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Question-1

Why is vapour pressure independent of initial amount?

Solution:
Vapour pressure of liquid is defined as pressure exerted by liquid at equilibrium at a given temperature. Different amounts of liquids exert same equilibrium pressure at a given temperature.

At equilibrium, rate of evaporation = rate of condensation

Different amounts of liquid will give same ratio of amount of evaporation to amount of condensation.

Question-2

Write Kp for CaCO3 decomposition.

Solution:

Kp = p CO2 where pCO2 - Partial vapor pressure.

Question-3

At 500 k, 1.04 g of hydrogen, and 12,060 g of I2 are in equilibrium with 5.058 g of hydrogen iodide in a vessel of capacity 1lit. Calculate the equilibrium constant.

Solution:
[H2] = g/mole [H I] = g/mole (H2 = 2; I2 = 254; HI = 128)

[I2] = g/mole H2 + I2 2HI

Kc = = = 6.48

Kc = 6.48.

Question-4

If the value of Kc be 50.21, for the reaction H2 + I2 2HI how much HI would be present at the equilibrium, if we start with 5.30 mol of I2 and 7.94 mol of H2?

Solution:
At equilibrium of 2x mol HI is formed, x mol H2 and x mole of I2 should be consumed.

At equilibrium [HI] = 2x; [H2] = [7.94 - x]

[I2] = [5.30 - x]
Kc =
50.21 =

4x2 = 2112.93 - 664.78x + 50.21 x2

46.21 x2 664.78 x + 2112.93 = 0;

 x2 - 14.3 x + 46 = 0

x =

x =

x =

2x = 9.72 moles

At equilibrium [HI] = 9.72 moles.

Question-5

Kc for N2 (g) + 3H2 (g) 2NH3 (g) is 61. What is Kc for 2NH3 N2(g) + 3H2 (g)?

Solution:
Equilibrium constant for the reverse reaction is the inverse of the equilibrium constant for the reaction in the forward reaction.

K'c for 2NH3 N2 (g) + 3H2 (g) is .

Question-6

For the equilibrium N2 (g) + 3H2 (g) 2NH3 (g) Kp is 6.8 ×105 at 298 K. What is the Kc value for the same equilibrium?

Solution:
For the reaction Δn = 2 - (1 + 3) = -2

Kp = Kc (RT)Δn;

6.8 ×105 = Kc ×(0.0831×298)-2

6.8 ×105 = Kc ×

(i.e.) Kc = 6.8 ×105 ×(0.0831×298)2 = 4170.1 ×.

Question-7

If the reaction between iron and steam proceeds as: 3Fe + 4H2O Fe3O4 + 4H2 and the partial pressure of steam be 50 mm and of hydrogen 940 mm at 250o. Calculate the pressure of steam at equilibrium when the partial pressure of hydrogen is 1800 mm.

Solution:
3Fe (s) + 4H2O (g) Fe3O4 (s) + 4H2 (g)

Kp = =

When pH2 = 1800 mm

Kp = = where x is pH2O

x = = 95.7.
Partial preesure of steam is 95.7 mm.

Question-8

Why is NaCl precipitated by passing HCl gas in a NaCl saturated solution?

Solution:
NaCl Na+ Cl-, when HCl is passed, [Cl-] is more. To decrease the concentration of Cl- backward reaction takes place. Hence NaCl is precipitated. This could also be attributed to common ion effect.

Question-9

What is effect of pressure for the following equilibrium?
(i) 2SO2 (g) + O2 (g) 2SO3 (g); Kc = 1.7 1026
(ii) C (s) + CO2 (g) 2CO (g)

Solution:
(i) 2SO2 (g) + O2 (g) 2SO3

|When equilibrium pressure is increased, to annul the effect, the reaction in which pressure is consumed will take place. Pressure is used to decrease the volume (i.e.) forward reaction will take place where 3 moles are reduced to 2 moles. The equilibrium is shifted to forward direction.

(ii) C (s) + CO2 (g) 2CO (g)

When pressure increased the reaction in which number of moles reduced will take place. (i.e.) equilibrium is shifted to reverse direction.

Question-10

N2 (g) + 3H2 (g) 2NH3 (g) ; ΔHo = -92.38 kJ mole-1, what happens temp is increased at equilibrium?

Solution:
When temperature is increased at equilibrium, the reaction in which heat is consumed will take place to annul the effect of increase in temperature. Backward reaction is a heat consuming reaction. Hence, decomposition of ammonia takes decomposition when temp is increased.

Question-11

Explain the dynamic equilibrium between ice and water.

Solution:
The dynamic equilibrium consists of a forward reaction in which ice melts to give water and a reverse reaction in which water solidifies to ice.

Ice water,

At 273K, both ice and water are in equilibrium. Both forward and backward reactions occur at the same rate. At T > 273K, more water is present and ΔG < O. At T < 273K, more ice is present and ΔG > 0.

Question-12

Explain why the equilibrium constant for a gaseous reaction can be written in terms of partial pressures instead of concentrations.

Solution:
The molar concentration of a gas is directly proportional to its partial pressure at a fixed temperature. Therefore, it is convenient to express the composition of a gaseous reaction mixture in terms of the partial pressures of the components rather than in terms of their molar concentrations.

Question-13

The equilibrium constant of a reaction decrease with increase in temperature. Is the reaction exothermic or endothermic?

Solution:
Exothermic because with increase in temperature, the equilibrium shifts in the backward direction.

Question-14

(a) Predict the direction of reaction when chlorine gas is added to an equilibrium mixture of PCl3, PCl5 and Cl2. The reaction is
PCl3(g) + Cl2(g) PCl5(g)

(b) What is the direction of reaction when chlorine gas is removed from an equilibrium mixture of these gases?

Solution:
(a) PCl3(g) + Cl2(g) PCl5(g)

(a) When Cl2 is added to the reaction mixture, increasing its concentration, the reaction goes in the forward direction (more PCl5 is formed).
PCl3 (g) + Cl2(g) PCl5(g)

(b) When Cl2 is removed from the reaction mixture, lowering its concentration, the reaction goes in the reverse direction (more PCl5 dissociates to PCl3 and Cl2) to partially restore the Cl2 that was removed.
PCl5(g) PCl3(g) + Cl2(g)

Question-15

What is the effect of reducing the volume on the system described below?
2C(s) + O2(g) 2CO(g)

Solution:
The forward reaction is accompanied by increase in volume. Hence according to Le-Chatelier’s principle, reducing the volume will shift the equilibrium in the forward direction.

Question-16

Consider the reaction:
2SO2(g) + O2(g) 2SO3(g) + 189.4 kJ
Indicate the direction in which the equilibrium will shift when
(i) a catalyst is added

(ii) pressure is decreased

Solution:
(i) A catalyst is added: The equilibrium will not shift in any direction because both the forward and backward rates are increased to the same extent.

(ii) Pressure is decreased: The equilibrium will shift to the left to produce more number of moles of the reactants.

Question-17

What is the effect of catalyst on the equilibrium? Give an example.

Solution:
A catalyst is a substance that increases the rate of a reaction but is not consumed by it. However, the presence of a catalyst speeds the reverse reaction as well as the forward reaction, leaving the equilibrium constant unchanged For example,

N2(g) + 3H2(g) 2NH3(g)

Question-18

Classify conjugate pairs among the following:
Q. HCl, HS-, HSO42-, H2O, HCO3-, H2S, HPO42-, NH3, Cl-, H3O+, NH4+, SO42-.

Solution:
HCl and Cl-, H2S and HS-, HSO42- & SO42-, H3O+ and H2O.

Question-19

The solubility of magnesium oxalate, MgC2O4, in water is 00093 mol/L. Calculate Ksp.

Solution:
Mg2C2O4(s) Mg+2(aq) + C2O4-2(aq)
Starting                         0               0
Equilibrium               9.3
× 10-3    9.3 × 10-3

[Mg+2] = 9.3 × 10-3 mol L-3. [C2O4-2] = 9.3 × 10-3 mol L-1

Ksp = [Mg+2][C2O4-2] = (9.3 × 10-3) (9.3 × 10-3) = 8.65 × 10-5 M2.

Question-20

What is the solubility of strontium sulphate, SrSO4, in 0.15 M sodium sulphate, Na2SO4? Ksp of SrsO4 = 2.5 × 10-7

Solution:
The equation for the dissolving SrSO4 is

SrSO4(s) Sr+2(aq) + SO4-2(aq)

In this solution there are two sources of SO4-2(aq). The soluble Na2SO4 is completely dissociated into Na+ ions and SO4-2(aq) ions, and the small amount of SrSO4 that dissolves also provides SO4-2(aq) ions. The SO4-2(aq) ion is common to both the insoluble electrolyte SrSO4 and the soluble salt NazSO4. There is, however, only one source of Sr+2 ions, namely the dissolution of solid SrSO4

Let

S = molar solubility of SrSO4 in 0.15 M Na2SO4

Then, [Sr+2] = S and [SO] = 0.15 + S

Ksp(SrSO4) = [Sr+2] [SO] = S(0.15 + S)

2.5 × 10-7 = 0.15S + S2

Assuming S to be small compared to 0.15 and that

2.5 × 10-7 = 0.15S

or S = 1.67 × 10-6 M.

Question-21

The solubility product of Fe(OH)3 is 1 × 10-36. What is the minimum concentration of OH- ions required to precipitate Fe(OH)3 from a 0.001 M solution of FeCl3?

Solution:
Kw = [Fe3+] [OH-]3

Precipitation will occur when ionic product, [Fe3+] [OH-3] becomes greater than Ks

[Fe3+] = [FeCl3] = 0.001M

The concentration of OH- ions required to start the precipitation is

  [OH-]3 = = = 1 × 10-11

[OH-] = (1 × 10-33)1/3 = 1 × 10-11 molL-1.

Question-22

Calculate the pH of 0.5 molar solution of sulphuric acid.

Solution:
H2SO4 2H+ + SO4-2

In a 0.5 m H2SO4 solution

[H+] = 2 × 0.5 mol L-1 = 1 mol L-1

pH = - log [H+] = -log (1) = 0

Question-23

The pH of 0.05M aqueous solution of diethyl amine is 12.0 calculate its Kb.

Solution:
pH = 12.0; pOH = 14 – 12.00 = 2.00

[OH-] = 10-2M ; Kb =
kb = = 0.2
× 10-2  = 2 × 10-3.

Question-24

The solubility of lead iodide in water is 0.63 gm/litre. Calculate the solubility product of lead iodide (At mass of Pb = 207; I = 127).

Solution:
Solubility of lead iodide in moles / litre

= = = 1.36 × 10-3 moles / litres

PbI2 Pb2+ + 2I-

KSP = [Pb2+][I-]2
       = [1.36
× 10-3][2 × 1.36 × 10-3]2
       = 4
× (1.36 × 10-3)3
       = 4
× (10-3)3 (1.36)3
       = 4
× 10-9 × 1.4 × 1.4 × 1.4
       = 4
× 10-9 × 1.4 × 1.4 × 1.4
       = 10.9
× 10-9
KSP = 1.09
× 10-8 mol-3 L-3.

Question-25

How many moles of AgBr (KSP = 5 × 10-13 mol-2 L-2) will dissolve in 0.01M NaBr solution?
AgBr [Ag+] [Br-]
KSP = [x] [x + 0.01] where x = solubility of Ag+ or Br-.

Solution:
5 × 10-13 (x) (10-2) (compare to 0.01, x is small)

(i.e.)    x = = 5 × 10-1 mol L-1.

Question-26

Two moles of HI when heated at 44.40C until equilibrium is reached, were found to be 22% dissociated. Calculate the equilibrium constant for the reaction.

Solution:
2HI H2 + I2

x = 22% (i.e.) that = 0.22

2 HI H2 + Ι2

0.22 dissociate to give 0.11 moles of H2 and 0.11 mole of Ι2

Equilibrium concentration = (1 – 0.22) = 0.11 moles / litre

Equilibrium concentration of HI = 0.78 moles / litre.


Kc = =

Kc = 0.0198.

Question-27

How does the solubility of AgCl change in the presence of NaCl solution?
(S = 1.3
× 10-5M)

Solution:
AgCl(S)Ag+(aq) + Cl-(aq)

KSP = [Ag+] [Cl-]

Addition of NaCl to saturated solution of AgCl increases the concentration of Cl- ions. According to Le Chatelier’s AgCl(s), there by reducing the Ag+ concentration by precipitating AgCl. Thus solubility of AgCl in NaCl solution is lower than water, due to common ion effect.

Question-28

A 50.0 L reaction vessel contains 1.00 mol N2, 3.00 mol H2, and 0.055 mol NH3. Will more ammonia, (NH3), be formed or will it dissociate when the mixture goes to equilibrium at 4000C? The equation is N2(g) + 3H2(g) 2NH3(g)  Kc is 0.500 at 4000 C.

Solution:
N2(g) + 3H2(g) 2NH3(g)

Concentrations = 0.0200 M, 0.0600 M, 0.0011 M

Qc = = = 0.2801

 

Because Qc = 23.1 is greater than Kc = 0.500, the reaction will go to the left as it approaches equilibrium. Therefore, ammonia will dissociate.

Question-29

Consider the exothermic formation of sulphur trioxide from sulphur dioxide and oxygen in the gas-phase.
                         2SO2(g) + O2(g) 2SO3(g)
At 900 K, Kp for this reaction is 40.5 atm-1 and
Δ H = -198 kJ
(a) Write the expression for the equilibrium constant for this reaction.

(b) Will the equilibrium constant for this reaction at room temperature (-300 K) be greater than, less than, or equal to the equilibrium constant at 900 K? Explain your answer.

(c) How will the equilibrium be affected if the volume of the vessel containing the three gases is reduced, keeping the temperature constant: what happens?

(d) What is the effect of adding one mole of He(g) to a flask containing SO2 , O2 and SO3 at equilibrium at constant temperature?

Solution:
(a) Kp =

(b) The equilibrium constant at 300 K will be greater than the equilibrium constant at 900 K. This is an exothermic reaction. If we decrease the temperature from 900 to 300 K, the equilibrium will be displaced to the right, releasing heat, more SO3 is produced and SO2 and O2 are used up. The equilibrium constant increases.

(c) Decreasing the volume of the vessel will increase the partial pressure of each gas and therefore increase the total pressure. The system will shift to the side with fewer numbers of moles of gas. Since there are two moles of gas on the right, but three on the left, the system will shift to the right. More SO3 will be produced, and some SO2 and O2 will be used up.

(d) Adding He(g) has no effect at all. The partial pressures of SO2 and O2 and SO3 are unchanged by the addition of helium. The total pressure in the container increases, but as the partial pressure of the gases involved in the equilibrium are unaffected, the equilibrium does not shift.

Question-30

Suppose we mix 25.0 cm3 of 0.001 M AgNO3(aq) with 75.0 cm3 of 0.001 M Na2CO3(aq). Does a precipitate of Ag2CO3 form? The Ksp of Ag2CO3 is 6.2 × 10-12 at 298 K.

Solution:
At the instant of mixing,

    Conc. Of [Ag+] = = 2.5 × 10-4 M

Conc. Of [CO-23 ] = = 7.5 × 10-4 M

We are considering the reaction

AgCO3(s) 2Ag+(aq) + CO3 -2

Ki = [Ag+]2 [CO3-2] = (2.5 × 10-4)2 (7.5 × 10-4)

Ki = 4.69 × 10-11

Since, Ki > Ksp, the solution is saturated and Ag2CO3 will form.

Question-31

Which of the following are Lewis bases?
H2O, BF3, H+, Ag+, NH3 & CO.

Solution:
H2O, NH3 and CO are Lewis bases as they have lone pairs of electrons to donate to Lewis acids.

Question-32

Which of the following act both as Bronsted acids and bases?
HSO4-, NH3, CO, H2O, SO4-.

Solution:
HSO4-, NH3, H2O can donate as well as accept a proton. Hence these can act as Bronsted acids and bases.

CO, SO42- can only accept proton, hence they are Bronsted's bases.

 

Question-33

A solution has a hydroxide concentration of 1.0 ×10-5 M at 25o C. Is the solution acidic, natural or basic?

Solution:
[OH-] = 1.0 ×10-5 M;

[H+][OH-] = 1 ×10-14;

[H+] = = 1.0 ×10-9 M

as [H+] is less than 1 ×10-7 M, the solution is basic.

Question-34

Calculate the pH value of mixture containing 50cc M-Hcl and 30cc M-NaOH solution assuming both to be completely ionised.

Solution:
Total volume after mixing      = 50 + 30 = 80 cc

Molarity of HCl after mixing      = M

Molarity of NaOH after mixing   = M

Net molarity of HCl after mixing = - = 0.25 M

                                      [H+] = 0.25 = 2.5 ×10-1

                                         pH = -log (2.5 ×10-1) = 0.6021.

Question-35

The dissociation constant of an acid HA is 1.6 × 10-5. Calculate H3O+ ion concentration of its 0.01 M solution.

Solution:
Ka = ; Ka = C α2

α =

α = = = = 4 ×10-2

[H+] = a ×C = (4 ×10-2) ×10-2 = 4 ×10-4 mol. L-1.

Question-36

Calculate the concentration of H3O+ ion in a mixture of 0.02 m acetic acid and 0.2 m sodium acetate. (Given Ka for acetic acid is 1.8 10-5)

Solution:
CH3COOH CH3COO- + H3O+

Ka =

    = 0.2 M + x 0.2 M

[CH3COOH] = 0.02 M - x 0.02 M

1.8 ×10-5 =

(i.e.) H3O+ = = 1.8 ×10-6 mol. L-1.

Question-37

Calculate pH of features 10-2 M H2SO4.

Solution:
H2SO4 2H+ + SO42-

One mole of H2SO4 gives 2 moles of H+(aq)(aq) ions.

[H+] = 2[H2SO4]

[H+] = 2[10-2 M]

   pH = - log [2 10-2]

        = 2 - log 2 = 2 - 0.3010

   pH = 1. 699.

Question-38

Calculate pH of 0.001 M Ba (OH)2.

Solution:
Ba(OH)2 Ba2+ + 2OH-

Each mole of Ba(OH)2 gives 2[OH-] moles.

0.001 m Ba(OH)2 gives 2 ×[0.001 m] OH-1 ions

[H+] = = = 0.5 10-11

[H+] = 5 ×10-12

   pH = - log10 [5 ×10-12]

   pH = - [log10 5 + log 10-12]

        = - (-12) log 10 - log 5

        = 12 ×1 - 0.6990

   pH = 11. 301.

Question-39

How many grams of NaOH must be dissolved in one litre of solution to give it a pH of 12?

Solution:
PH = 12; - log [H+] = 12; [H+] = antilog of -12 = 1 ×10-12.

[OH-] = = = 1×10-2 = 0.01 moles/litre

We know, weight per litre molarity ×m.wt.

= 0.01 ×40 = 0.4 gms/litre.

Weight of NaOH present = 0.4 gms/litre.

Question-40

A buffer solution is prepared by mixing 6 g. of acetic acid and 13.6 g of sodium acetate CH3COONa.3H2O and making the total volume 250 ml. Calculate the

(i) pH of the solution and

(ii) pH change on addition of 1 ml of 1m HCl to it. Ka = 1.8
×10-5.

Solution:
[CH3COOH] = = = 0.1 moles ; 0.1/.250 = 0.004

[CH3COONa] = = = 0.1 moles = 0.004

               pH = pKa + log10

              pKa = - log10 Ka = - log 1.8 ×10-5 5 - 0.2553 = 4.74

               pH = 4.74 + log [0.004]/[0.004]
   
               pH = 4.74 + log101 = 4.74 + 0 = 4.74.

Question-41

Calculate the amount of (NH4)2 SO4 in gms which must be added to 500 ml of 0.200 MNH3 to yield a solution with pH 9.35 (Kb for NH3 = 1.78 ×10-5)

Solution:
pOH = pKb + log

  pH = 9.35

pOH = 14 - 9.35 = 4.65; Kb = 1.78 ×10-5

pKb = 4.7447

Number of moles of NH3 added: one litre of 0.2 M NH3 contain 0.2 mole

500 ml contains 0.2 ×= 0.1 mole

    Moles of (NH4)2 SO4 = x

4.65 = 4.7447 + log

log = - 0.0947 or log = 0.0947

        = 1.2436; x = = 0.08 mole

 M. wt. of (NH4)2 SO4 = mole ×m.wt = 0.08 ×132 = 10.56 grams.

Question-42

Calculate the pH of a solution made by diluting 25 cc of N/100 HCl to 500 cc assuming complete ionisation of HCl.

Solution:
N1V1 = N2V2

N1 500 = ×25

N1 = = 0.0005

Since HCl is completely ionised

HCl = H+Cl-

Conc of H+ ions conc HCl = 0.0005 g ions litre-1

                   pH = - log [H+] = - log 0.0005 = 3.3.

Question-43

What would be the pH of a solution that contains 100 ml of 0.1 N HCl and 9.9 ml of 1 N NaOH solution?

Solution:
9.9 ml of 1 N solution of NaOH should be converted into 0.1 N NaOH

9.9 1 N x ×0.1; (i.e.) x = = 99 ml.

9.9 ml of 1.0 N NaOH = 99 ml of 0.1 N, NaOH

Volume of HCl left unneutralised (100 - 99 ml) of 0.1 N 1 ml of 0.1 N

Total volume of solution after mixing 100 + 9.9 = 109.9 ml 110 ml (app)

Conc. HCl after mixing =9.09 ×10-4 m
                         [H+] = 9.09 ×10-4
                            pH = - log [9.09 ×10-4] = 4 - 0.9546 = 3.0454.

Question-44

Calculate the degree of hydrolysis of 0.1M solution of sodium acetate at 25o C ; Ka=1.8× 10-5.

Solution:
Since sodium acetate is a salt of strong base and weak acid

Kb = = = 5.5 ×10-10

 Since solution is 0.1, h =

                                  = = 7.452 ×10-5.

Question-45

Ka for butyric acid is 2.0 ×10-5. Calculate pH and hydroxyl ion concentration of 0.2 M aqueous solution of sodium butyrate.

Solution:
For salt of a weak acid with strong base

   pH = pKw + pKa + log C

         = ×14 + log + log 0.2

         = 7 + 2.3494 - 0.3494

         = 9

  pOH = 14 - 9 = 5

[OH-] = 10-5 M.

Question-46

The pH of blood stream is maintained by a proper balance of H2CO3 and NaHCO3 concentrations. What volume of 5 M NaHCO3 solution should be mixed with a 10 ml sample of blood which is 2 M in H2CO3 in order to maintain a pH of 7. 4? Ka for H2CO3 in blood is 7. 8 ×10-7.

Solution:
Let the amount of NaHCO3 mixed = x ml

Number of moles of NaHCO3 in x ml of 5 m NaHCO3 = = 0. 005 x

Number of moles of H2CO3 in 10 ml of 2 m H2CO3 = = 0. 02

  pH = - log Ka + log

7. 4 = - log 7. 8 ×10-7 + log

   x = 78.36 ml.

Question-47

To a buffer solution containing 0. 02 mole of propionic acid and 0.015 mole of sodium propionate per litre, 0.01 mole / litre of HCl is added. What is the pH of the solution? Ka propionic acid = 1.34 × 10-5.

Solution:
[propionic acid] = (0.02 + 0.01)M

    [propionate] = (0.015 + 0.010)M
   
    pH = pka + log [salt] / [Acid]

    pH = - log (1.34 ×10-5) + log = 4.9528.

Question-48

A buffer solution of acetic acid and sodium acetate of 1 M concentration was prepared. Calculate pH of the system when 0.01 mole of NaOH is added per litre of the above solution. Ka = 1.8 × 10-5.

Solution:
NaOH ionises fully: Hence OH- ions added will react with CH3COOH of the buffer system.

Molar conc. before adding NaOH

      1             0         1
CH3COOH + OH
CH3COO- + H2O

Molar conc. after adding 0.01 mole of (1 - 0.01) = 0.99

PH = pKa + log
     = - log 1.8 ×10-5 + log

     = 4.74 + 0.01= 4.75.

Question-49

The solubility product of AgCl in water is 1.5 × 10-10. Calculate its solubility in 0.01 M NaCl aqueous.

Solution:
Let the solubility of AgCl in 0.01 M NaCl solution = x g mole L-1

[Ag+] in NaCl solution = x mol L-1

[Cl-] in 0.01 M NaCl solution x + 0.01 M L-1

[Ag+] [Cl-] = Ksp

(x) (x + 0.01) = 1.5 ×10-10

   0.01 x + x2 = 1.5 ×10-10

Since value of x is small x2 is neglected.

0.01 x = 1.5 ×10-10

       x = 1.5 ×10-8 mol L-1.





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