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Q1. What are the factors which affect chemical equilibria?

Solution

Equilibrium is affected by the following three factors : 1. Concentration: By adding or removing products or reactants to the reaction vessel. 2. Pressure: By changing partial pressure of gaseous reactants and products. 3. Temperature: By changing the temperature.

Q2. 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

Q3. 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%

Q4. 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

Q5. 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

Q6. Calculate the pH of a solution that has [HO^-] concentration of 0.0026 mol/L.

Solution

[HO^-] = 0.0026 = 2.6 x 10^-3 M  [H+] = K_w / [HO^-] = 1 x 10^-14/ 2.6 x 10^-3 M                           = 3.85 x 10^-12 M pH = - log [H₃O⁺] = 12.58

Q7. 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⁺

Q8. What is Le Chatelier's principle?

Solution

Le Chatelier's principle states that when a system in chemical equilibrium is disturbed by a change of temperature, pressure, or a concentration, the system shifts in equilibrium composition in a way that tends to counteract this change of variable.

Q9. The K_sp for AgCl is 1.8 x 10¹⁰. What is the molar solubility of AgCl in pure water?

Solution

Let x be the molar solubility, then AgCl  Ag⁺ + Cl^-              x        x   x = (1.8 x 10¹⁰)^1/2    = 1.3 x 10^-5 M

Q10. 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.

Q11. Give three examples of physical equilibria.

Solution

The three types of physical equilibria are:   1)Solid-Liquid Equilibrium e.g. Rate of melting of ice = Rate of freezing of water   2) Liquid- Gas Equilibrium e.g. Rate of condensation of water vapour = Rate of vaporization of water liquid   3) Solid- Gas Equilibrium   e.g. NH₄Cl_(s)  NH₄Cl_(g)

Q12. 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.

Q13. What are general characteristics of physical equilibria?

Solution

The general characteristics of physical equilibria are as below: At the equilibrium point, the measurable properties of the system become constant. The equilibrium is of dynamic nature; means that both the forward and backward processes are taking place even at the equilibrium point but with the same speed. The equilibrium involving gases can be attained only in closed containers. At equilibrium, the concentrations of different substances involved in the process become constant at constant temperature.The equilibrium involving a solid dissolved in a liquid at a particular temperature is represented by the saturated solution.

Q14. Explain the effect of change in concentration on systems at equilibrium with the help of Le-Chatelier's principle.

Solution

The effect of change in concentration on systems at equilibrium can be explained as follows: 1) When the concentration of reactant(s) is increased, the system tries to reduce their concentration by favoring the forward reaction. 2) When the concentration of product(s) is increased, the system tries to reduce their concentration by favoring the backward reaction. 3) When the concentration of reactant(s) is decreased, the system tries to increase their concentration by favoring the backward reaction. 4) When the concentration of product(s) is decreased, the system tries to increase their concentration by favoring the forward reaction.  

Q15. 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.

Q16. Iron (III) nitrate has a solubility of 0.15 M. Find concentration of the ions in solution.

Solution

                                                Fe(NO₃)₃ → Fe3⁺_(aq) + 3 NO₃^-_(aq)   The concentration of the ions can be determined from the balancing coefficients from the equation:                                                   [Fe³⁺] = 1× [Fe(NO₃)₃] = 1 × 0.15 = 0.15 M                           [NO₃^-] = 3× [Fe(NO₃)₃] = 3 × 0.15 = 0.45 M

Q17. State and explain the law of mass action.

Solution

The law of mass action is defined as at constant temperature , the rate of a forward reaction is directly proportional to the product of the molar concentration of the reactants each raised to a power equal to its stoichiometric coefficient that appear in the balanced chemical equation.   For example , consider the following general reaction;                          aA + bB → Products Hence, according to the law, rate of reaction = k [A]^a [B]^b, where 'a' and 'b' represents the stoichiometric coefficient of reactants A and B respectively and k is the rate constant.

Q18. How is solubility different from solubility Product?

Solution

The solubility of a solute is the maximum quantity of solute that can dissolve in a certain quantity of solvent or quantity of solution at a specified temperature. Unlike the solubility of a substance, the solubility product is independent of what else is dissolved in solution. The solubility of a substance can be calculated from its solubility product.

Q19. 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.

Q20. 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

Q21. The K_sp for Cr(OH) ₃ is 1.2 x 10^-15. What is the molar solubility of Cr(OH)₃ in pure water?

Solution

Let x be the molar solubility of Cr(OH)₃, then you have        Cr(OH)₃  Cr³⁺ + 3 OH^-                             x      3 x Thus,        x (3 x)³ = 1.2 x 10^-15

Q22. What do you mean by solubility product?

Solution

The Solubility Product, K_s, is the equilibrium constant for the equilibrium that exists between a slightly soluble salt and its ions in a saturated solution. For example, AgCl_(s) = Ag⁺_(aq) + Cl^-_(aq) K_sp = [Ag⁺_(aq)][Cl^-_(aq)] K_sp = 1.74 x 10^-10 mol² lit^-2

Q23. Define degree of Ionization (α).

Solution

The strength or ionization ability of an electrolyte is expressed in terms of degree of Ionization. It is denoted by α. It is defined as “the ratio of number of molecules which dissociate into ions to the total number of molecules of the electrolyte”.

Q24. Calculate the pH of a solution that has a [OH^1-] = 1 X 10^-5 M

Solution

First determine pOH = -log[OH^- ] = -log [1 X 10^-5 ] = 5 Now we know, Ionic product of water = pH + pOH = 14                              pH + 5 = 14               Or,    pH = 14 -5 = 9

Q25. Calculate the solubility (in moles per liter) of Cu(OH)₂ at 25 °C. The solubility of Cu(OH)₂ is 1.72 x 10^-6 gram per 100ml of solution at 25 °C.

Solution

Moles of Cu(OH)₂: 1.72 x 10^-6 g / 97.57 g/mol = 1.763 x 10^-8 mol   Solubility of Cu(OH)₂  in Molarity:  1.763 x 10^-8 / 0.1 L = 1.76 x 10^-7 moles per liter

Q26. What is a buffer solution?

Solution

A buffer solution is one which resists changes in pH when small quantities of an acid or an alkali are added to it.

Q27. The pH of a soil sample is 8.3. What is the value of [HO^-] of the sample?

Solution

pH = -log[H₃O⁺] 1/ [H₃O⁺] = antilog pH antilog 8.3 = 6.76x 10⁸ or, [H₃O⁺] = 1/6.76x 10⁸ = 1.47 x 10^-9 M [HO^-] = K_w / [H⁺] = 10^-14/1.47 x 10^-9 M = 6.76 x 10^-6 M                                                            

Q28. 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

Q29. How do buffer solutions work?

Solution

A buffer solution has to contain things which will remove any hydrogen ions or hydroxide ions that you might add to it - otherwise the pH will change. Acidic and alkaline buffer solutions achieve this in different ways.

Q30. What are limitations of Arrhenius theory?

Solution

It fails to explain the acidic nature of substances like CO₂, SO₂,SO₃ etc which do not have hydrogen and similarly it also fails to explain the basic nature of substances like NH₃, CaO, MgO etc which do not have -OH group.   It fails to explain the acidic and basic behaviour of substances in solvents other than water such as ammonia,alcohol.etc   The theory fails to explain why strong electrolytes like nitric acid (HNO₃) are completely ionized in aqueous solution whereas weak electrolytes like acetic acid (CH₃COOH) are ionized to a small extent only.

Q31. What is the nature of chemical equilibrium?

Solution

Chemical equilibrium is dynamic in nature, which means that both the forward and the backward reactions continue to take place even after establishment of equilibrium but with the same speed.

Q32. What do you mean by strong and weak electrolytes?

Solution

Strong and weak electrolytes are distinguished on the basis of their ability to get ionized in water or any other polar solvents. Strong electrolytes are almost completely ionized while weak electrolytes are those substances which get ionized only to a small extent in their solutions.   Examples of strong electrolytes: HCl, NaOH, NaCl etc. Examples of weak electrolytes: CH₃COOH, H₂O, NH₃OH etc.

Q33. What is Ostwald’s Dilution Law?

Solution

The Ostwald’s Dilution Law is defined for a weak electrolyte as “ the degree of ionization is proportional to the square root of the molar concentration or directly proportional to square root of the volume of the solution which contains one mole of the electrolyte.” Mathematically, we can write Ostwald’s Dilution Law as below:e                                              α = √ k_a/C   = √ k_aV                                        Or, α = √ k_b/C   = √ k_bV

Q34. The solubility of a gas in a liquid is given by Henry's law. What kind of equilibria is shown by Henry's law?

Solution

In this is it gives idea about physical equilibrium between a gas and its solution in a liquid.

Q35. Define degree of hydrolysis (h).

Solution

The fraction of the total salt that is hydrolyzed at equilibrium is called degree or extent of hydrolysis. It is denoted by h.

Q36. What do you mean by equilibrium state?

Solution

Equilibrium state refers to the static nature of a reversible system. At equilibrium point the rate of change in concentration of product becomes equal to that of reactants after initiation and continuation of the process for some time.

Q37. Define the term salt hydrolysis?

Solution

Salt hydrolysis is a process in which a salt reacts with water to form acid and base. Salts of strong acids and bases when dissolved in water get completely ionized into cations and anions. Whereas salts of weak acids and bases as well as salts of weak acid base with strong acids and bases when dissolved in water get hydrolyzed and acid-base equilibrium is established. The nature of the solution after hydrolysis depends upon the relative strength of the acid and base formed.

Q38. Calculate the value of the solubility-product constant, K_sp, for Cu(OH)₂ at 298 K. The solubility of Cu(OH) ₂ is 1.72 x 10^-6 gram per 100ml of solution at 298 K.

Solution

[ Cu²⁺] = 1.76 x 10^-7 M [OH^-] = 2 x (1.76 x 10^-7 M ) = 3.52 x 10^-7 K_sp = [Cu²⁺] [[OH^-] ² = (1.76 x 10^-7) (3.52 x 10^-7)² = 2.18 x 10^-20

Q39. 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

Q40. How solubility and solubility products are are related?

Solution

For a general reaction                        AxBy x A ^y+ + y B^x-                           Solubility                xS       yS                                         K_sp = [x S ]^x [y S¯]^y = x ^x y ^y[S ] ^x+y                                       For AB type salt K_sp = S² or S = (K_sp)^1/2,       For AB₂/A₂B type salt, K_sp = 4S³ or, S = (K_sp/4)^1/3    , Where S represents Molar solubility of the salt.