JEE Advanced (Subjective Type Questions): Equilibrium- 1 | Chapter-wise Tests for JEE Main & Advanced PDF Download

Q.1. A solution contains Na₂CO₃ and NaHCO₃. 10 ml of solution requires 2.5 ml of 0.1 M H₂SO₄ for neutralisation using phenolphthalein as an indicator. Methyl orange is then added when a further 2.5ml of 0.2M H₂SO₄ was required. Calculate the amount of Na₂CO₃ and NaHCO₃ in one litre of the solution.

Ans. 5.3 g/l; 4.2 g/l;

Solution. Phenolphthalein indicates half neutralization.

Na₂CO₃ + H⁺ → NaHCO₃ + Na⁺             ...(i)

Methyl orange indicates complete neutralisation

NaHCO₃ + H⁺ → Na⁺ + H₂O + CO₂        ...(ii)

∴ Volume of 0.1M H₂SO₄ required for complete neutralisation = 2 × 2.5 = 5.0 ml

0.1 M H₂SO₄ ≡ 0.2 NH₂SO₄

[For H₂SO₄ molarity = 2 × normality]

(∵ Mol. wt. of H₂SO₄ = 98, and eq. wt. of H₂SO₄ = 49)

∴ 0.2 M H₂SO₄ ≡ 0.4 NH₂SO₄

N₁ = normality of Na₂CO₃,  V₁ = volume of Na₂CO₃ = 10 ml,

N₂ = normality of H₂SO₄ = 0.2, V2 = volume of H₂SO₄ = 5.0 ml

∴ Eq. wt. of 

Strength of Na₂CO₃ = 53 × 0.1 = 5.3 g/l

[∴ strength = normality × Eq. wt]

For neutralization with methyl orange, volume of 0.2 M

H₂SO₄ used = 2.5 ml = 2.5 ml of 0.4 N H₂SO₄

= 5 ml of 0.2 N H₂SO₄ [∴ N₁V₁ = N₂V₂]

From 5 ml of 0.2 N H₂SO₄, 2.5 ml is used for neutralising NaHCO₃ formed during first half neutralization Na₂CO₃

∴ Volume of 0.2N H₂SO₄ used for neutralisation of NaHCO₃ present in original solution = 5.0 – 2.5 = 2.5 ml

∴ N₁V₁ = N₂V₂
where N₁ = Normality of NaHCO₃,

N₂ = Normality of H₂SO₄ = 0.2,

V1 = Volume of NaHCO₃ = 10 ml, V₂ = Volume of H₂SO₄ = 2.5 ml

N₁V₁ = N₂V₂ ⇒ N₁ × 10 = 0.2 × 2.5

Eq. wt. of NaHCO₃ = 84

∴ Strength of NaHCO₃ = 84 × 0.05 = 4.2 g/l

Q.2. How many moles of sodium propionate should be added to one litre of an aqueous solution containing 0.020 mole of propionic acid to obtain a buffer solution of pH 4.75? What will be pH if 0.010 mole of hydrogen chloride is dissolved in the above buffer solution. Compare the last pH value with the pH of 0.010 molar HCl solution. Dissociation constant of propionic acid, K_a at 25ºC = 1.34 × 10^–5.

Ans. 1.5072 × 10^–2 mol, 4.09, 2

Solution. Suppose the number of moles of sodium propionate = x

When 0.01 mole of HCl is added, there is (0.01 + 0.02) M of propionic acid and (0.015 – 0.010) M of propionate. Therefore

The pH of a 0.010 molar HCl solution = –log 10^–2 = 2

Q.3. One mole of nitrogen is mixed with three moles of hydrogen in a 4 litre container. If 0.25 per cent of nitrogen is converted to ammonia by the following reaction

calculate the equilibrium constant (Kc) in concentration units. What will be the value of Kc for the following equilibrium ?

Ans. 1.48 × 10^–5 l2 mol^–2, 3.82 × 10^–3 litre mol^–1

Solution. 

Now we know that 

Since 0.0025 and 0.0075 are very small, 1–0.0025 and 3–0.0075 may be taken as 1 and 3 respectively.

Substitute the various values

= 1.48 × 10^–5 litre² mol^–2

For the equilibrium,

Q.4. Twenty ml. of 0.2 M sodium hydroxide is added to 50 ml. of 0.2 M acetic acid to give 70 ml of the solution. What is the pH of this solution? Calculate the additional volume of 0.2 M NaOH required to make the pH of the solution 4.74. The ionization constant of acetic acid is 1.8 × 10^–5.

Ans. 4.5686, 4.86 ml

Solution. TIPS/Formulae :

(i) Find the moles of each species after reaction.

Given, NaOH  0.2 M, 20 ml;  CH₃COOH  0.2 M, 50 ml K_a = 1.8 × 10^–5

V of 0.2M NaOH required to make pH = 4.74 =?

From the chemical reaction

It is evident that 70 ml of the product will contain

(i) 30 ml of 0.2 M unused CH3COOH

[unused CH₃COOH = 50 – 20 = 30 ml]

(ii) 20 ml of CH₃COONa.

∴ No. of moles of CH₃COOH in solution

Similarly, No. of moles of CH₃COONa solution

Substituting the values of the various values

Calculation of the additional volume of 0.2 M NaOH required to make pH of solution 4.74.

NOTE THIS STEP : Let x ml. be the volume of additional 0.2 M NaOH added to make the pH of the solution 4.74. This will further neutralise x ml. of 0.2 M CH₃COOH and produce x ml. of 0.2 M sodium acetate. The resulting solution (70 + x) will now contain
(i) (30 – x) ml of 0.2 M acetic acid.
(ii) (20 + x) ml of 0.2 M sodium acetate.

Number of moles of acetic acid in (70 + x) ml. solution

Number of moles of CH₃COONa in (70 + x) ml. solution

or 1.001 x + x = 30 – 20.22; 2.011x = 9.78 or x = 4.86

Therefore, the additional volume of 0.2 M NaOH required to make the pH of the solution 4.74 is 4.86 ml.

Q.5. The dissociation constant of a weak acid HA is 4.9 × 10^–8. After making the necessary approximations, calculate (i) percentage ionization, (ii) pH and (iii) OH^– concentration in a decimolar solution of the acid. Water has a pH of 7.

Ans. (i) 7 × 10^–2%, (ii) 4.1549, (iii) 1.43 × 10^–10 mol/l

Solution. (i) From the dissociation of weak acid HA,

It α is the degree of ionization of the acid HA,

then [H⁺] =  0.1 α   [∵ the acid is decimolar]

[A^–]= 0.1 α; [HA] = 0.1 (1 – α)

Therefore,

(since acid is weak, 1 – α = 1)

∴  Percentage ionization = 100 × 7 × 10^–4 = 7 × 10–2%

(ii) Calculation of pH

[H⁺] = 0.1a = 0.1 × 7 ×  10^–4 mole/litre  [∵ λ = 7 x 10^-4]

= 7 × 10^–5 mole/litre

Now since pH = – log [H⁺] = – log [7 × 10^–5]                

= 5 – log 7 = 5 – 0.8451 = 4.1549

(iii) Concentration of OH^– in decimolar solution

[H⁺]  = 7 × 10^–5 mole per litre

Now, K_w = [H+] [OH–] or   1.0 × 10^–14 = 7 × 10^–5 × [OH^–]

 mole per litre

Q.6. A solution contains a mixture of Ag (0.10 M) and Hg₂⁺⁺ (0.10 M) which are to be separated by selective precipitation.
 Calculate the maximum concentration of iodide ion at which one of them gets precipitated almost completely. What percentage of that metal ion is precipitated?

[K_sp : AgI = 8.5 × 10^–17; Hg₂I₂ = 2.5 × 10^–26]

Ans. 5.0 × 10^–13 M, 99.83%

Solution. TIPS/Formulae :

For precipitation to occur ionic product > K_sp.
Mixture solution contains 0.1 M Ag+ and 0.1 Mg Hg₂²⁺.
K_sp of Hg₂I₂ = 2.5 × 10^–26 is much smaller than K_sp of AgI which is 8.5 × 10^–17.

[I^–] concentration needed to precipitate Hg₂I₂ is calculated as :

Similarly, [I^–] concentration needed to precipitate AgI is :

NOTE : Since [I^–] concentration needed to ppt. AgI is smaller than that needed to ppt. Hg₂I₂, AgI is completely precipitated first. AgI starts precipitation with [I^–] = 8.5 × 10^–16 M.
However, Hg₂I₂ starts precipitating with AgI only when molar concentration of I^– reaches 5.0 × 10^–13 M.

[Ag⁺] left when Hg₂I₂ begins to ppt. is given by

Thus % [Ag⁺] left unprecipitated

= 0.17%

Hence % Ag+ precipitated = 99.83%

Q.7. One mole of Cl₂ and 3 moles of PCl₅ are placed in a 100 litre vessel heated to 227ºC. The equilibrium pressure is 2.05 atmosphere. Assuming ideal behaviour, calculate the degree of dissociation for PCl₅ and K_p for the reaction :

Ans. 33.3%, 0.41

Solution. Let x be the degree of dissociation of PCl₅(g), then

∴ Total number of moles at equilibrium

= 3 (1 – x) + 3x + 1 + 3x = 3 (1 + x) + 1

Using the gas equation :  PV = nRT

Here,   P = 2.05 atm., V = 100 litres,  R = 0.082 atm/deg.,

T = 273 + 227 = 500 K

or 3 + 3x +  1 = 5 or 3x = 5 – 4 or

Hence perecentage dissociation of PCl₅ = 0.333 × 100 = 33.3%

Calculation of K_P for the reaction :

Substituting, x = 1/3 and P = 2.05 atm., we get

Q.8. Arrange the following in :

(i) increasing bond length : F₂,   N₂,   Cl₂,   O₂

(ii) increasing acid strength : HClO₃, HClO₄, HClO₂, HClO

(iii) increasing basicity : H₂O, OH^–, CH₃OH, CH₃O^–

(iv) Arrange the following oxides in the decreasing order of Bronsted basicity :

BaO, SO₃, CO₂, Cl₂O₇, B₂O₃

Ans. (i) N₂ < O₂ < F₂ < Cl₂,  
(ii) HClO < HClO₂ <  HClO₃ < HClO₄
(iii) H₂O < CH₃OH < OH– < CH₃O^–
(iv) BaO > B₂O₃ > CO₂ > SO₃ > Cl₂O₇

Solution. (i) N₂ < O₂ < F₂ < Cl₂

i.e., N ≡ N ;O <O ; F,F ; Cl,Cl

NOTE :
As the number of bonds increases the bond length decreases.
So N₂ < O₂ < halogens. Among F₂ and Cl₂ bond length of Cl₂ will be higher because of higher atomic radii.

(ii) Among oxyacids of the same element, acidic nature increases with its oxidation number, e.g.,

Weaker the base stronger is its conjugate acids

 (Decreasing acidic order of the conjugate bases.)

(iv) BaO > B₂O₃ > CO₂ > SO₃ > Cl₂O₇

Basicity increases with increase in oxidation state

(Ba = +2, B = +3, C = +4, S = +6, Cl = +7).

Q.9. The [H⁺] in 0.2 M solution of formic acid is 6.4 × 10^–3 mole litre^–1. To this solution sodium formate is added so as to adjust the concentration of sodium formate to one mole litre^–1. What wil be pH of this solution? K_a for HCOOH is 2.4 × 10^–4 and degree of dissociation of HCOONa is 0.75.

Ans. 4.19

Solution. TIPS/Formulae :

Calculation of concentration of HCOOH.

Here, c = 0.2 M; [H⁺] = 6.4 × 10^–3

NOTE : Thus the degree of dissociation of HCOOH is very low which on addition of sodium formate is further suppressed due to common ion effect.

Since the degree of dissociation is very low (3.2 × 10^–2), it can be neglected and hence [HCOOH] can be taken as 0.2 M.

Calculation of concentration of HCOO^–, [HCOO^–]

It can be obtained in the following manner :

For acidic buffer 

Q.10. The equilibrium constant of the reaction

at 100ºC is 50. If a one litre flask containing one mole of A₂ is connected to a two litre flask containing two mole of B₂, how many mole  of AB will be formed at 373ºC?

Ans. 1.886

Solution. 

On solving we get,   23x² – 75x + 50 = 0; x = 2.317  or  0.943

The value 2.317 is inadmissable because initial concentration of reactants is 2 moles and so x = 0.943

∴ Moles of AB formed = 2 × 0.943 = 1.886

Q.11. The solubility of Mg(OH)₂ in pure water is 9.57×10^–3 g/litre. Calculate its solubility (in g/litre) in 0.02 M Mg(NO₃)₂ solution.

Ans. 8.7 × 10^–4 g/litre

Solution. Solubility of Mg(OH)₂ in water

S = 9.57 × 10^–3 g/litre 

K_sp = (S) (2S)² = 4S3 = 4 (1.65 × 10^–4)³ = 1.8 × 10^–11 approx.

Calculation of solubility of Mg(OH)₂, say, x, in Mg(NO₃)₂ or [Mg²⁺] = x + 0.02; [OH^–] = x

K_sp = [Mg²⁺] [2OH^–]²  or 1.8 × 10^–11 = (x + 0.02) (2x)²

Neglecting x in comparison to 0.02 (common ion effect)

x = 1.5 × 10^–5 moles/litres  

= 1.5 × 58 × 10^–5 = 8.7 × 10^–4 g/litre.

Q.12. What is the pH of the solution when 0.20 mole of hydrochloric acid 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.
 K_a for acetic acid = 1.8 × 10^–5.

Ans. 4.5686, 1

Solution. (i) Amount of HCl added = 0.20 mole

[H⁺] = 0.2 g litre^–1

NOTE : Added H⁺ ions will combine with the acetate ions forming acetic acid with the result concentration of acetate ions will decrease while that of acetic acid will increase.

∴ Concentration of acetate ions after adding 0.20 mole of HCl.

[CH₃COO^–] = 1.0 – 0.2 = 0.8 mole

Similarly, concentration of acetic acid,

[CH₃COOH] = 1.0 + 0.2 = 1.2 mole

pH = 4.7447 + 0.3010 – 0.4771 = 4.5686

(ii) Amount of HCl added = 0.20 mole

Out of 0.2 mole of [H⁺] added, 0.1 mole will combine with 0.1 mole of CH₃COO^– forming 0 mole of CH₃COOH.

∴ Total concentration of acetic acid [CH₃COOH]

= 0.1 + 0.1 = 0.2 mole

In presence of [H⁺], CH₃COOH will not ionize.
Therefore, pH of the solution will be due to the presence of H⁺ of HCl, i.e. 0.2 – 0.1  =  0.1 mole HCl

pH = –log [H⁺] = –log [0.1] = 1

Q.13. At a certain temperature equilibrium constant (K_c) is 16 for the reaction.

If we take one mole each of all the four gases in a one litre container, what would be the equilibrium concentrations of NO(g) and NO₂(g)?

Ans. 1.6 moles, 0.4 moles

Solution. Initial concentration of each gas = 1 mole

Let the No. of moles of NO₂ reacted at equilibrium = x

∴ Thus the concentration of NO at equilibrium

= 1 + x = 1 + 0.6 = 1.6 moles

Concentration of NO₂ at equilibrium

= 1 – x = 1 – 0.6 = 0.4 moles

Q.14. N₂O₄ is 25% dissociated at 37ºC and one atmosph er e pressure. Calculate (i) K_p and (ii) the percentage dissociation at 0.1 atmosphere and 37ºC.

Ans. 0.266 atm, 64%

Solution. (i) 

where P is total pressure

  [∵ a = 0.25]

∴ Percentage dissociation = 63 %

Q.15. 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? 

K dissociation (HCN) = 4.1 × 10^–10.

Ans. 8.85 × 10–3 M

Solution. TIPS/Formulae :

If x moles of HCl are added then they will combine with

NaCN to form x moles of very weak acid HCN.

     [log 4.1 = 0.6128]

x = 8.85 × 10^–3 M = 8.85 × 10–3 moles of HC

Q.16. The equilibrium constant K_p of the reaction :

is 900 atm. at 800 K. A mixture containing SO₃ and O₂ having initial partial pressure of 1 and 2 atm. respectively is heated at constant volume to equilibrate. Calculate the partial pressure of each gas at 800 K.

Ans. = 0.0236 atm,  = 2.0118 atm,  = 0.9764 atm

Solution. NOTE : Since the reaction is carried out at constant volume, change in partial pressure of a species will be directly proportional to the change in its amount. Hence, we can write

Where 2 x is th e change in par tial pr essure of SO₃ at equilibrium.

Substituting the expression of partial pressure in the expression. For K_b, we get

Assuming x is very small as compared to 1

On usual calculations, x = 0.0118 atm

Thus  = 2x = 2 × 0.0118 atm = 0.0236 atm

 = 2 atm + x = 2 + 0.0118 = 2.0118 atm

 = 1 atm – 2x = 1 – 0.0236 = 0.9764 atm

Q.17. Freshly precipitated aluminium and magnesium hydroxides are stirred vigorously in a buffer solution containing 0.25 mole/l of ammonium chloride and 0.05 mole/l of ammonium hydroxide. Calculate the concentration of aluminium and magnesium ions in solution: 

K_b[NH₄OH] = 1.80 × 10^–5
 K_sp [Mg(OH)₂] = 6 × 10^–10
 K_sp [Al(OH)₃] = 6 × 10^–32

Ans. 46.29 mol ion/l, 1.286 × 10^–15 mol ion/l

Solution. TIPS/Formulae :

p(OH) for basic buffer 

We know that 

pOH = 5 – log 1.8 + log 5 = 5.6989 – 0.2552

– log [OH^–] = 5.4437; log [OH^–] = –5.4437

[OH^–] = 3.5999 × 10^–6            [Taking antilog]

 = 46.29 mole ion per litre

K_sp for Al(OH)₃ = [Al³⁺] [OH^–]3

6 × 10^–32 = [Al³⁺] (3.5999 × 10^–6)³

  = 1.286 × 10^–15 mol ion/l

Q.18. For the reaction : 

hydrogen gas is introduced into a five litre flask at 327ºC, containing 0.2 mole of CO(g) and a catalyst, until the pressure is 4.92 atm. At this point 0.1 mole of CH₃OH(g) is formed. Calculate the equilibrium constant, K_p and K_c.

Ans. 277.78 mol^–2 l², 0.1144 atm^–2

Solution. Let the total number of moles of all gases at equilibrium point = n

P = 4.92 atm.     V = 5l

R = 0.0821 atm. l mol^–1 K^–1    T = 273 + 327 = 600K

By applying the formula   PV = nRT

(i) Calculation of the number of moles of the individual gases at equilibrium point.

No. of moles of CH3OH formed = 0.1 (Given)

∴  No. of moles of CO (also) = 0.1

[∴  moles of CO = moles of CH₃OH formed]

Hence No. of moles of H₂ = 0.5 – (0.1 + 0.1) = 0.3

∴ Molar concentration of various species will be

(ii) Calculation of K_p. We know that

K_p = K_c × (RT)^Δn = 277.78 × (0.0821 × 600)^–2 (Δn = 1 – 3 = –2)

Q.19. What is the pH of 1.0 M solution of acetic acid? To what volume must one liter 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. 2.3724, 2.78 × 10⁴ litres

Solution. 

Thus pH = – log H⁺ = –log 4.24 × 10^–3 = 2.3724

Case II. pH after dilution  = 2 × original pH

= 2 × 2.3724 = 4.7448

Let conc. after dilution = c₁

and degree of dissociation = α₁

Since pH = – log [H⁺]

4.7448 = – log [H⁺]

[H⁺] = 1.8 × 10^–5 = c₁α₁ ∴ c₁α₁ = 1.8 × 10^–5

Dissociation constant

Substituting the value of α₁ in the following relation

Since the number of moles of CH₃COOH before and after dilution will be same

∴  Mole of  CH₃COOH before dilution = Mole of CH₃COOH after dilute

Q.20. The solubility product of Ag₂C₂O₄ at 25ºC is 1.29 × 10^–11 mol³ l^–3. A solution of K₂C₂O₄ containing 0.1520 mole in 500 ml water is shaken at 25ºC with excess of Ag₂CO₃ till the following equilibrium is reached :

At equilibrium the solution contains 0.0358 mole of K₂CO₃.
 Assuming the degree of dissociation of K₂C₂O₄ and K₂CO₃ to be equal, calculate the solubility product of Ag₂CO₃.

Ans. 3.794 × 10^–12 mol³ l^–3

Solution. 

Molar concentration of  left unreacted

[∵  500 ml = 0.5 L ]

Given that K_sp for Ag₂C₂O₄ = 1.29 × 10^–11 mol³ l^–3 at 25ºC