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Q1. Discuss different types of salt hydrolysis.

Solution

Salts of following types exhibit the phenomena of salt hydrolysis. In salts of weak acids and strong bases anions get hydrolyzed to give more of OH^- and the solution is alkaline. It is called anionic hydrolysis. In salts of strong acids and weak bases Cations get hydrolyzed to give more of H⁺ and the solution is acidic. It is called cationic hydrolysis. In salts of weak acids and weak bases both anions and cations get hydrolyzed the acidity or basicity of the solution depends upon K_a and k_b. When k_a = k_b the solution is neutral. It is called anionic hydrolysis.

Q2. Define ionization Constant? Show how can we derive expression for K_a for a weak acid like CH₃COOH?

Solution

Ionization constant is the product of Equilibrium constant K and concentration of water term in an ionic equilibrium of a weak electrolyte. We can derive an expression from the following ionic equilibrium of CH₃COOH in water.                                                          CH₃COOH +  H₂O     CH₃COO^- +  H^{+ Initial molar concentration                                  C                                                0                      0 Molar conc. at equilibrium point                          C(1 – α )                                   c α                    c α} Applying the law of chemical equilibrium   K = [CH₃COO^-] [H⁺] [CH₃COOH] [H₂O] K _[H2O]    = cα²  (1–α )                                                                                                           cα²                                    Or,                Ka =                                                                               (1 – α )   Where K is equilibrium constant and K_a is known as ionization or dissociation constant of the acid, similarly a K_b expression can be derived for a weak base.

Q3. How will you differentiate between physical and chemical equilibria?

Solution

Physical equilibria do not involve changes to the chemical properties of the substances involved. For instance, equilibrium between water vapor and liquid water in a partly filled sealed container. It is a physical equilibrium since the water molecules have only changed from liquid to vapor. Chemical equilibria on the other hand involve changes in the chemical composition of substances. Bond breaking and bond formation is involved. An example would be the dissociation of acetic acid. There is an interchange of particles between ions and molecules whose rate of exchange becomes equal when equilibrium is established. CH₃COOH + H₂O  H₃O ⁺¹ + CH₃COO^-1

Q4. Calculate the dissociation constant of a monobasic acid having the degree of dissociation 2% in its 0.01M solution.

Solution

Use Ionization constant expression   Substitute the respective values; we get K_a =  (0.01)(0.02 x 0.02)/ (1 - 0.02)     = 4.08 x 10^-6  M/L

Q5. Calculate the value of K_a for the following reaction having the equilibrium concentration of H⁺ 7.1 x 10^-5 mol/L in a 0.05 molar solution.                                                    H₂S  H⁺ + HS^-

Solution

[H+]  = C?      Or, ? = 7.1 x 10^-5/ 0.05 = 1.42 x 10^-3   Ka = C ?²        = .05 x 1.42 x 10^-3 x 1.42x 10^-3 = 1.0 x 10^-7

Q6. The K_sp for Ag₂CrO₄ is 9x10^-12 M³. What is the molar solubility of Ag₂CrO₄ in pure water?

Solution

Let x be the molar solubility of Ag₂CrO₄, then       Ag₂CrO₄ = 2 Ag⁺ + CrO₄^2-                             2 x        x       (2 x)² (x) = K_sp         x = {(9x10^-12)/4}^(1/3)           = 1.3x10^-4 M   Hence, the molar solubility of Ag₂CrO₄ is 1.3x10^-4 M.

Q7. What is the concentration of barium and sulfate ions in a saturated barium sulfate solution at 25°C?

Solution

K_sp = [Ba²⁺][SO₄^2-] = 1.1 x 10^-10   In this reaction,                      BaSO4(s) <---> Ba²⁺(aq) + SO₄^2-(aq)   [Ba²⁺] = [SO₄^2-] So we can substitute one concentration for the other!  [Ba²⁺][Ba²⁺]=[Ba²⁺] ² = 1.1 x 10^-10  [Ba²⁺] = 1.05 x 10^-5

Q8. Explain that chemical equilibria can be achieved from either side.

Solution

Q9. A crystal of common salt of given mass is kept in aqueous solution. After 12 hours, its mass remains the same. Is the crystal in equilibrium with the solution?

Solution

Yes, as the solution becomes saturated the mass of the crystal remains same because of establishment of a physical equilibrium between the crystal of the common salt and its saturated solution.

Q10. Give the conjugate acid for each of the following: OH^-, CH₃COO^-, Cl^-, CO₃^-2, CH₃NH₂

Solution

The conjugate acids are respectively as below: H₂O, CH₃COOH, HCl , HCO₃^-1 CH₃NH₃⁺

Q11. The pOH of a solution of NaOH is 11.00. What is the [H⁺] for this solution?

• 1) 1.0 x 10^-3
• 2) 4.0 x 10^-12
• 3) 5.0 x 10^-12
• 4) 2.5 x 10^-3

Solution

  pOH = 14 - pH 11 = 14 - pH pH = 14 - 11 = 3 pH = - log [H⁺] -3 = log [H⁺] [H⁺] =Anti [- 3] [H⁺] = 0.001  

Q12. Chemical equilibrium is dynamic in nature. Explain.

Solution

A chemical equilibrium is dynamic in nature which in other words means that reactions continue to occur in both forward as well as backward direction with the same speed. As a result, the amount of product formed immediately reacts backward to give reactants and so there is no change in concentration of reactant or product with the passage of time.

Q13. The calculated value of K_b of NH₃ is 1.8 x 10^-5. Calculate the pH of 1.5 M solution of ammonia.

Solution

Q14. For the equilibrium system described by 2 SO_{2 (g)} + O_{2 (g)} ↔ 2 SO_{3 (g)} at a particular temperature the equilibrium concentrations of SO₂, O₂ and SO₃ were 0.75 M, 0.30 M, and 0.15 M, respectively. Calculate the equilibrium constant, K_eq, for the reaction.

Solution

Equilibrium constant expression for the balanced equation:                                                     substitute the known values, and solve for the unknown K_eq                                    

Q15. The pH of lemon juice is 2.32 at 298 k. What will be its [H₃O⁺]?

Solution

pH = -log [H₃O⁺] 2.32 = -log [H₃O⁺] antilog (- 2.32) = [H₃O⁺] [H₃O⁺] = 4.786 x 10^-3 [OH^-] = 10^-14 / 4.786 x 10^-3 = 2.09 x 10^-12

Q16. The following reaction is a reversible reaction. The value of K_c for this reaction is 31.4 at 588K. CO_(g) + H₂O _(g) CO_{2 (g)} + H_{2 (g)} If a 10.00L vessel has 2.50 mol CO₂ and H₂O, and 5.00 mol CO₂ and H₂ gas at 588K, which way will the reaction proceed ?

Solution

[CO] = .25 M [H₂O] = .25 M [CO₂] = .50 M [H₂] =.50 M   Hence, Qc = [CO₂] [H₂]/ [CO] [H₂O] = 4   Since Q_c < K_c, it will proceed in the forward direction.

Q17. An aqueous solution of Sr(NO₃)₂ is added slowly to 1.0 liter of a well-stirred solution containing 0.020 mole F^- and 0.10 mole SO₄^2- at 25 °C( ΔV = O) Which salt precipitates first? What is the concentration of strontium ion (Sr²⁺), in the solution when the first precipitate begins to form? The solubility product constant, K_sp for strontium sulfate, SrSO₄, is 7.6 x 10^-7. The solubility product constant for strontium fluoride, SrF₂, is 7.9 x 10^-10

Solution

Calculate [Sr²⁺] required for precipitation in each salt   Case-1 Let the [Sr²⁺] be x   K_sp = [Sr²⁺][F^-]² = 7.9 x 10 ^-10   = (x) (0.020 mole / 1.0 L)2 = 7.9 x 10^-10   x = 2.0 x 10^-6 M   Case-2 Let the [Sr²⁺] be y   K_sp = [Sr²⁺][SO₄^2-¯] = 7.6 x 10^-7   = (y) (0.10 mole/1.0 liter) = 7.6 x 10^-7   y = 7.6 x 10^-6 M   Since 2.0 x 10^-6 M < 7.6 x 10^-6 M,   SrF₂ must precipitate first.   When SrF₂ precipitates, [Sr²⁺] = 2.0 x 10^-6 M

Q18. The K_sp for AgCl is 1.8 x 10^-10. If Ag⁺ and Cl^- are both in solution and in equilibrium with AgCl. What is [Ag⁺] if [Cl^-] = .020 M?

Solution

[Ag⁺] = K_sp /[Cl^-]          = 1.8 x 10^-10/.020  [Ag⁺] = 9.0 x 10^-9M

Q19. A solution is formed by combining 0.025 mol CaCl₂ and 0.015 mol Pb(NO₃)₂ producing one litre of solution. Will PbCl₂ precipitate in this solution?

Solution

0.015 mol of Pb(NO₃)₂ gives 0.015 mol of Pb²⁺ ions or ,0.015 M) 0.025 mol of CaCl₂ yields 0.05 mol Cl^- ions or. 0.05 M   Hence , [Pb²⁺][Cl]² = (0.015)(0.050)² = 3.75 x 10^-5   K_sp of PbCl₂ = 1.6 x 10^-5   Since the ion product is larger than the K_sp, the solution precipitates.

Q20. CO₂ gas is more soluble in aq. NaOH solution than in water, explain.

Solution

CO₂ dissolves in water to form carbonic acid (H₂CO₃). As their reaction is reversible carbonic acid (H₂CO₃) dissociates to give CO₂ and H₂O in the backward reaction.   However, with aqueous NaOH solution , H₂CO₃ reacts as below:                          H₂CO_3(aq) + NaOH_(aq)  → Na₂CO_3(aq)  + H₂O As a result, more of CO₂ dissolves in water to form carbonic acid. This means that the gas is more soluble in aqueous NaOH than in water.

Q21. A solution is made by dissolving 0.63 gm of nitric acid in 200 ml of water, assuming that the acid is completely dissociated, calculate its pH value.

Solution

Molarity of the solution = w/m/V in litre = 0.63/63/.2 = .05 M Hence [H₃O⁺] = 5.0 X 10^-2 (assuming that the acid is completely dissociated) pH = - log[H₃O⁺]      = 1.30

Q22. The following reaction is at equilibrium: 4NH_{3 (g)} + 3O_{2 (g)}  6H₂O _(g) + 2N_{2 (g)} How will the equilibrium shift if: a) The volume is increased, b) Helium gas is added

Solution

a) Since the volume is being increased, the pressure is dropping. In response, the equilibrium will shift right to attempt to make more moles of gas and increase the pressure.   b) The equilibrium will not change. To change the gaseous equilibrium, the partial pressures of at least one gas need to be changed. Since He is un-reactive with all of the gases present, the total pressure will change but the partial pressures will not be affected.

Q23. What is the common ion effect?

Solution

The solubility of slightly soluble substances can be decreased by the presence of a common ion. This effect is known as common ion effect. Addition of a common ion to a slightly soluble salt solution will add up to the concentration of the common ion. According to Le Chatelier's Principle that will place a stress upon the slightly soluble salt equilibria. Thus, the equilibrium will shift so as to undo the stress of added common ion.

Q24. The ionization constant of propionic acid is 1.32 x 10^-5. Calculate the degree of ionization of acid in its 0.05 M solution.

Solution

Use Oswald’s Dilution Law expression                                 α = √ k_a/C   Substitute the respective values; we get                                             α = √ 1.32x 10^-5/0.05                                                 = 16.24 X10^-2                                                 = 0.16                                                   =16%

Q25. Equal volumes of 0.1 M MgSO₄ and 0.1 M Ba Cl₂ are mixed together. Predict the formation of BaSO₄ precipitate. Given K_sp of BaSO₄ = 1.5 x 10^-9

Solution

BaSO₄ Ba²⁺  + SO₄^-   K_sp = [Ba²⁺][ SO₄^-2] = 1.5 x 10^-9   Concentration of[Ba²⁺]and [ SO₄^-2] in the final solution [Ba²⁺] = 0.1 M/2 = 0.05 M [ SO₄^-2] = 0.1 M/2 = 0.05 M Now Ionic product = 0.05 x 0.05 = 2.5 x 10 ^-3  which is greater than K_sp = [Ba²⁺][ SO₄^-2] = 1.5 x 10^-9 Hence the precipitation of BaSO₄ will occur.

Q26. For the equilibrium system described by: PCl_{5 (g)}  PCl_{3 (g)} + Cl_{2 (g)} K_eq equals 35 at 487°C. If the concentrations of the PCl5 and PCl3 are 0.015 M and 0.78 M, respectively, what is the concentration of the Cl₂?

Solution

Let x = the unknown, [Cl₂].  Substitute in known values and solve for c:                                      

Q27. Calculate the concentration of hydroxyl ion in 0.2 M solution of NH₃OH having K_b = 1.8 x 10^-4.

Solution

Use Oswald’s Dilution Law formula;                                              α = √ k_b/C    Substitute the respective values; we get                                             α =√ k_b/C = √1.8 x 10^-4 /0.2 = .03 = 3%   [OH^-]   = 0.2 x .03 = 6.0 x 10 ^-3 M