03_Ionic Equilibrium (2)

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Chapter 3

Ionic Equilibrium Gilbert Newton Lewis

Born

23 October 1875

Died

23 March 1946

Nationality

American

Field

Chemistry

Known for

Acid, Base Theory

There is no scientist in American history who has contributed more extensively to all fields in chemistry than Gilbert Newton Lewis. His thinking was far ahead of his time and his theories have had profound influence on modern chemistry Lewis researched standard electrode potentials, conductivity, free energy and other thermodynamic constants for the elements. These tables are still being used. His ability to organize and apply the scattered laws of thermodynamics brought about the evolution of physical chemistry into the science as it is known today. Lewis once defined physical chemistry as encompassing "everything that is interesting. Lewis work on acid base theory is most generalized theory. Lewis did not believe only that an electron completely transfers from one atom to another, as in the positive-negative theory. He describes the partial transfer of two electrons, one from each of the two bonding atoms, so that there is a shared pair of electrons between them. This eliminates the need for the formation of oppositely charged atoms when there was no indication of individually charged atoms (ions) in a compound. This was the first description of covalent bonding. Lewis' research on isotopes is an example of his wide-ranging and prolific interests. He published twenty-six papers on heavy hydrogen and heavy water, isotopes of lithium, and neutron physics. He predicted the existence of naturally occurring heavy water before he isolated it.

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Chapter-3 / Ionic Equilibrium

3.1 Concepts of Acid and Base Arrhenius concept Acid is a substance which is capable of furnishing H + ions in aqueous solution. e.g. - HCl, H 2SO4 etc. While base is a substance which furnishes OH– ions e.g. - NaOH, NH4OH etc. Limitations. Arrhenius theory failed to explain. (i) behaviour of acids/bases in non aqueous solutions. (ii) neutralization reaction giving salt in absence of a solvent. (iii) acid character of certain salts like AlCl 3,BF3 etc. (iv) existence of H+ in water. Bronsted Lowery Concept Acid is a substance which is capable of donating a proton while base is a substance which is capable of accepting a proton. This is also called protonic theory of acids-bases. HCl

H2O

acid

base

H2O +

NH3

‡ˆ ˆˆ †ˆ

H3O+ conjugate acid

‡ˆ ˆˆ †ˆ

+

Cl–

conjugate base

NH +4

+

OH–

acid base conjugate acid conjugate base conjugate base of an acid is a species formed by the loss of a proton from acid. Acid → H+ + conjugated base Similarly conjugate acid is formed from a base by gain of H +. Base + H+ → conjugated acid Weak acid has a strong conjugate base and vice-versa. A Bronsted - Lowery acid base reaction always proceeds in the direction from the stronger to the weaker acid base combination. e.g. HI + OH– → H2O + I– Strong Strong Weak Weak acid base acid base Lewis concept Acid is a substance which can accept a pair of electrons while base is a substance which can donate a pair of electrons. Hence Lewis acids are (i) Molecules in which central atom has incomplete octet e.g. BF 3, AlCl3 and FeCl3 etc. (ii) (iii)

Simple cations like Ag+, H+ etc. Molecules in which central atom has vacant d-orbitals e.g.- SiF4, SnCl4 etc.

(iv)

Molecules in which atoms of different electronegativities are joined by multiple bond. e.g. CO 2, SO3 etc.

(v)

In carbonyl complexes, metal atoms act as Lew’s acids e.g. Ni in Ni(CO) 4 etc.

And Lewis bases are (i) Neutral molecules like NH3, RNH2 etc. (ii)

Negatively charged anions. e.g. CN–, Cl– etc.

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(iii)

Molecules with carbon-carbon multiple bonds can act as lewis base in some cases - e.g. C2 H 4 in  Ag ( C 2 H 4 ) 

(iv)

+

In complex compounds, the ligands act as Lewis bases e.g. CO in Ni(CO) 4 etc.

3.2 Strong and weak electrolytes Electrolytes which dissociate completely into ions in aqueous solutions are called strong electrolytes e.g. HCl, H2SO4, NaCl etc. Whereas electrolytes which dissociate to a lesser extent are called weak electrolytes eg. CH3COOH, NH4OH etc.

3.3 Ionisation of Weak Acids Let us consider the ionization of weak acid HA, having initial concentration ‘c’ in mol/litre H A ‡ˆ ˆˆ †ˆ c c(1–x )

H + + A− 0 0 cx cx

At equ ilibriu m

It Ka is ionization constant for acid, then Ka =

[H + ][A − ] [HA] cx.cx cx 2 = c(1 − x) (1 − x)

=

for very weak acids. x < < 1 Q

Ka =

cx 2 (∴ 1 − x ; 1) 1

or

x=

Ka C

or

x=

Ka × V  n c = ÷ n V 

or

x α V (Ostwald’s dilution law)

From above calculation, we can calculate Ka = Ka × c c

(i)

[H + ] = cx = c

(ii)

[A − ] = cx = Ka × c

(iii)

[H A] = c(1 − x) ≅ c (x 7

(c) < 7

(d) 0

Ans. (c) 3.

If temperature is raised degree of dissociation of water will (a)

increase

(b) decrease

(c) remain same

(d) none of these

(c) expand

(d) none of these

(c) lesser than 14

(d) depends on temp.

Ans. (d) 4.

On increasing temperature, PH range will (a)

remain same

(b) shrink

Ans. (b) 5.

Value of PH + POH at any temp. is (a)

always 14

(b) greater than 14

Ans. (d) Passage – 2 PH value of solution is – log [H+]. If solution is very dilute than, concentration of H + from water should also be considered. While at very higher concentration (above 1M). It should be taken O. 1. PH value of 10–5 N – H2SO4 is (a)

5

(b) 5/2

(c) 5 × 2

(d) 4.7

(c) Just above 7

(d) Just below 7

Ans. (d) 2. PH value of 10–8M – HCl should be (a)

8

(b) 6

Ans. (d) 3.

If 100 ml of 10–4M – HCl and 200 ml of 10–4M HNO3 are mixed together. Resultant PH will be (a)

4

(b) above 4

(c) below 4

(d) None of these

(c) 11.7

(d) 12.3

Ans. (a) 4. PH value of 10–2M – Ca(OH)2 solution is (a)

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2

(b) 4

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Ans. (d) 5.

When 1 or mol of

M N M HCl, 200 ml of H 2SO4 and 200 ml of NaOH are mixed together. Resultant PH will 10 10 10

be (a)

0

(b) 7

(c) 1

(d) 2

Ans. (b) Passage – 3 Consider an ionic solid that dissolves in water according to the equation

ˆˆ† MnXm(s) ‡ ˆ ˆ nMm +(aq) + mXn –(aq) The equilibrium constant for this reaction, Ksp = [Mm +]n [Xn –]m is known as the solubility product of M nXm. The form of this equilibrium is important in understanding effects such as the influence of pH, complex formation and common ion effect. Equilibrium constants in solution should be written correctly using activities and not concentrations. The difference between these quantities is large in concentrated ionic solutions and K sp is quantitatively reliable as a guide of solubilities only for very dilute solutions, if solubility product of AB type salt is 4 × 10– 10 at 18°C. 1.

If mol. wt. of AB is 143.5 g/mol., then solubility in g/lit of AB is (a)

14.35 gm/lit

(b) 2.87 × 10– 3 gm/lit

(c) 1.43 gm/lit

(d) 28.7 gm/lit

Ans. (b) 2.

If ppt. of AB is washed with 5 lit water, loss in wt. of ppt. of AB is (a)

10– 4 mol/lit

(b) 10– 4 gm/mol

(c) 10– 4 mol

(d) 10– 4 gm

Ans. (c) 3.

The % error in washing of 2 mol of ppt. of AB with 100 cc of N/100 solution of BC2 is (a)

4 × 10–7%

(b) 4 × 10– 2%

(c) 14 × 10– 3%

(d) 2 × 10– 5%

Ans. (a) 4.

The solubility of Calomel in water at 25°C is s moles/lit. Its solubility product is (a)

4s3

(b) 12s3

(c) 10s5

(d) s2

Ans. (a) Passage – 4 A solution which remains in equilibrium with undissolved solute, in contact, is said to be saturated. The concentration of a saturated solution at a given temperature is called solubility. The product of concentration of ions in a saturated solution of an electrolyte at a given temperature, is called solubility product (K sp). For the electrolyte AxBy with solubility S,

ˆˆ† AxBy(s) ‡ ˆ ˆ xAy + + yBx – The solubility product (Ksp) is given as, Ksp = xx × yy × Sx + y. While calculating the solubility of a sparingly soluble salt in the presence of some strong electrolyte containing a common ion, the common ion concentration is practically equal to that of strong electrolyte. If in a solution, the ionic product of an electrolyte exceeds its K sp value at a particular temperature, then precipitation occurs. 1.

The solubility of PbSO4 in water is 0.0303 g/l at 25°C, its solubility product at that temperature is (a)

10– 4 M2

(b) 9.18 × 10– 4 M2

(c) 10– 8 M2

(d) 9.18 × 10– 8 M2

Ans. (c) 2.

The solubility of BaSO4 in 0.1 M BaCl2 solution is (Ksp of BaSO4 = 1.5 × 10– 9) Sri Chaitanya IIT-JEE Head Office: 11, Central Market, West Punjabi Bagh , New Delhi-110 026. Phones: 25226309/10. Web: srichaitanya.edu.in, E-mail: [email protected]

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1.5 × 10– 9 M

(a)

Chapter-3 / Ionic Equilibrium

(b) 1.5 × 10– 8 M

(c) 2.25 × 10– 16 M

(d) 2.25 × 10– 18 M

Ans. (b) 3.

If S0, S1, S2 and S3 are the solubilities of AgCl in water, 0.01 M CaCl 2, 0.01 M NaCl and 0.05 M AgNO 3 solutions respectively, then (a)

S0 > S1 > S2 > S3

(b) S0 > S2 > S1 > S3

(c) S0 > S1 = S2 > S3

(d) S0 > S2 > S3 > S1

Ans. (b) pH of 0.1 M (NH4)2HPO4 solution is approximately equal to (Kb of NH4OH = 10– 5, K a1 , K a 2 and K a 3 of H3PO4

4.

are 10– 4, 10– 8 and 10– 10 respectively) (a)

8.52

(b) 4.52

(c) 9.52

(d) 7.52

Ans. (a)

Passage – 5 In case of a weak acid, a H + is calculated using Ostwald’s dilution law, i.e., a H + =

Ka × C

where Ka is the dissociation constant of weak acid with concentration C. In case of a mixture of two or more strong acids/bases, the pH (or pOH) is calculated by using the following relation,

∑ NV (for mixture of acids) ∑V ∑ NV or a − = (for mixture of bases) OH ∑V where, ∑ NV is the total equivalents of strong acids or bases and ∑ V is the total volume of the solution. a

= H+

In a mixture of a strong acid (concentration C1) and a weak acid (concentration C2 and dissociation constant K a 2 ), HA →

H+ + A–

C1

C1

ˆˆ† HB ‡ ˆ ˆ C2 – x

(strong acid)

C1

H+ + B– (weak acid)

C1 + x

x

Total a H + for this mixture can be calculated from the following expression:

Ka2 =

( C1 + x ) x ( C2 − x )

where x is the number of moles of HB dissociated at equilibrium. In a mixture of two or more weak acids with concentrations C1 and C2 and dissociation constants K a1 and K a 2 respectively, total a H + can be calculated by using the following expression:

a -50-

H+

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pH of a buffer is calculated by using Henderson’s equation. In the titration of a weak acid with a strong base, one uses the concept of buffer before the point of neutralization and at the point of neutralization, concept of salt hydrolysis is used. 1.

At any temperature, a solution will be neutral if its pH is equal to (a)

7

(b)

1 pKa 2

(c)

1 pKw 2

(d)

1 pKb 2

Ans. (c) 2.

pH of 10– 9 M CH3COOH at 25°C is (Ka for CH3COOH = 1.8 × 10– 5, log 1.67 = 0.2041) (a)

2.8

(b) 6.8

(c) 7.8

(d) 4.8

Ans. (b) 3.

pH of a mixture obtained by mixing 100 ml of 0.1 M CH 3COOH (Ka = 1 × 10– 5) with 100 ml of 0.2 M HCOOH (Ka = 1 × 10– 4) is approximately (log 1.05 = 0.0212 and log 3.25 = 0.5119) (a)

2.9

(b) 3.9

(c) 4.9

(d) 5.9

Ans. (a) 4.

100 ml of a 0.1 M solution of acetic acid is titrated with 0.1 M Ca(OH) 2 solution pH of the solution in the titration flask at the titre values of 25 and 50 ml, respectively, are (pKa of CH3COOH = 4.74) (a)

2.37, 4.74

(b) 4.74, 8.98

(c) 2.37, 8.98

(d) 4.74, 2.37

Ans. (b)

Matching Type Questions 1. A. B. C.

List I 100 ml 1 M HCl + 100 ml 1 M NH4OH 100 ml 1 M NH4OH + 100 ml 1 M CH3COOH 100 ml 1 N Ca(OH)2 + 100 ml 1 M HCl

p. q. r.

D.

100 ml 1 N CH3COOH + 100 ml ½ N NaOH

s.

List II pH = 1/2 pKw pH = 7 – 1/2 (pKb + log C) pH = 7 + 1/2 (pKa – pKb) [salt ] pH = pKa + log [ Acid]

(a) (c) Ans (a)

A – q, B – r, C – p, D – s (b) A – r, B – s, C – q, D – p (d)

A – p, B – s, C – r, D – q A – s, B – p, C – q, D – r

2.

Compound Nature MnO2 p. Acidic NaHCO3 q. Basic NaCl r. Neutral NH4Cl s. Salt A – p, q, s; B – p, q; C – p, r; D – r, s A – p, q; B – p, q, s; C – r, s; D – p, r

(b) (d)

A – q, s; B – q, r; C – r, s; D – p, s A – p, q, r; B – p, q, s; C – p, r; D – q, r, s

(b) (d)

Nature p. pH > 7 q. pH < 7 r. Acidic s. Neutral A – p, s; B – q, r; C – p, r; D – p, s A – r, s; B – p, s; C – q, s; D – p, q

A. B. C. D. (a) (c) Ans. (c) 3. A. B. C. D. (a) (c)

Solution Pure H2O at 35°C Solution with [CH3COO–] = [CH3COOH] 10– 7 N HCl Pure H2O at 0°C A – q, t; B – p, q; C – q, r; D – r, s A – q, t; B – q, r; C – q, r; D – p, s

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Chapter-3 / Ionic Equilibrium

Ans. (c)

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Exercise - V Subjective Type 1.

For an organic monoprotic acid solution of concentration c 0 moles per litre, if Ka has a value comparable to Kw show that the hydronium concentration is given by

K K a c0  [H + ] =  w +  + +  [H ] K a + [H ]  where H+ stands for the hydronium ion. If [H+] = 10– 3 M & c0 = 10– 1 M in a solution of some organic monoprotic acid, what according to the above equation must be the order of magnitude of Ka? Ans. K ~−10−5 a 2.

The Kw of water at two different temperatures is T

25°C 50°C

Kw

1.08 × 10– 14

5.474 × 10– 14

Assuming that ∆H of any reaction is independent of temperature, calculate the enthalpy of neutralization of a strong acid and strong base. Ans. – 51.96 kJ/eq. 3.

The equilibrium constant of the reaction

ˆˆ† 2Ag(s) + 2I– + 2H2O ‡ ˆ ˆ 2AgI(s) + H2(g) + 2OH– is 1.2 × 10– 23 at 25°C. Calculate the pH of a solution at equilibrium with the iodine ion concentration = 0.10 and the pressure of H2 gas = 0.60 atm. Ans. 1.65 4.

Ethylenediamine, H2N – C2H4 – NH2, can interact with water in two steps, giving OH– in each step. Calculate the concentration of OH– and [H3N – C2H4 – NH3]2 + in a 0.15 M aqueous solution of the amine. K 1 = 8.5 × 10– 5, K2 = 2.7 × 10– 8 for the base.

Ans. [OH–] = 3.57 × 10– 3 M, [H2en]2 + = 2.7 × 10– 8 M 5.

What is the solubility of AgCl in 0.20 M NH3? Given : Ksp(AgCl) = 1.7 × 10–

10

M2, K1 = [Ag(NH3)+]/[Ag+] [NH3] = 2.33 × 103 M–

1

and K2 = [Ag(NH3 ) 2+ ] /

[Ag(NH3] = 7.14 × 103 M– 1. Ans. 9.66 × 10– 3 6.

Show that solubility of a sparingly soluble salt M2 + A2 – in which A2 – ions undergoes hydrolysis is given by

  S= K sp 1 +  

2

H +  H +   +  K2 K1K 2

 ÷ ÷ ÷ 

where K1 and K2 are the dissociation constant of acid H2A. Ksp is solubility product of MA. 7.

Potash alum, KAl(SO4)2 . 12 H2O is a strong electrolyte and it is 100% dissociated into K +, Al3 + and SO 24 ions. The solution is acidic because of the hydrolysis of Al 3 +, but not so acidic as might be expected, because the SO 24 can Sri Chaitanya IIT-JEE Head Office: 11, Central Market, West Punjabi Bagh , New Delhi-110 026. Phones: 25226309/10. Web: srichaitanya.edu.in, E-mail: [email protected]

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Chapter-3 / Ionic Equilibrium

sponge up some of the H 2O+ by forming HSO −4 . A solution of alum was made by dissolving 11.4 gm of KAI(SO 4)2 . 12H2O in enough water to make 0.1 litre of solution. Calculate the pH of this solution.

ˆˆ† Given : Al3 + + 2H2O ‡ ˆ ˆ Al(OH)2 + + H3O+;

Kh = 1.4 × 10– 5 M

HSO −4 + H2O ‡ˆ ˆˆ †ˆ H3O+ + SO 24− ;

K2 = 1.26 × 10– 2 M

Ans. 3.53 8.

A solution of volume V contains n1 moles of QCl and n2 moles of RCl where QOH and ROH are two weak bases of dissociation constants K1 and K2 respectively. Show that the pH of the solution is given by

pH =

 K K   1 V log  1 2 ÷  2  K w  ( n1K 2 + K1n 2 ) 

State assumptions, if any. 9.

(

2The ‘fixing’ of photographic film involves dissolving unexposed silver bromide in a thiosulphate S2 O3

) solution:

3−

2ˆ ˆ † Ag ( S2 O3 ) 2 AgBr(s) + 2S2 O3 (aq) ‡ ˆ ˆ (aq) + Br– (aq)

Using Ksp = 5.4 × 10– 13 for AgBr and Kf = 2.9 × 1013 for Ag ( S2 O3 )

3− , calculate the equilibrium constant K for the 2

above reaction and calculate the molar solubility of AgBr in 0.10 M Na2S2O3. Ans. K = 15.7, x = 0.044 10. An acid base indicator has a Ka of 3 × 10– 5. The acid form of the indicator is red & the basic form is blue. By how much must the pH change in order to change the indicator form 75% red to 75% blue? Ans. – 0.95 Ka for butyric acid is 2.0 × 10 –5. Calculate pH and hydroxyl ion concentration in 0.2M aqueous solution of sodium butyrate. Ans. 9 12. Calculate the pH of an aqueous solution of 1.0M ammonium formate assuming complete dissociation. (pK a of formic 11.

acid = 3.8 and pKb of ammonia = 4.8). Ans. 6.5 Calculate the concentrations of [Ag+] ion, [Br–] ion, [Cl–] ion, [Ag(NH3)2]+ ion, NH + 4 ion and hydroxide ion in a solution which results from shaking excess AgCl and AgBr with 0.02 M ammonia solution. Assume that no monoamine complex is formed. Given, KSP of AgCl = 1 × 10– 10, KSP of AgBr = 5 × 10– 13, Kb of NH3 = 1.8 × 10– 5 and Kf of [Ag(NH3)2]+ = 1 × 108. Ans. [Ag+] = 6 × 10– 8 M, [Br–] = 8 × 10– 6 M, [Cl–] = 1.7 × 10– 3 M, [Ag(NH3)2]+ = 1.7 × 10– 3 M, 13.

–4 – –4 [ NH + 4 ] = 6 × 10 M, [OH ] = 6 × 10 M

14. The vapour pressure of 0.01 molal solution of weak base BOH in water at 20°C is 17.356mm. Calculate K b for base. Aqueous tension at 20°C is 17.540mm. Assume molality and molarity same. Ans. 9.74 × 10–4 15. Predict whether or not AgCl will be precipitated from a solution which is 0.02 M in NaCl and 0.05 M in KAg(CN) 2.

(

)

– 19 2 – 10 2 Given K inst Ag( CN ) − 2 = 4.0 × 10 M and KSP(AgCl) = 2.8 × 10 M .

Ans. Precipitate will occur.

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Chapter-3 / Ionic Equilibrium

Exercise - VI IIT-JEE Problems 1.

500 mL of 0.2 M aqueous solution of acetic acid is mixed with 500 mL of 0.2 M HCl at 25°C. (i)

Calculate the degree of dissociation of acetic acid in the resulting solution and pH of the solution.

(ii)

If 6 g of NaOH is added to the above solution, determine the final pH (assuming there is no change in volume

on mixing, Ka of acetic acid is 1.75 × 10– 5 mol/L. Ans. α = 1.75 × 10– 2%; pH = 1 2.

The average concentration of SO2 in the atmosphere over a city on a certain day is 10 ppm, when the average temperature is 298 K. Given that the solubility of SO 2 in water at 298 K is 1.3653 mol/L and pK a of H2SO3 is 1.92, estimate the pH of rain on that day.

Ans. 0.5 3.

The solubility of Pb(OH)2 in water is 6.7 × 10– 6 M. Calculate the solubility of Pb(OH) 2 in a buffer solution of pH = 8.

Ans. 1.2 × 10– 3 M 4.

What will be the resultant pH when 200 mL of an aqueous solution of HCl (pH = 2.0) is mixed with 300 mL of an aqueous solution of NaOH (pH = 12.0)?

Ans. pH = 11.3 5.

Given, Ag(NH3 ) 2+  → Ag + + 2NH3 , Kc = 6.2 × 10– 8 and Ksp of AgCl = 1.8 × 10 – 10 at 298 K. If ammonia is added to a water solution containing excess of AgCl(s) only. Calculate the concentration of the complex in 1.0 M aqueous ammonia.

Ans. [Ag(NH3)2]+ = 0.054 M 6.

An acid type indicator, H/n differs in colour from its conjugate base (In –). The human eye is sensitive to colour differences only when the ratio [In –] / [H/n] is greater than 10 or smaller than 0.1. What should be the minimum change in the pH of the solution to observe a complete colour change? (Ka = 1.0 × 10– 5)

Ans. 4 to 6 7.

A sample of AgCl was treated with 5.00 mL of 1.5 M Na 2CO3 solution to give Ag 2CO3. The remaining solution contained 0.0026 g of Cl– ions per litre. Calculate the solubility product of AgCl. [K sp (Ag2CO3) = 8.2 × 10– 12]

Ans. 96 gm 8.

What is the pH of a 0.50 M aqueous NaCN solution? pKb of CN– = 4.70

Ans. 11.5 9.

Calculate the pH of an aqueous solution of 1.0, M ammonium formate assuming complete dissociation (pK a of formic acid = 3.8 and pKb of ammonia = 4.8) -56-

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Ans. 6.5 10.

An aqueous solution of a metal bromide MBr2 (0.05 M) is saturated with H2S. What is the minimum pH which MS will precipitate. Ksp for MS = 6.0 × 10– 21, conc. of saturated H2S = 0.1 M and K1 = 10– 7 and K2 = 1.3 × 10– 13, for H2S.

Ans. 0.983 11.

The pH of blood stream is maintained by a proper balance of H 2CO3 and NaHCO3 concentrations. What volume of 5 M NaHCO3 solution should be mixed with a 10 mL maintain a pH of 7.4? Ka for H2CO3 in blood is 7.8 × 10– 7

Ans. 78.36 12.

The solubility product (Ksp) of Ca(OH)2 at 25°C is 4.52 × 10– 5. A 500 mL of saturated solution of Ca(OH) 2 is mixed with equal volume of 0.4 M NaOH. How much Ca(OH)2 in milligrams is precipitated.

Ans. 7.58 13.

A 40.0 mL solution of a weak base, BOH is titrated with 0.1 N HCl solution. The pH of the solution is found to be 10.04 and 9.14 after the addition of 5.0 mL and 20.0 mL of the acid respectively. Find out the dissociation constnat of the base.

Ans. 1.81 × 10– 5 14.

The solubility product of Ag2C2O4 at 25°C is 1.29 × 10– 11 mol3 L– 3. A solution of K2C2O4 containing 0.1520 mole in 500 mL water is shaken at 25°C with excess of Ag2CO3 till the following equilibrium is reached: Ag2CO3 + K2C2O4  → Ag2C2O4 + K2CO3 At equilibrium, the solution contains 0.0358 mole of K2CO3. Assuming the degree of dissociation of K 2C2O4 and K2CO3 to be equal, calculate the solubility product of Ag2CO3.

Ans. 3.9 × 10– 12 15.

What is the pH of a 1.0 M solution of acetic acid? To what volume must one litre of this solution be diluted so that the pH of the resulting solution will be twice the original value? Given : Ka = 1.8 × 10–5

Ans. 2.37; V = 2.77 × 104 L 16.

Freshly precipitated aluminium and magnesium hydroxides are stirred vigorously in a buffer solution containing 0.25 mol/L of NH4Cl and 0.05 M of ammonium hydroxide. Calculate the concentration of aluminium and magnesium ions in solution. Kb [NH4OH] = 1.8 × 10– 5 Ksp [Mg(OH)2] = 8.9 × 10– 12 Ksp [Al(OH)3] = 6 × 10– 32

Ans. [Al+ 3] = 1.28 × 10– 15; [Mg2 +] = 0.686 17.

How many gram-mole of HCl will be required to prepare one litre of buffer solution (containing NaCN and HCl) of pH 8.5 using 0.01 gram formula weight of NaCN? KHCN = 4.1 × 10– 10 Sri Chaitanya IIT-JEE Head Office: 11, Central Market, West Punjabi Bagh , New Delhi-110 026. Phones: 25226309/10. Web: srichaitanya.edu.in, E-mail: [email protected]

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Chapter-3 / Ionic Equilibrium

Ans. 8.85 × 10– 3 18.

What is the pH of the solution when 0.20 moles of HCl is added to one litre of a solution containing (i)

1 M each of acetic acid and acetate ion

(ii)

0.1 M each of acetic acid and acetate ion

Assume the total volume is one litre. Ka for acetic acid = 1.8 × 10– 5. Ans. (i) 4.57; (ii) 1 19.

The solubility of Mg(OH) 2 in pure water is 9.57 × 10 – 3 g/L. Calculate its solubility (in g/L) in 0.02 M Mg(NO 3)2 solution.

Ans. 8.7 × 10– 4 gm/L 20.

The concentration of hydrogen ions is a 0.2 molar solution of formic acid is 6.4 × 10 – 3 mol/L. To this solution, sodium format is added so as to adjust the concentration of sodium formate to one mole per litre. What will be the pH of this solution? The dissociation constant of formic acid is 2.4 × 10 – 4 and the degree of dissociation of sodium formate is 0.75.

Ans. pH = 4.2 21.

A solution contains a mixture of Ag(0.10 M) and Hg 2 + (0.10 M) which are to be separated by selective precipitation. Calculate the maximum concentration of iodine ion at which one of them gets precipitated almost completely. What percentage of that metal ion is precipitated? Ksp : AgI = 8.5 × 10– 17, HgI2 = 2.5 × 10– 26

Ans. [I–] = 5 × 10– 13 M 22.

The dissociation constant of a weak acid HA is 4.9 × 10– 8. After making the necessary approximations, calculate (i)

percentage ionization

(ii)

pH

(iii) OH– concentration in a decimolar solution of the acid. Water has a pH of 7. Ans. (i) 0.07%; (ii) 4.15; (iii) 1.4 × 10– 10

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Answers Exercise - I True/False 1. 1. True

2.

True

Fill in the Blanks 1. 4.

3.

False

4.

1. pKw neutralization

Assertion and Reason 1. 1. (a) (c)

2.

(a)

False

5.

False

5.

2. decrease 3. anionic

strong

3.

(d)

4.

(b)

5.

Exercise - II Only One Option is correct Level - I 1. 1. (a) 2. 6. (a) 7. (c) 11. (a) 12. (d) 16. (d) 17. (d) 21. (d) 22. (a) Level - II 26. (d) 27. 31. (c) 32. 36. (b) 37. 41. (c) 42. 46. (c) 47. (a) Level - II 26.

(d) 8. 13. 18. 23.

3. (d) (a) (d) (c)

(a) 9. 14. 19. 24.

4. (d) (c) (a) (a)

(c) 10. 15. 20. 25.

5. (b) (c) (b) (a)

(c)

(c) (d) (d) (c) 48.

28. 33. 38. 43. (d)

(c) (c) (c) (d) 49.

29. 34. 39. 44. (a)

(b) (b) (d) (d) 50.

30. 35. 40. 45. (a)

(a) (d) (b) (b)

5.

Exercise - III More Than One Choice Correct 1. 1. (a, c, d) (b) 6. (a, b) 7. (b, c) 11. (c, d) 12. (b, c, d) 16. (a, b, c) 17. (a, c, d) 21. (a, b, d)

2.

(a, b)

3.

(c, d)

4.

(c)

8. 13. 18.

(b, c) (a, b, c) (a, b, c)

9. 14. 19.

(b, d) (a, b, c) (a, b, c)

10. 15. 20.

(a, b) (a, b) (b, c, d)

Exercise - IV Passage – 1 1. Passage – 1 1. (c)

2.

(c)

3.

(d)

4.

(b)

5.

(d)

Passage – 2 1. (d)

2.

(d)

3.

(a)

4.

(d)

5.

(b)

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Chapter-3 / Ionic Equilibrium

Passage – 3 1. (b)

2.

(c)

3.

(a)

4.

(a)

Passage – 4 1. (c)

2.

(b)

3.

(b)

4.

(a)

Passage – 5 1. (c)

2.

(b)

3.

(a)

4.

(b)

3.

(c)

Matching Type Questions 1.

(a)

2.

(c)

Passage – 2 1. Passage – 3 1. Passage – 4 1. Matching Type Questions 1.

Exercise - V 1.

K a ~−10−5

2. 3. 4. 5. 7. 9. 10. 11. 12. 13.

– 51.96 kJ/eq. 1.65 [OH–] = 3.57 × 10– 3 M, [H2en]2 + = 2.7 × 10– 8 M 9.66 × 10– 3 3.53 K = 15.7, x = 0.044 – 0.95 9 6.5 [Ag+] = 6 × 10– 8 M, [Br–] = 8 × 10– 6 M, [ N H 4+ ] = 6 × 10– 4 M, [OH–] = 6 × 10– 4 M 14. 9.74 × 10–4 15. Precipitate will occur.

[Cl–] = 1.7 × 10–

3

M,

[Ag(NH3)2]+ = 1.7 × 10–

3

M,

Exercise - VI 1.

α = 1.75 × 10 %; pH = 1

2.

0.5

3.

1.2 × 10– 3 M

4.

pH = 11.3

–2

+

5.

[Ag(NH3)2] = 0.054 M

6.

4 to 6

7.

96 gm

8.

11.5

9.

6.5

10.

0.983

11.

78.36

12.

7.58

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IIT/TYP/U2/Chemistry/07

13.

1.81 × 10– 5

14.

3.9 × 10– 12

15.

2.37; V = 2.77 × 104 L

16.

[Al+ 3] = 1.28 × 10– 15; [Mg2 +] = 0.686

17.

8.85 × 10– 3

18.

(i) 4.57; (ii) 1

20.

pH = 4.2

22.

(i) 0.07%; (ii) 4.15; (iii) 1.4 × 10– 10

19. 21.

–4

8.7 × 10 gm/L –

[I ] = 5 × 10

– 13

M

Exercise - V Subjective Type 1.

Exercise - VI IIT-JEE Level Problem

1.

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Chapter-1 3 / Ionic Equilibrium

Exercise - I General Type (Fill in the blanks/ True or False/ Assertion & Reason ) True/False 1. 2. 3. 4. 5.

True True False False False

Fill in the Blanks 1. 2. 3. 4. 5.

pKw decrease strong neutralization anionic

Assertion and Reason 1. 2. 3. 4. 5.

(a) (a) (d) (b) (c)

Exercise - II Level – I 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18.

(a) (d) (a) (c) (c) (a) (c) (d) (d) (b) (a) (d) (a) (c) (c) (d) (d) (d) -2-

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IIT/OYPTYP/U1U2/Chemistry/07

19. 20. 21. 22. 23. 24. 25.

(a) (b) (d) (a) (c) (a) (a)

Level - II 26. 27. 28. 29. 30. 31. 32. 33. 34. 35. 36. 37. 38. 39. 40. 41. 42. 43. 44. 45. 46. 47. 48. 49. 50.

(d) (c) (c) (b) (a) (c) (d) (c) (b) (d) (b) (d) (c) (d) (b) (c) (c) (d) (d) (b) (c) (a) (d) (a) (a)

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Chapter-1 3 / Ionic Equilibrium

Exercise - III 1.

(a, c, d)

2.

(a, b)

3.

(c, d)

4.

(c)

5.

(b)

6.

(a, b)

7.

(b, c)

8.

(b, c)

9.

(b, d)

10.

(a, b)

11.

(c, d)

12.

(b, c, d)

13.

(a, b, c)

14.

(a, b, c)

15.

(a, b)

16.

(a, b, c)

17.

(a, c, d)

18.

(a, b, c)

19.

(a, b, c)

20.

(b, c, d)

21.

(a, b, d)

Exercise - IV Passage – 1 1.

(c)

2.

(c)

3.

(d)

4.

(b)

5.

(d)

Passage – 2 1.

(d)

2.

(d)

3.

(a)

4.

(d) -4-

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IIT/OYPTYP/U1U2/Chemistry/07

5.

(b)

Passage – 3 1.

(b)

2.

(c)

3.

(a)

4.

(a)

Passage – 4 1.

(c)

2.

(b)

3.

(b)

4.

(a)

Passage – 5 1.

(c)

2.

(b)

3.

(a)

4.

(b)

Matching Type Questions 1. 2. 3.

(a) (c) (c)

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Chapter-1 3 / Ionic Equilibrium

Exercise - V

1.

K a ~−10−5

2.

– 51.96 kJ/eq. 3. 1.65 4. [OH–] = 3.57 × 10– 3 M, [H2en]2 + = 2.7 × 10– 8 M 5. 9.66 × 10– 3 7. 3.53 9. K = 15.7, x = 0.044 10. – 0.95 11. 9 12. 6.5 13. [Ag+] = 6 × 10– 8 M, [Br–] = 8 × 10– 6 M, [Cl–] = 1.7 × 10– 3 M, [Ag(NH3)2]+ = 1.7 × 10– 3 M, [ N H 4+ ] = 6 × 10– 4 M, [OH–] = 6 × 10– 4 M 14. 9.74 × 10–4 15. Precipitate will occur. Exercise - VI

1. 3.

α = 1.75 × 10– 2%; pH = 1

2.

0.5

–3

1.2 × 10 M 4. pH = 11.3 + 5. [Ag(NH3)2] = 0.054 M 6. 4 to 6 7. 96 gm 8. 11.5 9. 6.5 10. 0.983 11. 78.36 12. 7.58 –5 13. 1.81 × 10 14. 3.9 × 10– 12 15. 2.37; V = 2.77 × 104 L 16. [Al+ 3] = 1.28 × 10– 15; [Mg2 +] = 0.686 –3 17. 8.85 × 10 18. (i) 4.57; (ii) 1 –4 19. 8.7 × 10 gm/L 20. pH = 4.2 21. [I–] = 5 × 10– 13 M 22. (i) 0.07%; (ii) 4.15; (iii) 1.4 × 10– 10

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