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Ionic Equilibrium | Chemistry - Answer the following questions | 12th Chemistry : UNIT 8 : Ionic Equilibrium

Chapter: 12th Chemistry : UNIT 8 : Ionic Equilibrium

Answer the following questions

Chemistry : Ionic Equilibrium : Book Back Questions and Answers with Solutions: Answer the following questions

Chemistry : Ionic Equilibrium

Answer the following questions:


1. What are Lewis acids and bases? Give two example for each.

A Lewis acid is a positive ion (or) an electron deficient molecule Example: BF3, CO2, Fe3+

A Lewis base is anion (or) neutral molecule with at least one lone pair of electrons

Example: NH3, F, Acetylene


2. Discuss the Lowry – Bronsted concept of acids and bases.

An acid is defined as a substance that has a tendency to donate a proton to another substance (a proton donor)

A base is a substance that has a tendency to accept a proton form other substance (a proton acceptor)

Hydrogen chloride is dissolved in water, HCl behaves as an acid and H2O is base.

HCl + H2O H3O+ + Cl

Proton donar (acid) + Proton acceptor (Base)Proton donar (acid) + Proton acceptor (Base)



3. Indentify the conjugate acid base pair for the following reaction in aqueous solution

i) HS- (aq) + HF  F- (aq) + H2 S(aq)

ii) HPO42- + SO32- ↔ PO43- + HSO3-

iii) NH4+ + CO32- ↔ NH3 + HCO3

i) HS(aq) + HF F(aq) + H2S(aq)

ii) HPO42− + SO32− PO43− + HSO3

iii) NH4+ + CO32− NH3 + HCO3

i) HS(aq) + HF F(aq) + H2S(aq) 

HF donates a proton to HS and gives F. HF is acid. Hence the conjugate base of HF is F.

HS accepts a proton from HF and forms H2S. HS is base and its conjugate acid is H2S.

ii) HPO42− + SO32− PO43− + HSO3

HPO42− donates a proton to SO32−. HPO42− is acid. Hence its conjugate base is PO43−

SO32− accepts a proton from HPO42−. SO32− is base. Hence its conjugate acid is HSO3.

iii) NH4+ + CO32− NH3 + HCO3

NH4+ donates a proton to CO32−. NH4+ is acid. Hence its conjugate base is NH3.

CO32− accepts a proton from NH4+. CO32− is base. Hence its conjugate acid is HCO3.


4. Account for the acidic nature of HClO4 in terms of Bronsted – Lowry theory, identify its conjugate base.

HClO4 having the tendency to donate a proton. Hence it is acidic in nature.

HClO4 + H2O H3O+ + ClO4

perchloric acid + water Hydronium ion + chlorate ion


HClO4 donates a proton to H2O and forms H3O+ and ClO4

ClO4 is the conjugate base of HClO4.

 

5. When aqueous ammonia is added to CuSO4 solution, the solution turns deep blue due to the formation of tetramminecopper (II) complex, [Cu(H2 O)4 ](aq)2+ +4NH3 (aq)  [Cu(NH3)4 ](aq)2+ , among H2O and NH3 Which is stronger Lewis base.

Ammonia was added to the aqueous CuSO4 solution and forms tetramine Cu (II) complex.

Water present in the coordination sphere was replaced by NH3 and the complex formed. Moreover the higher electronegativity of oxygen also responsible for less availability of lone pair of electrons on the oxygen atom. [Hence NH3 is strong and Lewis base than H2O]


6. The concentration of hydroxide ion in a water sample is found to be 2.5 ×10-6 M .. Identify the nature of the solution.

[OH] = 2.5 × 10-6 M

 pOH = -ℓog (2.5 × 10−6)

= −ℓog102.5 + 6 ℓog10l0

 = −0.3979 + 6 = 5.6021

pOH = 5.6021

pH + pOH = 14

pH = 14 − 5.6021 = 8.3979 = 8.4

Hence the solution is Basic


7. A lab assistant prepared a solution by adding a calculated quantity of HCl gas at 25oC to get a solution with [H3O+ ] = 4 ×10-5M . Is the solution neutral (or) acidic (or) basic.

[H30+] = 4 × 10−5 M

 pH = −ℓog [H3O+]

= −log (4 × l0−5)

= −log 4 + 5 log 10

 pH = −0.6020 + 5 = 4.3979

 pH = 4.4

The solution is acidic


8. Calculate the pH of 0.04 M HNO3 Solution.

Solution:

Concentration of HNO3 = 0.04M

[H3+] = 0.04 mol dm-3

pH = -log[H3O+]

 = -log(0.04)

 = -log(4 ×10-2 )

= 2-log4

=2-0.6021

=1.3979 = 1.40


9. Define solubility product

The solubility product of a compound is defined as the product of the molar concentration of the constituent ions, each raised to the power of its stoichiometric co-efficient in a balanced equilibrium equation.



10. Define ionic product of water. Give its value at room temperature.

2H2O H3O+ + OH

According to law of mass action,

K eq = ( [H3O+] [OH] ) / ( [H2O]2)

K eq = [H2O]2 = [H3O+] [OH]

Kw = l × 10−14 mol2 dm− 6

The constant Kw is known as ionic product of water and it is given by the product of concentrations of [H30+] and [OH] ions at 298 K.

 

11. Explain common ion effect with an example

When a salt of a weak acid is added to the acid itself, the dissociation of the weak acid is suppressed further.

Example: Addition of sodium acetate to acetic acid solution.

The acetic acid dissociated weakly. The common ion, CH3COO suppress the dissociation of acetic acid

CH3COOH(aq) H +(aq) + CH3 COO(aq)

acetic acid Hydrogen ion + acetate ion

The added sodium acetate, salt completely dissociates to produce Na+ and CH3COO ion.

To maintain the equilibrium, the excess CH3COO ions combines with H+ ions to produce much more unionized CH3COOH and dissociation of CH3COOH suppressed.

 

12. Derive an expression for Ostwald’s dilution law

Ostwald dilution law relates the dissociation constant of the weak electrolyte with degree of dissociation and the concentration of the weak electrolyte

CH3COOH CH3COO + H+

Ka = ( [CH3 COO][H+] ) / [CH3COOH]

= degree of dissociation


Ka = Cα.Cα / C(l-α) = α2.c2 / C(l−α) = α2.C / l−α

If α is too small Ka = α2c

Calculation for degree of dissociation (α):

α2 = kc/c , α = √[ka/c]

Calculation for concentration:

[H+] = [CH3COO] = Cα

C. √[ka/c] = √C . √Ka = √[Ka. C]

Ka = α2C / 1− α is known as the Ostwald’s dilution law.

 

13. Define pH

pH of a solution is defined as the negative logarithm of base 10 of the molar concentration of the hydronium ions present in the solution.

pH = −log10[ H30+]


14. Calculate the pH of 1.5×10-3M solution of Ba (OH)2

Answer:

Ba(OH)2 → Ba2+ + 2OH

1.5 × 10−3 M → 2 × 1.5 × 10−3 M.

[OH] = 3 × 10−3M

[ pH + pOH = 14]

pH = 14 − pOH

pH = 14 − (−log[OH])

= 14 + log [OH]

= 14 + log (3 × l0−3)

= 14 + 0.4771−3

= 11 + 0.4771

 pH = 11.48


15. 50ml of 0.05M HNO3 is added to 50ml of 0.025M KOH . Calculate the pH of the resultant solution.

Answer:

Number of moles of HNO3 = 0.05 × 50 × l0−3

= 2.5 × 10−3

Number of moles of KOH = 0.025 × 50 × 10−3

= 1.25 × 10−3

Number of moles of HNO3 after mixing

= 2.5 × 10−3 − 1.25 × 10−3

= 1.25 × 10−3

Concentration of HNO3

= Number of moles of HNO3 / Volume in litre

After mixing, total volume = 100 ml

= 100 × 10−3 L

[H+] = (1.25 × 10−3 moles) / (100 × 10−3L)

= 1.25 × 10−2 moles L−1

pH = −log [H+]

pH = −log10(1.25 × 10−2)

= −log10 l.25 + 21og10l0              [log10 l0 = l ]

pH = −0.0969 + 2 = 1.9031


16. The Ka value for HCN is 10-9 . What is the pH of 0.4M HCN solution?

Given:

Ka =10−9

c = 0.4 M

pH = −log[H+]

[H+] = √[Ka × c] = √[10-9 × 0.4] = 2 × 10-5

  pH = −log10(2 ×l0−5)

= log102 + 5 log10 10 = −0.3010 + 5 = 4.699

 pH = 4.699


17. Calculate the extent of hydrolysis and the pH of 0.1 M ammonium acetate Given that Ka =Kb =1.8 ×10-5

Answer:

h = √Kh = √[ Kw / (KaKb) ] = √[(1x10−14)/(1.8 × 10-5 × 1.8 × 10- 5)]

= √ [ (l/1.8) × 10-4 ] = √0.5555×10-4

= 0.7453×10-2

pH = 1/2 pKw + 1/2 PKa – 1/2 PKb

 Given that Ka = Kb = 1.8 × 10−5

 If Ka = Kb , then pKa = pKb

pH = l/2 pKw = 1/2 (14) = 7


18. Derive an expression for the hydrolysis constant and degree of hydrolysis of salt of strong acid and weak base

Let us consider the reactions between a strong acid, HCl, and a weak base, NH4OH, to produce a salt, NH4Cl and water

 HCl(aq) + NH4OH(aq) NH4Cl(aq)+H2O(ℓ)

For strong acid

[H+] > [OH]; the solution is acidic and the pH is less than 7.

Kh.Kb = Kw

Kh = Kw/Kb

Kh = h2C

[H+] =√[Kh.C]

substitute Kh value in the above equation

[H+] = √{Kw/Kb . C}

pH = −ℓog10[H+]

 = − log (Kw.C/ Kb)1/2

= − 1/2 log Kw−1/2 log C + 1/2 log Kb

PH = 7 - l/2pKb − l/2 log C 


19. Solubility product of Ag2CrO4 is 1 ×10-12 . What is the solubility of Ag2 CrO4 in 0.01M AgNO3 solution?

Answer:

Given that Ksp =1 × 10-12

Ag2 CrO4 (s) ↔ 2 Ag+(aq) + CrO4−2 (aq)

 s  2s + s

AgNO3 (s)  Ag+(aq) + NO3- (aq)

 0.01M ↔ 0.01M + 0.01M

[Ag+] = 2s + 0.01     

0.01>>2S 

 [Ag+ ]= 0.01M

[CrO42−]= s

Ksp = [Ag+]2 [CrO42- ]

1 × 10-12 = (0.01)2 (s)

(s) = 1 × 10-12 / (10-2)2 = 1 × 10−8 M


20. Write the expression for the solubility product of Ca3 (PO4 )2

Answer:

Ca3(PO4)2 3Ca2+ + 2PO43-

     s                    3s           2s

Ksp = [Ca2+]3 [PO43−]2

Ksp = (3s)3 (2s)2

Ksp = 27s3. 4s2

 Ksp = 108 s5


21. A saturated solution, prepared by dissolving CaF2 (s) in water, has [Ca2+] = 3.3 ×10-4M What is the Ksp of CaF2 ?

Answer:

CaF2(s) Ca(aq)2+ + 2F (aq)

[F] = 2 × 3.3 × 10-4 M

= 6.6 × 10−4 M

Ksp = [Ca2+][F]2

 = (3.3 × 10−4) (6.6 × 10−4)2

 = 1.44 × 10−10


22. Ksp of AgCl is 1.8 ×1010 . Calculate molar solubility in 1 M AgNO3

Answer:

AgCl(s) Ag+(aq) + Cl (aq)

x = solubility of AgCl in 1M AgNO3

AgNO3(aq) Ag+(aq) [1M] + NO3 (aq) [1M]

 [Ag+] = x + 1       1M ( x < < 1)

[Cl] = x

Ksp = [Ag+] [Cl]

1.8 × 10−10 = (1) (x)

 x = 1.8 × 10−10 M


23. A particular saturated solution of silver chromate Ag2 CrO4 has [Ag+ ]=5 ×10-5 and [CrO4 ]2 -=4.4 ×10 -4 M. What is the value of Ksp for Ag2 CrO4 ?

Answer:

Ag2CrO4(s) 2Ag+(aq) + CrO42−(aq)

Ksp = [Ag+]2 [CrO42−]

= (5 × l0-5)2 (4.4 × 10−4) = (25×10−10) 4.4 × l0−4

 = 1.1 × 10−12 = 110 × 10−14

 = 1.10 × 10−12


24. Write the expression for the solubility product of Hg2 Cl2 .

Answer:

Hg2Cl2 Hg22+ + 2Cl

     s            s         2s

Ksp = [Hg22+] [Cl]2

= (s) (2s)2

Ksp = 4s3


25. Ksp of Ag2 CrO4 is 1.1×10-12 . what is solubility of Ag2 CrO4 in 0.1M K2 CrO4.

Answer:

Ag2CrO4 2Ag+ + CrO42−

   x                2x           x

x is the solubility of Ag2CrO4 in 0.1 M K2CrO4

K2CrO4 2K+ + CrO42−

 0.1 M            0.2 M        0.1 M

[Ag+] = 2x

[CrO42−] = (x + 0.1) ≈ 0.1           x < < 0.1

Ksp = [Ag+]2 [CrO42−]

1.1 × 10−12 = (2x)2 (0.1)

1.1 × 10−12 = 0.4 x2

 x2 = [1.1×10−12] / 0.4 ; x = √[1.1 × 10−12] / [0.4]

 x = √(2.75 × 10−12)

 x = 1.65 × 10−6 M


26. Will a precipitate be formed when 0.150 L of 0.1M Pb(NO3)2 and 0.100L of 0.2 M NaCl are mixed? Ksp (PbCl2 )=1.2 ×10-5 .

Answer:

When two or more solutions are mixed, the resulting concentrations are different from the orignal.

Total volume = 0.150 L

Pb(NO3)2 Pb2+ + 2NO3

 0.1 M       0.1M      0.2 M

Number of moles

Pb2+ = molarity × Volume of the solution in liter

 = 0.1 × 0.15

[Pb2+]mix = (0.1 × 0.15) / 0.25 = 0.06 M

NaCl       Na+     +    Cl

0.2 M        0.2M    0.2M

No. of moles Cl= 0.2 × 0.1

[Cl]mix = (0.2 × 0.1) / 0.25 = 0.08 M

Precipitation of PbCl2 (s) occurs if

[Pb2+] [Cl]2 > Ksp

[Pb2+] [Cl]2 = (0.06) (0.08)2

= 3.84 × 10−4

Since ionic product [Pb2+] [Cl]2 > Ksp , PbCl2 is precipitated

 

27. Ksp of Al(OH)3 is 1 ×10-15M . At what pH does 1.0 ×10-3M Al3+ precipitate on the addition of buffer of NH4Cl and NH4OH solution?

Answer:

Al(OH)3 Al3+(aq) + 3OH(aq)

Ksp = [A𝑙3+] [OH]3

Al(OH)3 precipitates when 

[Al3+] [OH]3 > Ksp

(l ×10−3) [OH]3 > 1 × 10−15

[OH]3 > 1 × 10−12

[OH] > 1 × 10-4 M

[OH] = l× 10-4 M

pOH = −log10[OH]= −log(1×10−4) = 4

pH = 14 − 4 = 10

Thus, Al(OH)3 precipitates at a pH of 10


EVALUATE YOURSELF:

 

1. Classify the following as acid (or) base using Arrhenius concept

i) HNO3 ii) Ba(OH)2 iii) H3PO4 iv) CH3COOH

Answer:

acid: i) HNO3 iii) H3PO4 iv) CH3COOH

base: ii) Ba(OH)2

 

2. Write a balanced equation for the dissociation of the following in water and identify the conjugate acid-base pairs.

i) NH4+ ii) H2SO4 iii) CH3COOH

Answer:


NH4+   +       H2O             H3O+   +   NH3

acid l         base 2              acid 2      base 1

H2SO4 + H2O H3O+ + HSO4-

acid l    base 2   acid 2   base 1

CH3COOH + H2O H3O+ + CH3COO-

     acid l     base 2    acid 2      base 1

 

3. Identify the Lewis acid and the Lewis base in the following reactions.

a) CaO + CO2 —> CaCO3

 CH3 Cl

b) CH3 - O - CH3 + AlCl3

Answer:

i) CaO - Lewis base ; CO2 - Lewis acid

ii) 

AlCl3 - Lewis acid

 

4. H3BO3 accepts hydroxide ion from water as shown below

H3BO3(aq) + H2O(ℓ) B(OH)4- + H+

Predict the nature of H3BO3 using Lewis concept


 

5. At a particular temperature, the Kw of a neutral solution was equal to 4xl0−14. Calculate the concentration of [H3O+] and [OH-].

Given solution is neutral

[H3O+] = [OH]

Let [H3O+] = x; then [OH] = x

 KW = [H3O+] [OH]

4 × 10−14 = x . x

x2 = 4 × 10−14

x = √[4×10−14] = 2 × 10-7

 

6. a) Calculate pH of 10−8 H2SO4

b) Calculate the concentration of hydrogen ion in moles per litre of a solution whose pH is 5.4 

c) Calculate the pH of an aqueous solution obtained by mixing 50 ml of 0.2 M HCl with 50ml 0.1 M NaOH.

a)


In this case the concentration of H2SO4 is very low and hence [H3O+] from water cannot be neglected

  [H3O+] = 2 × 10−8 (from H2SO4) + 10-7 (from water)

= 10−8 (2 + 10)

= 12 × 10−8 = 1.2 × 10−7

pH = −log10[H3O+]

= −log10 (1.2 × l0−7)

= 7 − log10 1.2

pH = 7 − 0.0791 = 6.9209

b)

pH of the solution = 5.4

[H3O+] = antilog of (-pH)

= antilog of (−5.4)

= antilog of (−6 + 0.6)

= 3.981 × 10−6

[H3O+] = 3.98 × 10−6 mol dm−3

c)

No of moles of HCl = 0.2 × 50 × 10−3 = 10 × l0−3

No of moles of NaOH = 0.1 × 50 × l0−3 = 5 × l0−3

No of moles of HCl after mixing

= (10 × l0−3 ) – (5 × 10−3)

= 5 × 10−3

After mixing total volume = 100 mL

  Concentration of HCl in moles per litre

= [5 × 10−3 mole] / [100 × 10−3L]

[H3O] = 5 × 10−2 M

pH = −log10 (5 × l0−2) = −log105 + 21og1010

 = 2 − log5 = 0.6990 + 2

= 2 − 0.6990 = 1.3010

= 1.30

pH = 1.3

 

7. Kb for NH4OH is 1.8 × 10-5. Calculating the percentage of ionisation of 0.06 M ammonium hydroxide solution.

α = √[Kb / C ] = √[1.8 × 10−5 ] / [ 6 × 10−2 ] = √(3 × 10-4)

= 1.732 × 10−2

= 1.732 / 100 = 1.732 %

 

8. a) Explain the buffer action in a basic buffer containing equimolar ammonium hydroxide and ammonium chloride.

b) Calculate the pH of a buffer solution consisting of 0.4M CH3COOH and 0.4M CH3COONa. What is the change in the pH after adding 0.01 mol of HCl to 500 ml of the above buffer solution. Assume that the addition of HCl causes negligible change in the volume.

Given:

(Ka = 1.8 × l0-5).

a) Dissociation of buffer components

NH4OH (aq) NH4+ (aq) + OH(aq)

NH4Cl NH4+ + Cl

Addition of H+

The added H+ ions are neutralized by NH4OH and there is no appreciable decrease in pH.

NH4OH(aq) + H+ → NH4+ (aq) + H2O (ℓ)

Addition of OH

The added OH ions react with NH4+ to produce unionized NH4OH. Since NH4OH is a weak base, there is no appreciable increase in pH

NH4+ (aq) + OH (aq) → NH4OH (aq)

b)

pH of buffer

CH3COOH (aq) CH3COO (aq) + H+ (aq)

      0.4 – α                   α                     α

CH3COONa (aq) → CH3COO(aq) + Na+ (aq)

[H+] = Ka[CH3COOH] / [CH3C00]

[CH3COOH] = 0.4 − α 0.4

[CH3COO] = 0.4 − α 0.4

[H+] = Ka(0.4) / (0.4)

[H+] = 1.8 × 10-5

pH = −log (1.8 × l0-5) = 4.74

Addition of 0.01 mol HCl to 500 ml of buffer

Added [H+] = 0.01 mol / 500 ml. = 0.01 mol / ½ L

= 0.02 M

CH3COOH(aq) CH3COO(aq) + H+(aq)

0.4 - α                         α                    α

CH3COONa → CH3COO + Na+

0.4                      0.4            0.4

CH3COO + HCl → CH3COOH + C𝑙

(0.02)          0.02           0.02        0.02

[CH3COOH] = 0.4 − α + 0.02 = 0.42 − α 0.42

 [CH3COO] = 0.4 + α − 0.02 = 0.38 + α 0.38

[H+] = [ (l.8 × l0-5) (0.42) ] / (0.38)

[H+] = 1.99 × 10-5

pH = −log (1.99 × 10-5)

= 5 − log 1.99

 = 5 − 0.30 = 4.70

 

9. a) How can you prepare a buffer solution of pH 9. You are provided with 0.1M NH4OH solution and ammonium chloride crystals. (Given: pKa for NH4OH is 4.7 at 25° C).

b) What volume of 0.6 M sodium formate solution is required to prepare a buffer solution of pH 4.0 by mixing it with 100ml of 0.8M formic acid. (Given: pKa for formic acid is 3.75).

a) pOH = pKb + log ( [salt] / [base] )

pH + pOH = 14

9 + pOH = 14

pOH = 14 − 9 = 5

5 = 4.7 + log ( [NH4Cl] / [NH4OH] )

0.3 = log ( [NH4Cl] / 0.1)

[NH4Cl] / 0.1 = antilog of (0.3)

[NH4Cl] = 0.1 M × 1.995

= 0.1995 M

= 0.2 M

Amount of NH4Cl required to prepare 1 litre 0.2M solution

= strength of NH4Cl × molar mass of NH4Cl

 = 0.2 × 53.5

 = 10.70 g

10.70 g ammonium chloride is dissolved in water and the solution is made up of one litre to get 0.2 M solution. On mixing equal volume of the given NH4OH solution and the prepared NH4Cl solution will give a buffer solution with required pH value (pH = 9).

b)

PH = pKa + log ( [salt] / [base] )

4= 3.75 + log ([sodium formate] / [formic acid])

[sodium formate] = Number of moles of HCOONa / Volume of HCOONa

= 0.6 × V × 10−3

[Formic acid] = Number of moles of HCOOH / Volume of HCOOH

= 0.8 × 100 × 10−3

= 80 × 10−3

4 = 3.75 + log (0.6 V / 80)

0.25 = log (0.6 V / 80)

antilog of 0.25 = 0.6 V / 80

0.6 V = 1.778 × 80

= 142.4

V = 142.4 mL / 0.6 = 237.33 mL

 

10. Calculate the i) hydrolysis constant,

ii) degree of hydrolysis and iii) pH of 0.05M sodium carbonate solution pKa for HCO−3 is 10.26.

Sodium carbonate is a salt of weak acid, H2CO3 and a strong base, NaOH and hence the solution is alkaline due to hydrolysis.

Na2CO3(aq) → 2Na+ (aq) + CO32− (aq)

CO32− (aq) + H2O(ℓ) HCO3 + OH

i) h = √[Kw / (Ka×C) ]

 = √(1 × 10−14) / (1.8 × 10 −5 × 0.1)

h = 7.5 × 10−5

 Given that pKa = 4.74

pKa = −log Ka

ie., Ka = antilog of (−pKa)

= antilog of (−4.74)

= antilog of (−5 + 0.26)

= 10−5 × 1.8

[antilog of 0.26 = 1.82 1.8]

ii) Kh = Kw / Ka = 1×10−14 / 1.8×10-5

 Kh = 5.56 × 10−10

iii) pH = 7 + pKa /2 + logC/2

 pH = 7 + (4.74/2) + (log 0.1 / 2) = 7 + 2.37 − 0.5

 = 8.87

 

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