules evaporate from the liquid phase is equal to the rate at which molecules condense back into the liquid phase. As a result, there is no net change in the amount of liquid or vapor present.
Molecular Properties in Equilibrium
In this equilibrium state, the average properties of the molecules in the liquid phase are equal to the average properties of the molecules in the vapor phase. This implies that certain molecular properties are equal between the two phases.
Intermolecular Forces
Intermolecular forces are the attractive forces that exist between molecules. In the liquid phase, these forces are stronger due to the close proximity of molecules. In the vapor phase, the intermolecular forces are weaker since the molecules are more spread out. However, in equilibrium, the average intermolecular forces in both phases are equal. This ensures that the rate of evaporation is balanced by the rate of condensation.
Potential Energy
Potential energy is the energy possessed by an object due to its position or configuration. In the liquid phase, the molecules are closer together, resulting in higher potential energy. In the vapor phase, the molecules are more spread out, leading to lower potential energy. However, in equilibrium, the average potential energy of the molecules in both phases is equal.
Total Energy
Total energy refers to the sum of kinetic energy and potential energy. While the potential energy may differ between the liquid and vapor phases, the total energy remains constant in equilibrium. This is because the energy is conserved during the phase transition from liquid to vapor and vice versa.
Kinetic Energy
Kinetic energy is the energy possessed by an object due to its motion. In the liquid phase, the molecules have lower kinetic energy compared to the vapor phase. This is because the molecules in the liquid are more confined and have less freedom of movement. However, in equilibrium, the average kinetic energy of the molecules in both phases is equal.
Conclusion
In conclusion, at the boiling point, the liquid and its vapor are in equilibrium. The average intermolecular forces, potential energy, total energy, and kinetic energy of the molecules in both phases are equal. This equilibrium state ensures a balance between evaporation and condensation, resulting in no net change in the amount of liquid or vapor.
When a liquid is heated, its temperature gradually increases until it reaches a certain point called the boiling point. At this temperature, the liquid changes from its liquid phase to a gaseous phase, resulting in the formation of vapor. The process of liquid turning into vapor is known as boiling.
Equilibrium at Boiling Point
At the boiling point, the liquid and its vapor are in equilibrium with each other. This means that the rate at which molecules evaporate from the liquid phase is equal to the rate at which molecules condense back into the liquid phase. As a result, there is no net change in the amount of liquid or vapor present.
Molecular Properties in Equilibrium
In this equilibrium state, the average properties of the molecules in the liquid phase are equal to the average properties of the molecules in the vapor phase. This implies that certain molecular properties are equal between the two phases.
Intermolecular Forces
Intermolecular forces are the attractive forces that exist between molecules. In the liquid phase, these forces are stronger due to the close proximity of molecules. In the vapor phase, the intermolecular forces are weaker since the molecules are more spread out. However, in equilibrium, the average intermolecular forces in both phases are equal. This ensures that the rate of evaporation is balanced by the rate of condensation.
Potential Energy
Potential energy is the energy possessed by an object due to its position or configuration. In the liquid phase, the molecules are closer together, resulting in higher potential energy. In the vapor phase, the molecules are more spread out, leading to lower potential energy. However, in equilibrium, the average potential energy of the molecules in both phases is equal.
Total Energy
Total energy refers to the sum of kinetic energy and potential energy. While the potential energy may differ between the liquid and vapor phases, the total energy remains constant in equilibrium. This is because the energy is conserved during the phase transition from liquid to vapor and vice versa.
Kinetic Energy
Kinetic energy is the energy possessed by an object due to its motion. In the liquid phase, the molecules have lower kinetic energy compared to the vapor phase. This is because the molecules in the liquid are more confined and have less freedom of movement. However, in equilibrium, the average kinetic energy of the molecules in both phases is equal.
Conclusion
In conclusion, at the boiling point, the liquid and its vapor are in equilibrium. The average intermolecular forces, potential energy, total energy, and kinetic energy of the molecules in both phases are equal. This equilibrium state ensures a balance between evaporation and condensation, resulting in no net change in the amount of liquid or vapor.
For a sparingly soluble salt ApBq, the relationship of its solubility product (LS) with its solubility (S) is- a)LS = S p+q.pp.qq
- b)LS = S p+q.pq.qp
- c)LS = S pq.pp.qq
- d)LS = Spq.(pq)p+q
Correct answer is option 'A'. Can you explain this answer?
For a sparingly soluble salt ApBq, the relationship of its solubility product (LS) with its solubility (S) is
a)
LS = S p+q.pp.qq
b)
LS = S p+q.pq.qp
c)
LS = S pq.pp.qq
d)
LS = Spq.(pq)p+q
|
Aryan Dasgupta answered |
The correct option: (a) Ls = S^P + q . p^p.q^q
Explanation:
Explanation:
The molar solubility (in mol L–1) of a sparingly soluble salt MX4 is ‘s’. The corresponding solubility product is Ksp. ‘s’ is given in term of Ksp by the relation :- a)s = (256 Ksp)1/5
- b)s = (128 Ksp)1/4
- c)s = (Ksp / 128)1/4
- d)s = (Ksp / 256)1/5
Correct answer is option 'D'. Can you explain this answer?
The molar solubility (in mol L–1) of a sparingly soluble salt MX4 is ‘s’. The corresponding solubility product is Ksp. ‘s’ is given in term of Ksp by the relation :
a)
s = (256 Ksp)1/5
b)
s = (128 Ksp)1/4
c)
s = (Ksp / 128)1/4
d)
s = (Ksp / 256)1/5
|
Ananya Das answered |
Which one of the following is the strongest acid?- a)ClO3(OH)
- b)ClO2(OH)
- c)SO(OH)2
- d)SO2(OH)2
Correct answer is option 'A'. Can you explain this answer?
Which one of the following is the strongest acid?
a)
ClO3(OH)
b)
ClO2(OH)
c)
SO(OH)2
d)
SO2(OH)2
|
|
Rohit Jain answered |
TIPS/Formulae :
(i) Higher the electronegatively of central atom higher will be the acidic strength.
(ii) In case of same atom higher the value of oxidation state of the metal, higher will be its acidic strength.
The electronegativity of Cl > S.
Oxidation no. of Cl in ClO3(OH) = + 7 Oxidation no. of Cl in ClO2(OH) = + 5 Oxidation no. of S in SO(OH)2 = + 4 Oxidation no. of S in SO2(OH)2 = + 6
(i) Higher the electronegatively of central atom higher will be the acidic strength.
(ii) In case of same atom higher the value of oxidation state of the metal, higher will be its acidic strength.
The electronegativity of Cl > S.
Oxidation no. of Cl in ClO3(OH) = + 7 Oxidation no. of Cl in ClO2(OH) = + 5 Oxidation no. of S in SO(OH)2 = + 4 Oxidation no. of S in SO2(OH)2 = + 6
∴ ClO3(OH) is the strongest acid.
What is the conjugate base of OH - ?- a)O2-
- b)O-
- c)H2O
- d)O2
Correct answer is option 'A'. Can you explain this answer?
What is the conjugate base of OH - ?
a)
O2-
b)
O-
c)
H2O
d)
O2
|
|
Anand Kumar answered |
H2O = H+ OH-
OH- = H+ + O2-
OH- = H+ + O2-
The following equilibrium in established when hydrogen chloride is dissolved in acetic acid.
The set that characterises the conjugate acid-base pairs is- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
The following equilibrium in established when hydrogen chloride is dissolved in acetic acid.
The set that characterises the conjugate acid-base pairs is
a)
b)
c)
d)
|
|
Geetika Shah answered |
Since HCl is stronger than CH3COOH hence acts as acid. On the other hand Cl– is a stronger base than 
and is the conjugate base of HCl.
and is the conjugate base of HCl.PARAGRAPH 1When 100 mL of 1.0 M HCl was mixed with 100 mL of 1.0 M NaOH in an insulated beaker at constant pressure, a temperature increase of 5.7 °C was measured for the beaker and its contents (Expt.1).
Because the enthalpy of neutralization of a strong acid with a strong base is a constant (–57.0 kJ mol–1), this experiment could be used to measure the calorimeter constant. In a second experiment (Expt. 2), 100 mL of 2.0 M acetic acid (Ka = 2.0 × 10–5) was mixed with 100 mL of 1.0 M NaOH (under identical conditions to Expt. 1) where a temperature rise of 5.6 °C was measured. (Consider heat capacity of all solutions as 4.2 J g–1 K–1 and density of all solutions as 1.0 g mL–1)Q. Enthalpy of dissociation (in kJ mol–1) of acetic acid obtained from the Expt.2 is- a)1.0
- b)10.0
- c)24.5
- d)51.4
Correct answer is option 'A'. Can you explain this answer?
PARAGRAPH 1
When 100 mL of 1.0 M HCl was mixed with 100 mL of 1.0 M NaOH in an insulated beaker at constant pressure, a temperature increase of 5.7 °C was measured for the beaker and its contents (Expt.1).
Because the enthalpy of neutralization of a strong acid with a strong base is a constant (–57.0 kJ mol–1), this experiment could be used to measure the calorimeter constant. In a second experiment (Expt. 2), 100 mL of 2.0 M acetic acid (Ka = 2.0 × 10–5) was mixed with 100 mL of 1.0 M NaOH (under identical conditions to Expt. 1) where a temperature rise of 5.6 °C was measured. (Consider heat capacity of all solutions as 4.2 J g–1 K–1 and density of all solutions as 1.0 g mL–1)
Because the enthalpy of neutralization of a strong acid with a strong base is a constant (–57.0 kJ mol–1), this experiment could be used to measure the calorimeter constant. In a second experiment (Expt. 2), 100 mL of 2.0 M acetic acid (Ka = 2.0 × 10–5) was mixed with 100 mL of 1.0 M NaOH (under identical conditions to Expt. 1) where a temperature rise of 5.6 °C was measured. (Consider heat capacity of all solutions as 4.2 J g–1 K–1 and density of all solutions as 1.0 g mL–1)
Q. Enthalpy of dissociation (in kJ mol–1) of acetic acid obtained from the Expt.2 is
a)
1.0
b)
10.0
c)
24.5
d)
51.4
|
Avik Nair answered |
Let the heat capacity of insulated beaker be C.
Mass of aqueous content in expt. 1 = (100 + 100) × 1 = 200 g
⇒ ± Total heat capacity = (C + 200 × 4.2) J/K Moles of acid, base neutralised in expt. 1 = 0.1 × 1 = 0.1
⇒ Heat released in expt. 1 = 0.1 × 57 = 5.7 KJ = 5.7 × 1000 J
⇒ 5.7 × 1000 = (C + 200 × 4.2) × Δ´T.
5.7 × 1000 = (C + 200 + 4.2) × 5.7
⇒ (C + 200 × 4.2) = 1000
In second experiment,
Mass of aqueous content in expt. 1 = (100 + 100) × 1 = 200 g
⇒ ± Total heat capacity = (C + 200 × 4.2) J/K Moles of acid, base neutralised in expt. 1 = 0.1 × 1 = 0.1
⇒ Heat released in expt. 1 = 0.1 × 57 = 5.7 KJ = 5.7 × 1000 J
⇒ 5.7 × 1000 = (C + 200 × 4.2) × Δ´T.
5.7 × 1000 = (C + 200 + 4.2) × 5.7
⇒ (C + 200 × 4.2) = 1000
In second experiment,
Total mass of aqueous content = 200 g
⇒ Total heat capacity = (C + 200 × 4.2) = 1000
⇒ Heat released = 1000 × 5.6 = 5600 J.
Overall, only 0.1 mol of CH3COOH undergo neutralization.
Overall, only 0.1 mol of CH3COOH undergo neutralization.
⇒ Δ Hneutralization of CH3COOH = -5600/0.1
= – 56000 J/mol
= – 56 KJ/mol.
= – 56 KJ/mol.
⇒ Δ Hneutralization of CH3COOH = 57 – 56 = 1 KJ/mol
For the reaction :
The forward reaction at constant temperature is favoured by- a)introducing an inert gas at constant volume
- b)introducing chlorine gas at constant volume
- c)introducing an inert gas at constant pressure
- d)increasing the volume of the container
- e)introducing PCl5 at constant volume
Correct answer is option 'C,D,E'. Can you explain this answer?
For the reaction :
The forward reaction at constant temperature is favoured by
a)
introducing an inert gas at constant volume
b)
introducing chlorine gas at constant volume
c)
introducing an inert gas at constant pressure
d)
increasing the volume of the container
e)
introducing PCl5 at constant volume
|
Hansa Sharma answered |
Introduction of an inert gas at constant pressure causes the equilibrium to shift in a direction in which number of moles increases. The forward reaction is further accelerated by increase in the quantity of substrate, i.e., PCl5 and by the increase of space, i.e., volume of container.
Change in volume of the system does not alter which of the following equilibria?- a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
Change in volume of the system does not alter which of the following equilibria?
a)
b)
c)
d)
|
|
Smruti Sucharita answered |
bcoz in first rxn no of moles of product =no.of moles of reactants....
so this system is in eqbm ..
so volume has no effect on this.
The best indicator for detection of end point in titration of a weak acid and a strong base is :- a)methyl orange (3 to 4)
- b)methyl red (5 to 6)
- c)bromothymol blue (6 to 7.5)
- d)phenolphthalein (8 to 9.6)
Correct answer is option 'D'. Can you explain this answer?
The best indicator for detection of end point in titration of a weak acid and a strong base is :
a)
methyl orange (3 to 4)
b)
methyl red (5 to 6)
c)
bromothymol blue (6 to 7.5)
d)
phenolphthalein (8 to 9.6)
|
Ipsita Sen answered |
1. Then a weak acid is titrated against a strong base NIOS basic salt and exist present at the end.
2. Which makes and Slightly basic with PH around 8 and signup telling that changes its colour in this pH range would be the best choice of indicated to indicate the end.
2. Which makes and Slightly basic with PH around 8 and signup telling that changes its colour in this pH range would be the best choice of indicated to indicate the end.
The reaction which proceeds in the forward direction is- a)Fe2O3 + 6HCl → 2FeCl3 + 3H2O
- b)NH3 + H2O + NaCl → NH4Cl + NaOH
- c)SnCl4 + Hg2Cl2 → SnCl2 + 2HgCl2
- d)2CuI + I2 + 4K+ → 2Cu2+ + 4KI
Correct answer is option 'A'. Can you explain this answer?
The reaction which proceeds in the forward direction is
a)
Fe2O3 + 6HCl → 2FeCl3 + 3H2O
b)
NH3 + H2O + NaCl → NH4Cl + NaOH
c)
SnCl4 + Hg2Cl2 → SnCl2 + 2HgCl2
d)
2CuI + I2 + 4K+ → 2Cu2+ + 4KI
|
Rohit Yadav answered |
The correct option: (a) Fe2O3 + 6HCl = 2FeCl3 + 3H2O
Explanation:
Due to absence of hydrolysis of FeCl3, backward reaction will not take place.
Explanation:
Due to absence of hydrolysis of FeCl3, backward reaction will not take place.
The pH of a 0.1 molar solution of the acid HQ is 3. The value of the ionization constant, Ka of the acid is :- a)3 × 10–1
- b)1 × 10–3
- c)1 × 10–5
- d)1 × 10–7
Correct answer is option 'C'. Can you explain this answer?
The pH of a 0.1 molar solution of the acid HQ is 3. The value of the ionization constant, Ka of the acid is :
a)
3 × 10–1
b)
1 × 10–3
c)
1 × 10–5
d)
1 × 10–7
|
Siddharth Datta answered |
The pH of a solution is related to the ionization constant (Ka) of the acid through the equation:
pH = -log[H+]
In this case, the pH of the solution is given as 3. Therefore, [H+] = 10^-3.
Since the acid HQ is a monoprotic acid, it can be assumed that [HQ] = [H+].
The ionization constant (Ka) of the acid is defined as:
Ka = [H+][A-] / [HA]
Given that [HQ] = [H+] = 10^-3 and [A-] = 0.1 (since the acid is 0.1 M), we can substitute these values into the equation:
Ka = (10^-3)(0.1) / (10^-3)
The value of the ionization constant (Ka) is therefore 0.1.
pH = -log[H+]
In this case, the pH of the solution is given as 3. Therefore, [H+] = 10^-3.
Since the acid HQ is a monoprotic acid, it can be assumed that [HQ] = [H+].
The ionization constant (Ka) of the acid is defined as:
Ka = [H+][A-] / [HA]
Given that [HQ] = [H+] = 10^-3 and [A-] = 0.1 (since the acid is 0.1 M), we can substitute these values into the equation:
Ka = (10^-3)(0.1) / (10^-3)
The value of the ionization constant (Ka) is therefore 0.1.
An acidic buffer solution can be prepared by mixing the solutions of- a)ammonium acetate and acetic acid
- b)ammonium chloride and ammonioum hydroxide
- c)sulphuric acid and sodium sulphate
- d)sodium chloride and sodium hydroxide.
Correct answer is option 'A'. Can you explain this answer?
An acidic buffer solution can be prepared by mixing the solutions of
a)
ammonium acetate and acetic acid
b)
ammonium chloride and ammonioum hydroxide
c)
sulphuric acid and sodium sulphate
d)
sodium chloride and sodium hydroxide.
|
Maulik Khanna answered |
NOTE : Acidic buffer is mixture of weak acid and its salt with common anion.
(a) CH3COOH + CH3COONH4 is acidic buffer.
(b) NH4Cl + NH4OH is basic buffer.
(c) H2SO4 + Na2SO4 is not buffer because both the compounds are strong electrolytes.
(d) NaCl + NaOH is not buffer solution because both compounds are strong electrolytes.
(b) NH4Cl + NH4OH is basic buffer.
(c) H2SO4 + Na2SO4 is not buffer because both the compounds are strong electrolytes.
(d) NaCl + NaOH is not buffer solution because both compounds are strong electrolytes.
The solubility product of a salt having general formula MX2 , in water is : 4 × 10 -12. The concentration of M2+ ions in the aqueous solution of the salt is- a)4.0 x 10-10 M
- b)1.6 x 10-4 M
- c)1.0 x 10-4 M
- d)2.0 x 10- 6 M
Correct answer is option 'C'. Can you explain this answer?
The solubility product of a salt having general formula MX2 , in water is : 4 × 10 -12. The concentration of M2+ ions in the aqueous solution of the salt is
a)
4.0 x 10-10 M
b)
1.6 x 10-4 M
c)
1.0 x 10-4 M
d)
2.0 x 10- 6 M
|
Anirban Sharma answered |
The solubility product (Ksp) of a salt MX2 in water is given by the equation:
Ksp = [M]x[X]2
where [M] represents the molar concentration of the metal ion M and [X] represents the molar concentration of the anion X.
Since the salt has a general formula MX2, it means that for every mole of the salt that dissolves, one mole of metal ion M is released along with 2 moles of anion X. Therefore, the molar concentration of the metal ion M is equal to the initial concentration of the salt, and the molar concentration of the anion X is twice the initial concentration of the salt.
Thus, the solubility product of the salt MX2 in water is:
Ksp = [M]x[X]2 = [MX2]x[2X]2 = [MX2]x4[X]2
Therefore, the solubility product of a salt having general formula MX2 in water is 4 times the product of the initial concentration of the salt and the square of the initial concentration of the anion X.
Ksp = [M]x[X]2
where [M] represents the molar concentration of the metal ion M and [X] represents the molar concentration of the anion X.
Since the salt has a general formula MX2, it means that for every mole of the salt that dissolves, one mole of metal ion M is released along with 2 moles of anion X. Therefore, the molar concentration of the metal ion M is equal to the initial concentration of the salt, and the molar concentration of the anion X is twice the initial concentration of the salt.
Thus, the solubility product of the salt MX2 in water is:
Ksp = [M]x[X]2 = [MX2]x[2X]2 = [MX2]x4[X]2
Therefore, the solubility product of a salt having general formula MX2 in water is 4 times the product of the initial concentration of the salt and the square of the initial concentration of the anion X.
Solid Ba(NO3)2 is gradually dissolved in a 1.0 × 10– 4 M Na2CO3 solution. At what concentration of Ba2+ will a precipitate begin to form? (KSP for for BaCO3 = 5.1 × 10–9)- a)5.1 × 10 –5 M
- b)8.1 × 10 –8 M
- c)8.1 × 10 –7 M
- d)4.1 × 10 –5 M
Correct answer is option 'A'. Can you explain this answer?
Solid Ba(NO3)2 is gradually dissolved in a 1.0 × 10– 4 M Na2CO3 solution. At what concentration of Ba2+ will a precipitate begin to form? (KSP for for BaCO3 = 5.1 × 10–9)
a)
5.1 × 10 –5 M
b)
8.1 × 10 –8 M
c)
8.1 × 10 –7 M
d)
4.1 × 10 –5 M
|
|
Hiral Mukherjee answered |
M solution of Na2SO4. Will precipitation occur? If so, what will be the concentrations of Ba2+, NO3^-, and Na+ when precipitation is complete?
To determine if precipitation will occur, we need to compare the solubility product constant (Ksp) of Ba(NO3)2 with the concentrations of the ions in the solution.
The balanced equation for the dissociation of Ba(NO3)2 is:
Ba(NO3)2(s) -> Ba2+(aq) + 2NO3^-(aq)
The solubility product constant (Ksp) expression for Ba(NO3)2 is:
Ksp = [Ba2+][NO3^-]^2
The Ksp value for Ba(NO3)2 is 4.5 x 10^(-5).
In the solution, Ba(NO3)2 will dissociate into Ba2+ and NO3^- ions. Na2SO4 will dissociate into Na+ and SO4^2- ions.
Since Na2SO4 is a soluble salt, it will completely dissociate in water. Therefore, the concentration of Na+ ions will be 1.0 M.
When precipitation occurs, the concentrations of Ba2+ and NO3^- ions will reach their maximum solubility according to the Ksp value.
Let's assume x M is the concentration of Ba2+ and 2x M is the concentration of NO3^-.
Using the Ksp expression, we can write:
Ksp = [Ba2+][NO3^-]^2
4.5 x 10^(-5) = x * (2x)^2
4.5 x 10^(-5) = 4x^3
x^3 = 1.125 x 10^(-5)
x ≈ 0.022 M
Therefore, when precipitation is complete, the concentrations will be approximately:
[Ba2+] = 0.022 M
[NO3^-] = 2x = 2 * 0.022 M = 0.044 M
[Na+] = 1.0 M
To determine if precipitation will occur, we need to compare the solubility product constant (Ksp) of Ba(NO3)2 with the concentrations of the ions in the solution.
The balanced equation for the dissociation of Ba(NO3)2 is:
Ba(NO3)2(s) -> Ba2+(aq) + 2NO3^-(aq)
The solubility product constant (Ksp) expression for Ba(NO3)2 is:
Ksp = [Ba2+][NO3^-]^2
The Ksp value for Ba(NO3)2 is 4.5 x 10^(-5).
In the solution, Ba(NO3)2 will dissociate into Ba2+ and NO3^- ions. Na2SO4 will dissociate into Na+ and SO4^2- ions.
Since Na2SO4 is a soluble salt, it will completely dissociate in water. Therefore, the concentration of Na+ ions will be 1.0 M.
When precipitation occurs, the concentrations of Ba2+ and NO3^- ions will reach their maximum solubility according to the Ksp value.
Let's assume x M is the concentration of Ba2+ and 2x M is the concentration of NO3^-.
Using the Ksp expression, we can write:
Ksp = [Ba2+][NO3^-]^2
4.5 x 10^(-5) = x * (2x)^2
4.5 x 10^(-5) = 4x^3
x^3 = 1.125 x 10^(-5)
x ≈ 0.022 M
Therefore, when precipitation is complete, the concentrations will be approximately:
[Ba2+] = 0.022 M
[NO3^-] = 2x = 2 * 0.022 M = 0.044 M
[Na+] = 1.0 M
An amount of solid NH4 HS is placed in a flask already containing ammonia gas at a certain temperature and 0.50 atm pressure. Ammonium hydrogen sulphide decomposes to yield NH3 and H2S gases in the flask. When the decomposition reaction reaches equilibrium, the total pressure in the flask rises to 0.84 atm? The equilibrium constant for NH4HS decomposition at this temperature is- a)0.11
- b)0.17
- c)0.18
- d)0.30
Correct answer is option 'A'. Can you explain this answer?
An amount of solid NH4 HS is placed in a flask already containing ammonia gas at a certain temperature and 0.50 atm pressure. Ammonium hydrogen sulphide decomposes to yield NH3 and H2S gases in the flask. When the decomposition reaction reaches equilibrium, the total pressure in the flask rises to 0.84 atm? The equilibrium constant for NH4HS decomposition at this temperature is
a)
0.11
b)
0.17
c)
0.18
d)
0.30
|
|
Anand Kumar answered |
The initial rate of hydrolysis of methyl acetate (1M) by a weak acid (HA, 1M) is 1/100th of that of a strong acid (HX, 1M), at 25°C. The Ka of HA is- a)1 × 10–4
- b)1 × 10–5
- c)1 × 10–6
- d)1 × 10–3
Correct answer is option 'A'. Can you explain this answer?
The initial rate of hydrolysis of methyl acetate (1M) by a weak acid (HA, 1M) is 1/100th of that of a strong acid (HX, 1M), at 25°C. The Ka of HA is
a)
1 × 10–4
b)
1 × 10–5
c)
1 × 10–6
d)
1 × 10–3
|
Lakshmi Mishra answered |
As ester hydrolysis is first order with respect [H+].
The pKa of a weak acid (HA) is 4.5. The pOH of an aqueous buffer solution of HA in which 50% of the acid is ionized is- a)7.0
- b)4.5
- c)2.5
- d)9.5
Correct answer is option 'D'. Can you explain this answer?
The pKa of a weak acid (HA) is 4.5. The pOH of an aqueous buffer solution of HA in which 50% of the acid is ionized is
a)
7.0
b)
4.5
c)
2.5
d)
9.5
|
|
Athira Shah answered |
Given pKa = 4.5 and acid is 50% ionised.
[HA] = [A–] (when acid is 50% ionised)
[HA] = [A–] (when acid is 50% ionised)
∴ pH = pKa + log 1
∴ pH = pKa = 4.5
pOH = 14 – pH = 14 – 4.5 = 9.5
∴ pH = pKa = 4.5
pOH = 14 – pH = 14 – 4.5 = 9.5
Which of the following solutions will have pH close to 1.0?- a)100 ml of (M/10) HCl + 100 ml of (M/10) NaOH
- b)55 ml of (M/10) HCl + 45 ml of (M/10) NaOH
- c)10 ml of (M/10) HCl + 90 ml of (M/10) NaOH
- d)75 ml of (M/5) HCl + 25 ml of (M/5) NaOH
Correct answer is option 'D'. Can you explain this answer?
Which of the following solutions will have pH close to 1.0?
a)
100 ml of (M/10) HCl + 100 ml of (M/10) NaOH
b)
55 ml of (M/10) HCl + 45 ml of (M/10) NaOH
c)
10 ml of (M/10) HCl + 90 ml of (M/10) NaOH
d)
75 ml of (M/5) HCl + 25 ml of (M/5) NaOH
|
Devika Banerjee answered |
The correct option (d) 75mL of (M/5)HCI + 25mL of (M/5) NaOH
Explanation:
75mL of M/5HCl + 25mL of M/5 NaOH
Meq of HCl = 15 (∴ Mol. wt. and Eq. wt. of HCl or NaOH are same)
Meq of NaOH = 5
Hence, in solution 10meq of HCl are remained in 100mL solution.
So, concentration of HCl in mixture = 10/100M = M/10
Explanation:
75mL of M/5HCl + 25mL of M/5 NaOH
Meq of HCl = 15 (∴ Mol. wt. and Eq. wt. of HCl or NaOH are same)
Meq of NaOH = 5
Hence, in solution 10meq of HCl are remained in 100mL solution.
So, concentration of HCl in mixture = 10/100M = M/10
Three reactions involving H2PO4– are given below :(i) H3PO4 + H2O → H3O+ + H2PO4–
(ii) H2PO4– + H2O → HPO42– + H3O+
(iii) H2PO4– + OH– → H3PO4 + O2–In which of the above does H2PO4– act as an acid ? - a)(ii) only
- b)(i) and (ii)
- c)(iii) only
- d)(i) only
Correct answer is option 'A'. Can you explain this answer?
Three reactions involving H2PO4– are given below :
(i) H3PO4 + H2O → H3O+ + H2PO4–
(ii) H2PO4– + H2O → HPO42– + H3O+
(iii) H2PO4– + OH– → H3PO4 + O2–
(ii) H2PO4– + H2O → HPO42– + H3O+
(iii) H2PO4– + OH– → H3PO4 + O2–
In which of the above does H2PO4– act as an acid ?
a)
(ii) only
b)
(i) and (ii)
c)
(iii) only
d)
(i) only
|
|
Nilotpal Chavan answered |
1) The reaction of H2PO4 with a strong base such as sodium hydroxide (NaOH) will result in the formation of water and sodium dihydrogen phosphate (NaH2PO4):
H2PO4- + NaOH → H2O + NaH2PO4
2) H2PO4 can also react with a strong acid, such as hydrochloric acid (HCl), to form water and phosphoric acid (H3PO4):
H2PO4- + HCl → H2O + H3PO4
3) H2PO4 can undergo oxidation-reduction reactions, such as in the reaction with hydrogen peroxide (H2O2) and a catalyst, which leads to the production of water and phosphate ions (PO4^3-):
H2PO4- + H2O2 → H2O + PO4^3-
H2PO4- + NaOH → H2O + NaH2PO4
2) H2PO4 can also react with a strong acid, such as hydrochloric acid (HCl), to form water and phosphoric acid (H3PO4):
H2PO4- + HCl → H2O + H3PO4
3) H2PO4 can undergo oxidation-reduction reactions, such as in the reaction with hydrogen peroxide (H2O2) and a catalyst, which leads to the production of water and phosphate ions (PO4^3-):
H2PO4- + H2O2 → H2O + PO4^3-
The set with correct order of acidity is- a)HClO < HClO2 < HClO3 < HClO4
- b)HClO4 < HClO3 < HClO2 < HClO
- c)HClO < HClO4 < HClO3 < HClO2
- d)HClO4 < HClO2 < HClO3 < HClO
Correct answer is option 'A'. Can you explain this answer?
The set with correct order of acidity is
a)
HClO < HClO2 < HClO3 < HClO4
b)
HClO4 < HClO3 < HClO2 < HClO
c)
HClO < HClO4 < HClO3 < HClO2
d)
HClO4 < HClO2 < HClO3 < HClO
|
Anand Khanna answered |
Correct option: (A) HClO < HClO2 < HClO3 < HClO4
Explanation :
The acidic character of the oxoacids increases with increase in oxidation number of the halogen atom i.e., HClO < HClO2 < HClO3 < HClO4 This can be explained on the basis of relative stability of the anions left after removal of a proton. Since the stability of the anion decreases in the order : CIO4- > ClO3- > CIO2- > ClO- acid strength also decreases in the same order.
Explanation :
The acidic character of the oxoacids increases with increase in oxidation number of the halogen atom i.e., HClO < HClO2 < HClO3 < HClO4 This can be explained on the basis of relative stability of the anions left after removal of a proton. Since the stability of the anion decreases in the order : CIO4- > ClO3- > CIO2- > ClO- acid strength also decreases in the same order.
2.5 ml of (2/5) M weak monoacidic base (Kb = 1 × 10–12 at 25°) is titrated (2/15) M HCl in water at 25°C. The concentration of H+ at equivalence point is (Kw = 1 × 10–14 at 25°C)- a)3.7 × 10–14 M
- b)3.2 × 10–7 M
- c)3.2 × 10–2 M
- d)2.7 × 10–2 M
Correct answer is option 'D'. Can you explain this answer?
2.5 ml of (2/5) M weak monoacidic base (Kb = 1 × 10–12 at 25°) is titrated (2/15) M HCl in water at 25°C. The concentration of H+ at equivalence point is (Kw = 1 × 10–14 at 25°C)
a)
3.7 × 10–14 M
b)
3.2 × 10–7 M
c)
3.2 × 10–2 M
d)
2.7 × 10–2 M
|
Sanchita Patel answered |
The correct option (d) 2.7 x 10^-2M
Explanation:
Weak monoacidic base eg. BOH is neutralised as follows
BOH + HCI → BCl+ H2O
At equivalence point all BOH gets converted into salt and remember! the coucentration of H^+ (or pH of solution) is due to hydrolysis of resultant salt (BCl, cationic hydrolysis here)
Explanation:
Weak monoacidic base eg. BOH is neutralised as follows
BOH + HCI → BCl+ H2O
At equivalence point all BOH gets converted into salt and remember! the coucentration of H^+ (or pH of solution) is due to hydrolysis of resultant salt (BCl, cationic hydrolysis here)
(h should be estimated whether that can be neglected or not) on calculating h = 0.27 (significant. not negligible)
[H^+] = Ch = 0.1 x 0.27 = 2.7 x 10^-2M
[H^+] = Ch = 0.1 x 0.27 = 2.7 x 10^-2M
When NaNO3 is heated in a closed vessel, oxygen is liberated and NaNO2 is left behind. At equilibrium.- a)addition of NaNO2 favours reverse reaction
- b)addition of NaNO3 favours forward reaction
- c)increasing temperature favours forward reaction
- d)increasing pressure favours reverse reaction
Correct answer is option 'C,D'. Can you explain this answer?
When NaNO3 is heated in a closed vessel, oxygen is liberated and NaNO2 is left behind. At equilibrium.
a)
addition of NaNO2 favours reverse reaction
b)
addition of NaNO3 favours forward reaction
c)
increasing temperature favours forward reaction
d)
increasing pressure favours reverse reaction
|
Sagar Mukherjee answered |
According to Le-chatelier principle an increase in pressure always favours the reaction, where volume or moles decrease (i.e. reverse direction). As heat is added, i.e. reaction is endothermic and is supported in forward direction with increase in temperature.
A vessel at 1000 K contains CO2 with a pressure of 0.5 atm.Some of the CO2 is converted into CO on the addition of graphite. If the total pressure at equilibrium is 0.8 atm, the value of K is :- a)1.8 atm
- b)3 atm
- c)0.3 atm
- d)0.18 atm
Correct answer is option 'A'. Can you explain this answer?
A vessel at 1000 K contains CO2 with a pressure of 0.5 atm.
Some of the CO2 is converted into CO on the addition of graphite. If the total pressure at equilibrium is 0.8 atm, the value of K is :
a)
1.8 atm
b)
3 atm
c)
0.3 atm
d)
0.18 atm
|
|
Ruchi Shah answered |
Total P at equilibrium = 0.5 – x + 2x = 0.5 + x atm
0.8 = 0.5 + x
∴ x = 0.8 – 0.5 = 0.3 atm
0.8 = 0.5 + x
∴ x = 0.8 – 0.5 = 0.3 atm
= 1.8 atm
The following acids have been arranged in the order of decreasing acid strength. Identify the correct order.ClOH (I), BrOH(II), IOH(III)- a)I > II > III
- b)II > I >III
- c)III > II > I
- d)I > III > II
Correct answer is option 'A'. Can you explain this answer?
The following acids have been arranged in the order of decreasing acid strength. Identify the correct order.
ClOH (I), BrOH(II), IOH(III)
a)
I > II > III
b)
II > I >III
c)
III > II > I
d)
I > III > II
|
Poulomi Datta answered |
Among oxyacids of halogens, if there are same number of
oxygens bonded to central atom, higher the
electronegativity of halogen, stronger the acid. Hence,
IOH<BrOH<ClOH
IOH<BrOH<ClOH
An example of a reversible reaction is :- a)Pb(NO3)2aq + 2NaI(aq) → PbI2(s) + 2NaNO3(aq)
- b)AgNO3(aq) + HCl(aq) → AgCl(s) + NaNO3(aq)
- c)2Na(s) + H2O(l) → 2NaOH(aq) + H2(g)
- d)KNO3(aq) + NaCl(aq) → KCl(aq) + NaNO3(aq)
Correct answer is option 'D'. Can you explain this answer?
An example of a reversible reaction is :
a)
Pb(NO3)2aq + 2NaI(aq) → PbI2(s) + 2NaNO3(aq)
b)
AgNO3(aq) + HCl(aq) → AgCl(s) + NaNO3(aq)
c)
2Na(s) + H2O(l) → 2NaOH(aq) + H2(g)
d)
KNO3(aq) + NaCl(aq) → KCl(aq) + NaNO3(aq)
|
Ishan Chatterjee answered |
The reaction you provided is not a reversible reaction. It is a precipitation reaction that forms lead iodide (PbI2) and sodium nitrate (NaNO3) as products. In a reversible reaction, the products can react with each other to form the original reactants. A better example of a reversible reaction is the conversion of water (H2O) into its components hydrogen gas (H2) and oxygen gas (O2):
2H2O(l) ⇌ 2H2(g) + O2(g)
2H2O(l) ⇌ 2H2(g) + O2(g)
The oxidation of SO2 by O2 to SO3 is an exothermic reaction. The yield of SO3 will be maximum if- a)temperature is increased and pressure is kept constant
- b)temperature is reduced and pressure is increased
- c)both temperature and pressure are increased
- d)both temperature and pressure are reduced
Correct answer is option 'B'. Can you explain this answer?
The oxidation of SO2 by O2 to SO3 is an exothermic reaction. The yield of SO3 will be maximum if
a)
temperature is increased and pressure is kept constant
b)
temperature is reduced and pressure is increased
c)
both temperature and pressure are increased
d)
both temperature and pressure are reduced
|
Kritika Sarkar answered |
TIPS/Formulae :
(i) According is Le-Chateliers principle, exothermic, reactions are favoured at low temperature.
(ii) According to Le-Chateliers principle, the reaction in which n < 0, are favoured at high pressure.
(i) According is Le-Chateliers principle, exothermic, reactions are favoured at low temperature.
(ii) According to Le-Chateliers principle, the reaction in which n < 0, are favoured at high pressure.
∵ It is exothermic reaction
∴Yield of SO3 is maximum at low temperature
n = 2 – 3= – 1 or n < 0
∴ Yield of SO3 is maximum at high pressure.
∴Yield of SO3 is maximum at low temperature
n = 2 – 3= – 1 or n < 0
∴ Yield of SO3 is maximum at high pressure.
When equal volumes of the following solutions are mixed, precipitation of AgCl (Ksp = 1.8×10–10) will occur only with- a)10–4 M (Ag+) and 10–4 M (Cl–)
- b)10–5 M (Ag+) and 10–5 M (Cl–)
- c)10–6 M (Ag+) and 10–6 M (Cl–)
- d)10–10 M (Ag+) and 10–10 M (Cl–)
Correct answer is option 'A'. Can you explain this answer?
When equal volumes of the following solutions are mixed, precipitation of AgCl (Ksp = 1.8×10–10) will occur only with
a)
10–4 M (Ag+) and 10–4 M (Cl–)
b)
10–5 M (Ag+) and 10–5 M (Cl–)
c)
10–6 M (Ag+) and 10–6 M (Cl–)
d)
10–10 M (Ag+) and 10–10 M (Cl–)
|
Athul Patel answered |
Answer:
a) 10^-4 M (Ag⁺) and 10^-4 M (Cl⁻) is the correct answer.
Explanation:
Given data:
When equal volumes of the some solutions are mixed, precipitation of AgCl will occur
K{sp} = 1.8x10^-10
When two solutions are mixed having equal volumes, the concentrated will be reduced.
It will become half of the initial solution.
Also, AgCl will precipate only when
[Ag+][Cl-] > Ksp
Hence we will check all the options
a)
[Ag+]= 10^-4/2 M
[Cl-] = 10^-4/2 M
[Ag+][Cl-] = 10^-4/2 M*10^-4/2 M
= 10^-8/4
= 2.5*10^-9
Which is greater than Ksp
Thus (a) is the correct answer
Thus 10⁻⁴ M (Ag⁺) and 10⁻⁴ M (Cl⁻) is the correct answer.
a) 10^-4 M (Ag⁺) and 10^-4 M (Cl⁻) is the correct answer.
Explanation:
Given data:
When equal volumes of the some solutions are mixed, precipitation of AgCl will occur
K{sp} = 1.8x10^-10
When two solutions are mixed having equal volumes, the concentrated will be reduced.
It will become half of the initial solution.
Also, AgCl will precipate only when
[Ag+][Cl-] > Ksp
Hence we will check all the options
a)
[Ag+]= 10^-4/2 M
[Cl-] = 10^-4/2 M
[Ag+][Cl-] = 10^-4/2 M*10^-4/2 M
= 10^-8/4
= 2.5*10^-9
Which is greater than Ksp
Thus (a) is the correct answer
Thus 10⁻⁴ M (Ag⁺) and 10⁻⁴ M (Cl⁻) is the correct answer.
Species acting as both Bronsted acid and base is- a)(HSO4)–1
- b)Na2CO3
- c)NH3
- d)OH–1.
Correct answer is option 'A'. Can you explain this answer?
Species acting as both Bronsted acid and base is
a)
(HSO4)–1
b)
Na2CO3
c)
NH3
d)
OH–1.
|
Anjali Chakraborty answered |
(HSO4)– can accept and donate a proton
(HSO4)– + H+ → H2SO4 (acting as base)
(HSO4)– – H+ → SO42–. (acting as acid)
(HSO4)– + H+ → H2SO4 (acting as base)
(HSO4)– – H+ → SO42–. (acting as acid)
PARAGRAPH 1When 100 mL of 1.0 M HCl was mixed with 100 mL of 1.0 M NaOH in an insulated beaker at constant pressure, a temperature increase of 5.7 °C was measured for the beaker and its contents (Expt.1).
Because the enthalpy of neutralization of a strong acid with a strong base is a constant (–57.0 kJ mol–1), this experiment could be used to measure the calorimeter constant. In a second experiment (Expt. 2), 100 mL of 2.0 M acetic acid (Ka = 2.0 × 10–5) was mixed with 100 mL of 1.0 M NaOH (under identical conditions to Expt. 1) where a temperature rise of 5.6 °C was measured. (Consider heat capacity of all solutions as 4.2 J g–1 K–1 and density of all solutions as 1.0 g mL–1)Q. The pH of the solution after Expt. 2 is- a)2.8
- b)4.7
- c)5.0
- d)7.0
Correct answer is option 'B'. Can you explain this answer?
PARAGRAPH 1
When 100 mL of 1.0 M HCl was mixed with 100 mL of 1.0 M NaOH in an insulated beaker at constant pressure, a temperature increase of 5.7 °C was measured for the beaker and its contents (Expt.1).
Because the enthalpy of neutralization of a strong acid with a strong base is a constant (–57.0 kJ mol–1), this experiment could be used to measure the calorimeter constant. In a second experiment (Expt. 2), 100 mL of 2.0 M acetic acid (Ka = 2.0 × 10–5) was mixed with 100 mL of 1.0 M NaOH (under identical conditions to Expt. 1) where a temperature rise of 5.6 °C was measured. (Consider heat capacity of all solutions as 4.2 J g–1 K–1 and density of all solutions as 1.0 g mL–1)
Because the enthalpy of neutralization of a strong acid with a strong base is a constant (–57.0 kJ mol–1), this experiment could be used to measure the calorimeter constant. In a second experiment (Expt. 2), 100 mL of 2.0 M acetic acid (Ka = 2.0 × 10–5) was mixed with 100 mL of 1.0 M NaOH (under identical conditions to Expt. 1) where a temperature rise of 5.6 °C was measured. (Consider heat capacity of all solutions as 4.2 J g–1 K–1 and density of all solutions as 1.0 g mL–1)
Q. The pH of the solution after Expt. 2 is
a)
2.8
b)
4.7
c)
5.0
d)
7.0
|
Rohan Chakraborty answered |
Answer: (B)
200 m.eq. + 100 m.eq Base →100 m.eq. salt
pH=pKa+log Salt/Acid
=2⋅10−5+log 100/100=5−log2=5−0.3010=4.7
200 m.eq. + 100 m.eq Base →100 m.eq. salt
pH=pKa+log Salt/Acid
=2⋅10−5+log 100/100=5−log2=5−0.3010=4.7
0.1 mole of CH3NH2 (Kb = 5 x10–4)is mixed with 0.08 mole of HCl and diluted to one litre. What will be the H+ concentration in the solution?- a)8 x 10-2 M
- b)8 x 10-11 M
- c)1.6 x 10-11 M
- d)8 x 10-5 M
Correct answer is option 'B'. Can you explain this answer?
0.1 mole of CH3NH2 (Kb = 5 x10–4)is mixed with 0.08 mole of HCl and diluted to one litre. What will be the H+ concentration in the solution?
a)
8 x 10-2 M
b)
8 x 10-11 M
c)
1.6 x 10-11 M
d)
8 x 10-5 M
|
Shanaya Tiwari answered |
-4) is dissolved in 1.0 L of water. Calculate the hydroxide ion concentration (OH-) in the solution.
To find the hydroxide ion concentration, we need to determine the concentration of the CH3NH2 in the solution.
The Kb value for CH3NH2 is 5 x 10^-4, which means that in water, CH3NH2 will partially ionize according to the following equation:
CH3NH2 + H2O ⇌ CH3NH3+ + OH-
Let x be the concentration of the CH3NH2 that ionizes. Therefore, the concentration of CH3NH3+ and OH- formed will also be x.
The initial concentration of CH3NH2 is 0.1 mole/L, so the concentration of CH3NH3+ and OH- formed will be x mole/L.
Using the Kb expression for CH3NH2, we can write:
Kb = [CH3NH3+][OH-]/[CH3NH2]
5 x 10^-4 = x * x /0.1
Rearranging the equation:
x^2 = 5 x 10^-4 * 0.1
x^2 = 5 x 10^-5
x = sqrt(5 x 10^-5)
x = 7.07 x 10^-3
Therefore, the concentration of OH- in the solution is 7.07 x 10^-3 M.
To find the hydroxide ion concentration, we need to determine the concentration of the CH3NH2 in the solution.
The Kb value for CH3NH2 is 5 x 10^-4, which means that in water, CH3NH2 will partially ionize according to the following equation:
CH3NH2 + H2O ⇌ CH3NH3+ + OH-
Let x be the concentration of the CH3NH2 that ionizes. Therefore, the concentration of CH3NH3+ and OH- formed will also be x.
The initial concentration of CH3NH2 is 0.1 mole/L, so the concentration of CH3NH3+ and OH- formed will be x mole/L.
Using the Kb expression for CH3NH2, we can write:
Kb = [CH3NH3+][OH-]/[CH3NH2]
5 x 10^-4 = x * x /0.1
Rearranging the equation:
x^2 = 5 x 10^-4 * 0.1
x^2 = 5 x 10^-5
x = sqrt(5 x 10^-5)
x = 7.07 x 10^-3
Therefore, the concentration of OH- in the solution is 7.07 x 10^-3 M.
Solubility product constant (Ksp) of salts of types MX, MX2 and M3X at temperature T are 4.0 × 10–8, 3.2 × 10–14 & 2.7 × 10–15, respectively. Solubilities (mol dm–3) of the salts at temperature 'T' are in the order –- a)MX > MX2 > M3X
- b)M3X > MX2 > MX
- c)MX2 > M3X > MX
- d)MX > M3X > MX2
Correct answer is option 'D'. Can you explain this answer?
Solubility product constant (Ksp) of salts of types MX, MX2 and M3X at temperature T are 4.0 × 10–8, 3.2 × 10–14 & 2.7 × 10–15, respectively. Solubilities (mol dm–3) of the salts at temperature 'T' are in the order –
a)
MX > MX2 > M3X
b)
M3X > MX2 > MX
c)
MX2 > M3X > MX
d)
MX > M3X > MX2
|
Gayatri Patel answered |
The correct option (d) MX > M3X > MX2
Explanation:
Explanation:
The compound that is not a Lewis acid is :- a)BF3
- b)AlCl3
- c)BeCl2
- d)SnCl4
Correct answer is option 'D'. Can you explain this answer?
The compound that is not a Lewis acid is :
a)
BF3
b)
AlCl3
c)
BeCl2
d)
SnCl4
|
Nandini Nair answered |
BF3 is known to be a strong Lewis acid that reacts instantly with the weak Lewis bases and it has an incomplete octet for accepting a lone pair from base molecules.
BeCl2 is also a Lewis acid due to its incomplete octet and ability to accept electron pairs to complete this octet, for which it readily reacts with the strong Lewis bases.
Similarly, SnCl4 is a weak Lewis acid, as it can accept an electron pair from a Lewis base.
BeCl2 is also a Lewis acid due to its incomplete octet and ability to accept electron pairs to complete this octet, for which it readily reacts with the strong Lewis bases.
Similarly, SnCl4 is a weak Lewis acid, as it can accept an electron pair from a Lewis base.
The equilibrium constant (Kc) for the reaction N2(g) + O2(g) → 2NO(g) at temperature T is 4 × 10–4. Thevalue of Kc for the reaction 
at the same temperature is:- a)0.02
- b)2.5 × 102
- c)4 × 10–4
- d)50.0
Correct answer is option 'D'. Can you explain this answer?
The equilibrium constant (Kc) for the reaction N2(g) + O2(g) → 2NO(g) at temperature T is 4 × 10–4. Thevalue of Kc for the reaction
at the same temperature is:a)
0.02
b)
2.5 × 102
c)
4 × 10–4
d)
50.0
|
Krithika Deshpande answered |
For the reaction
Hence for the reaction
Read the following statement and explanation and answer as per the options given below :Statement -1 HNO3 is a stronger acid than HNO2 Statement -2 In HNO3 there are two nitrogen-to-oxygen bonds whereas in HNO2 there is only one.- a)If both Statement -1 and Statement -2 are correct, and Statement -2 is the correct explanation of the Statement -2.
- b)If both Statement -1 and Statement -2 are correct, but Statement -2 is not the cor rect explanation of the Statement -1.
- c)If Statement -1 is correct but Statement -2 is incorrect.
- d)If Statement -1 is incorrect but Statement -2 is correct.
Correct answer is option 'C'. Can you explain this answer?
Read the following statement and explanation and answer as per the options given below :
Statement -1 HNO3 is a stronger acid than HNO2
Statement -2 In HNO3 there are two nitrogen-to-oxygen bonds whereas in HNO2 there is only one.
a)
If both Statement -1 and Statement -2 are correct, and Statement -2 is the correct explanation of the Statement -2.
b)
If both Statement -1 and Statement -2 are correct, but Statement -2 is not the cor rect explanation of the Statement -1.
c)
If Statement -1 is correct but Statement -2 is incorrect.
d)
If Statement -1 is incorrect but Statement -2 is correct.
|
|
Vivek answered |
HNO3 on releasing H+ , gives NO3- ion which is greater resonance stabilized than NO2- ion which is formed when HNO2 releases H+.
Structure of HNO3 is ---->
+
H-O-N(O-)=O
Structure of HNO2 is ---->
+
H-O-N=O
It is evident from the above structures that in HNO3, N has 3 sigma bonds and 1 pi bond with O....whereas in HNO2, N has 2 sigma bonds and 1 pi bond with O.
Hence, statement 1 is correct and statement 2 is incorrect.
Plzz upvote if u find this useful.... :)
Pure ammonia is placed in a vessel at a temperature where its dissociation constant (a) is appreciable. At equilibrium :- a)Kp does not change significantly with pressure.
- b)does not change with pressure.
- c)concentration of NH3 does not change with pressure.
- d)concentration of hydrogen is less than that of nitrogen.
Correct answer is option 'A'. Can you explain this answer?
Pure ammonia is placed in a vessel at a temperature where its dissociation constant (a) is appreciable. At equilibrium :
a)
Kp does not change significantly with pressure.
b)
does not change with pressure.
c)
concentration of NH3 does not change with pressure.
d)
concentration of hydrogen is less than that of nitrogen.
|
Raksha Nambiar answered |
Correct answer is option: A
option KP doesn't change signify with press her because KP is constant and does not change with pressure because its KP value is changes with change in temperature.
Chapter doubts & questions for Equilibrium - 35 Years Chapter wise Previous Year Solved Papers for JEE 2025 is part of JEE exam preparation. The chapters have been prepared according to the JEE exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for JEE 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.
Chapter doubts & questions of Equilibrium - 35 Years Chapter wise Previous Year Solved Papers for JEE in English & Hindi are available as part of JEE exam.
Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.
35 Years Chapter wise Previous Year Solved Papers for JEE
347 docs|185 tests
|
Signup to see your scores go up within 7 days!
Study with 1000+ FREE Docs, Videos & Tests
10M+ students study on EduRev
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up
within 7 days!
within 7 days!
Takes less than 10 seconds to signup