JEE Advanced (Single Correct MCQs): Equilibrium Chapter MCQs & solutions

NOTE : Acidic buffer is mixture of weak acid and its salt with common anion.

(a) CH₃COOH + CH₃COONH₄ is acidic buffer.
(b) NH₄Cl + NH₄OH is basic buffer.
(c) H₂SO₄ + Na₂SO₄ is not buffer because both the compounds are strong electrolytes.
(d) NaCl + NaOH is not buffer solution because both compounds are strong electrolytes.

TIPS/Formulae :
(i) pH of acid cannot be more than 7.
(ii) While cal culating pH in such case, consider contribution [H⁺]  from water also.
Molar conc. of HCl = 10^–8. (given)
∴ pH = 8. But this cannot be possible as pH of an acidic solution can not be more than 7. So we have to consider [H⁺] coming from H₂O.

Let x be the conc. of [H+] from H₂O

[For quadratic equation

∴  Total [H⁺] = 10^–8 + 9.5 × 10^–8
= 10.5 × 10^–8 or  pH                  
= – log (10.5 × 10^–8) = 6.98
It is between 6 and 7.

TIPS/Formulae :
(i) According is Le-Chateliers principle, exothermic, reactions are favoured at low temperature.
(ii) According to Le-Chateliers principle, the reaction in which n < 0, are favoured at high pressure.

∵ It is exothermic reaction
∴Yield of SO3 is maximum at low temperature
n = 2 – 3= – 1 or n < 0
∴ Yield of SO₃ is maximum at high pressure.

Only temperature affects the equilibrium constant. Since here ΔH = 2 – 2 = 0 , so there is no change in K_p when total pressure changes.

TIPS/Formulae :

(i) Lower the oxidation state of central atom, weaker will be oxy acid.
(ii) Weaker the acid, stronger will be its conjugate base.
Oxidation state of Cl in HClO is + 1, in HClO₂ is + 3, in HClO₃ is + 5, and in HClO₄ is + 7
∴ HClO is the weakest acid and so its conjugate base ClO^– is the strongest Bronsted base.

For pure water,  [H₃O⁺] = [OH^–]
⇒ K_w = 10^–6 × 10–6 = 10^–12

TIPS/Formulae :
For precipitation to occur ionic product > solubility products

Ionic product of C aF₂ = [Ca²⁺] [F^-]² Calculate I.P. in each case

(a) I.P. of CaF₂ = (10^–4) × (10^–4)²  = 10^–12
(b) I.P. of CaF₂ = (10^–2) × (10^–3)²  = 10^–8
(c) I.P. of CaF₂  =  (10^–5) × (10^–3)²  = 10^–11
(d) I.P. of CaF₂ =  (10^–3) × (10^–5)2  = 10^–13

∵ I.P > solubility in choice (b) only..
∴ ppt of CaF₂ is obtained in case of choice (b) only..

Vapours and liquid are at the same temperature.

Statement (a) is correct and the rest statements are wrong. K_p depends only on temperature hence at constant temp. K_p will not change.

For a basic buffer, 

⇒ pH = 4

TIPS/Formulae :

The equilibrium constant for the nuetralization of a weak acid with a strong base is given by

As all the reactants and products are present in aqueous form in (d) so  it is a reversible reaction. In others either solid or gas is generated which is insoluble or volatile and hence makes the reaction unidirectional.

TIPS/Formulae :

The pH of the solution at the equivalence point will be greater than 7 due to salt hydrolysis. So an indicator giving colour in basic medium will be suitable.
Phenolphthalein is a good indicator if the base is strong because strong base immediately changes the pH at end point.

Base + H⁺ → (conjugate acid) -
(conjugate acid)

NOTE : Electron acceptors or elements having incomplete octet are Lewis acids.
(i) BF₃ (B has 6 e^– in valance shell), AlCl₃ (Al has 6 electrons in valance shell), BeCl₂ (Be has 4 e^– in valance shell) are electron defecient compounds and hence Lewis acids.
(ii) SnCl₄ has complete octet so it is Lewis base.

CaO, CaCO₃ and Ca(OH)₂ dissolve in CH₃COOH due to formation of  (CH₃COO)2Ca. But CaC₂O₄ does not dissolve as CH₃COO^– is a stronger conjugate base than 

(a) is a neutral solution due to both cationic and anionic hydrolysis (Ka = Kb = 1.8×10–5); (b) is acidic solution due to cationic hydrolysis; (c) is acidic solution due to cationic hydrolysis; (d) is basic solution due to anionic hydrolysis.
Alternate solution.

For a precipitation to occur Solubility product < Ionic product
Given  K_sp = 1.8 × 10^–10
Calculating ionic products in each

which is greater than K_sp (1.8 × 10^–10).

TIPS/Formulae :
In acidic medium weak acids are unionized due to common ion effect and they are completely ionised in alkaline medium.
Aspirin (or acetyl salicylic acid) is unionised in stomach (where pH is 2-3 ) and is completely ionised in small intestine (when pH is 8)

TIPS/Formulae :
(i) Higher the electronegatively of central atom higher will be the acidic strength.
(ii) In case of same atom higher the value of oxidation state of the metal, higher will be its acidic strength.
The  electronegativity of Cl > S.
Oxidation no. of Cl in ClO₃(OH) = + 7 Oxidation no. of  Cl in ClO₂(OH) = + 5 Oxidation no. of S in SO(OH)₂ = + 4 Oxidation no. of  S in SO₂(OH)₂ = + 6

∴ ClO₃(OH) is the strongest acid.

NOTE : In case of alkaline earth hydroxides solubility increases on moving down the group.
Be(OH)₂ has lowest solubility and hence lowest solubility product. [Be at tip of the group]

Due to absence of hydrolysation of FeCl₃ backward reaction will not take place.

Since HCl is stronger than CH₃COOH hence acts as acid. On the other hand Cl^– is a stronger base than   and is the conjugate base of HCl.

(a) It is not correct answer because 100 ml M/10 HCl will completely neutralise 100 ml M/10 NaOH and the solution will be neutral.
(b) After neutralisation resultant solution will be acidic due to presence of excess of HCl.
(c) After netralisation resultant solution will be basic due to presence of excess of NaOH.
(d) M. eq. of HCl = 75 N/5 = 15 Meq

TIPS/Formulae :

Salts of weak base and strong acid get hydrolysed in aqueous solution forming an acidic solution.

Among  oxyacids of the same type formed by different elements, acidic nature increases with increasing electronegativity. In general, the strength of oxyacids decreases as we go down the family in the periodic table.

HOCl  (I)     >    HOBr  (II)    >    HOI (III)
[In halogen groups Cl is above Br and I]

The characteristics of the given solutions are:
NaCl – neutral solution
NH₄Cl – slightly acidic due to the following reaction

NaCN – slightly alkaline due to the following reaction

HCl – highly acidic
The pH of the solution will follow the order highly acidic < slightly acidic < neutral < slightly alkaline i.e.
HCl < NH₄Cl < NaCl < NaCN

The given reaction will be exothermic in nature due to the formation of three  X - Y bonds from the gaseous atoms. The reaction is also accompanied with the decrease in the gaseous species (i.e. Dn is negative).
Hence, the reaction will be affected by both temperature and pressure. The use of catalyst does not affect the equilibrium concentrations of the species in the chemical reaction.

K_b = K_c .( RT)^Δn

(R in L.atm.K^–1 mole^–1).