Q: Find out the solubility of Ni(OH)2 in 0.1 M NaOH. Given that the ionic product of Ni(OH)2 is 2—10−15

2×10−15=s(0.1)2 s=2×10−13 Solubility of Ni(OH)2=2×10−13M

Q: Which will make basic buffer?

(1) CH3COOH + NaOH →CH3COONa+H2O Before 25 mL 50 mL 0 ×0.1 M × 0.1 M = 2.5 mmol = 5 mmol After 0 2.5 mmol 2.5 mmol This is basic solution due to NaOH.This is not basic buffer.(2) CH3COOH+NaOH→CH3COONa+H2O Before 100 mL 100 mL 0 × 0.1 M × 0.1 M =10 mmol = 10mm olAfter 0 0 10 mmolHydrolysis of salt takes place.This is not basic buffer.(3) HCl + NH4OH → NH4Cl + H2O Before 100 mL 200 mL 0 × 0.1 M × 0.1 M = 10 mmol = 20 mmolAfter 0 10 mmol 10 mmol(4) HCl + NaOH → NH4Cl + H2O Before 100 mL 100 mL 0 × 0.1 M × 0.1 M 10 mmol 10 mmolAfter 0 0 10 mmol ⇒ Neutral solution

Q: The solubility of BaSO4 in water is 2.42×10−3gL−1 at 298 K. The value of its solubility product (Ksp) will be :(Given molar mass of BaSO4=233 g mol−1)

Solubility of BaSO4, s=2332.42×10−3( mol L−1) BaSO4( s)⇌sBa2+(aq)+sSO42−(aq) Ksp=[Ba2+][SO42−]=s2 =(1.04×10−5)2=1.08×10−10 mol2 L−2

Q: Which one of the following conditions will favour maximum formation of the product in the reaction,A2( g)+B2( g)⇌X2( g)ΔrH=−XkJ?

A2(g)+B2(g)⇌X2(g);ΔH=−xkJ On increasing pressure equilibrium shifts in a direction where pressure decreases i.e. forward direction.On decreasing temperature, equilibrium shifts in exothermic direction i.e., forward direction.So, high pressure and low temperature favours maximum formation of product.

Q: MY and NY3, two nearly insoluble salts, have the same Ksp values of 6.2 × 10−13 at room temperature. Which statement would be true in regard to MY and NY3?

MY insoluble salt Ksp=s2 MY(s)⇌M(aq)++Y(aq)− NY3 insoluble salt Ksp=4s3 NYs(s)⇌N(aq)++3Y(aq)− ∴S(MY)=6.2×10−13=7.8×10−7 Solubility values∴S(NY3)=(46.2×10−13)31=5.2×10−5 ∴S(MY)<S(NY3)

Q: Which of the following salts will give highest pH in water ?

Na2CO3 will give highest pH in water because it,is salt of strono base and weak acid.

Q: For a given exothermic reaction, Kp and Kp′ are the equilibrium constants at temperatures T1 and T2, respectively. Assuming that heat of reaction is constant in temperature range between T1 and T2, it is readily observed that:

For exothermic reactions, the value of the equilibrium constant decreases when the temperature increases.

Q: Identify the correct order of solubility in aqueous medium:

IA group elements sulphide are highly soluble, Zn+2 is IV group radical and Cu+2 is II group radical ∴Na2 S>ZnS>CuS

Question 1

Find out the solubility of Ni(OH)2​ in 0.1 M NaOH. Given that the ionic product of Ni(OH)2​ is 2—10−15

Accepted Answer

2×10−15=s(0.1)2 s=2×10−13 Solubility of Ni(OH)2​=2×10−13M

Question 2

pH of a saturated solution of Ca(OH)2​ is 9. The solubility product (Ksp​) of Ca(OH)2​ is :

Accepted Answer

Ca(OH)⇋Ca2++2OH− pH = 9 Hence pOH = 14 – 9 = 5										[OH−]=10−5M Hence [Ca2+]=210−5​ Thus Ksp​=[Ca2+][OH−]2 =(210−5​)(10−5)2 =0.5×10−15

Question 3

Conjugate base for Bronsted acids H2​O and HF are :

Accepted Answer

HF on loss of H⊕ ion becomes F⊖  is the conjugate base of HFExample :HF  +     H2​O   ⇋   		F⊖     +  		H3​O⊕ Acid 		 Base	        Conjugate		Conjugate										base				acid

Question 4

Which will make basic buffer?

Accepted Answer

(1)		CH3​COOH  +  NaOH    →CH3​COONa+H2​O Before			25 mL				50 mL																								0					×0.1 M				× 0.1 M				= 2.5 mmol 			= 5 mmol	After		 		0 						2.5 mmol 			2.5 mmol	This is basic solution due to NaOH.This is not basic buffer.(2)		CH3​COOH+NaOH→CH3​COONa+H2​O Before 		100 mL 		100 mL																					0					× 0.1 M 		× 0.1 M				=10 mmol 		= 10mm olAfter 				0 						0 								10 mmolHydrolysis of salt takes place.This is not basic buffer.(3)			HCl 		+		NH4​OH 		→ 		NH4​Cl 	+  H2​O  Before 	100 mL 				200 mL																					0				× 0.1 M 				× 0.1 M				= 10 mmol 			= 20 mmolAfter 			0 						10 mmol 						10 mmol(4)			HCl		+			NaOH 		→ 		NH4​Cl 	+	H2​O Before 	100 mL 				100 mL																					0				× 0.1 M 				× 0.1 M				10 mmol 				10 mmolAfter 			0 							0 								10 mmol	 ⇒  Neutral solution

Question 5

The solubility of BaSO4​ in water is 2.42×10−3gL−1 at 298 K. The value of its solubility product (Ksp​) will be :(Given molar mass of  BaSO4​=233 g mol−1)

Accepted Answer

Solubility of BaSO4​, s=2332.42×10−3​( mol L−1) BaSO4​( s)⇌sBa2+​(aq)+sSO42−​​(aq) Ksp​=[Ba2+][SO42−​]=s2 ​=(1.04×10−5)2=1.08×10−10 mol2 L−2​

Question 6

Which one of the following conditions will favour maximum formation of the product in the reaction,A2​( g)+B2​( g)⇌X2​( g)Δr​H=−XkJ?

Accepted Answer

A2​(g)+B2​(g)⇌X2​(g);ΔH=−xkJ On increasing pressure equilibrium shifts in a direction where pressure decreases i.e. forward direction.On decreasing temperature, equilibrium shifts in exothermic direction i.e., forward direction.So, high pressure and low temperature favours maximum formation of product.

Question 7

MY and NY3​, two nearly insoluble salts, have the same Ksp​ values of 6.2 × 10−13 at room temperature. Which statement would be true in regard to MY and NY3​?

Accepted Answer

MY insoluble salt Ksp​=s2 MY(s)​⇌M(aq)+​+Y(aq)−​ NY3​ insoluble salt Ksp​=4s3 NYs(s)​⇌N(aq)+​+3Y(aq)−​ ∴S(MY)​=6.2×10−13​=7.8×10−7 Solubility values∴S(NY3​)​=(46.2×10−13​)31​=5.2×10−5 ∴S(MY)​<S(NY3​)​

Question 8

Which of the following salts will give highest pH in water ?

Accepted Answer

Na2​CO3​  will give highest pH in water because it,is salt of strono base and weak acid.

Question 9

For a given exothermic reaction, Kp​ and Kp′​  are the equilibrium constants at temperatures T1​ and T2​,  respectively. Assuming that heat of reaction is constant in temperature range between T1​ and T2​,  it is readily observed that:

Accepted Answer

For exothermic reactions, the value of the equilibrium constant decreases when the temperature increases.

Question 10

Identify the correct order of solubility in aqueous medium:

Accepted Answer

IA group elements sulphide are highly soluble, Zn+2 is IV group radical and Cu+2  is II group radical ∴Na2​ S>ZnS>CuS

Question 11

Which of these is least likely to act as a Lewis base?

Accepted Answer

BF3​  is electron deficient so act as Lewis acid.

Question 12

Concentration of theAg+ ions in a saturated solution of Ag2​C2​O4​  is2.2×10−4 mol L−1 . Solubility product of Ag2​C2​O4​  is:

Accepted Answer

Ag2​C2​O4​( s)⇌2Ag+(aq)+C2​O42−​(aq) ​KSP​=[Ag+]2[C2​O42−​][Ag+]=2.2×10−4M∴[C2​O42−​]=22.2×10−4​M=1.1×10−4M∴KSP​=(2.2×10−4)2(1.1×10−4)=5.324×10−12​

Question 13

The equilibrium constants of the following are.N2​+3H2​ N2​+O2​H2​+21​O2​⟶H2​O​⇌2NH3​⇌2NOK3​​ K1​K2​​ The equilibrium constant (K) of the reaction:2NH3​+25​O2​⇌k​2NO+3H2​O , Will be

Accepted Answer

2NH3​⇌N2​+3H2​,1/K1​
 N2​+O2​⇌2NO,K2​
3H2​+23​O2​⟶3H2​O,(K3​)3
Adding All three equation
k=K1​K2​K33​​

Question 14

If the equilibrium constant for N2​(g)+O2​(g)⇌2NO(g) is K, the the equilibrium constant for 21​N2​(g)+21​O2​(g)⇌NO(g) will be :

Accepted Answer

N2​(g)+O2​(g)⇌2NO(g);K 21​N2​(g)+21​O2​(g)⇌NO(g);K′ K=[N2​][O2​][NO]2​ K′=[N2​]1/2[O2​]1/2NO​ ∴K′=K​

Question 15

Which one of the following pairs of solution is not an acidic buffer?

Accepted Answer

Strong acid with its salt cannot form buffer solution. HClO4​ and NaClO4​  cannot act as an acidic buffer.

Question 16

Aqueous solution of which of the following compounds is the best conductor of electric current?

Accepted Answer

Aqueous solution of HCl is the best conductor of electric current because HCl is strong acid, so it dissociates completely into ions.

Question 17

What is the pH of the resulting solution when equal volumes of 0.1 m NaOH and 0.01 M HCl are mixed?

Accepted Answer

N1​V1​−N2​V2​= N.V. 0.1 × 1 − 0.01 × 1 = N × 2 [OH−]=NR​=20.09​ = 0.045 NpOH = −log (0.045) = 1.35∴ pH = 14 − pOH = 14 − 1.35 = 12.65

Question 18

A liquid is in equilibrium with its vapours in a sealed container at a fixed temperature. The volume of the container is suddenly increased, (i) What is the initial effect of the change on the vapour pressure? (ii) How do the rates of evaporation and condensation change initially? (iii) What happens when equilibrium is restored finally and what will be the final vapour pressure?

Accepted Answer

(i) On increasing the volume of the container, the vapour pressure will initially decrease because the same amount of vapours are now distributed over a larger space.(ii) On increasing the volume of the container, the rate of evaporation will increase initially because now more space is available. Since the amount of the vapours per unit volume decrease on increasing the volume, therefore, the rate of condensation will decrease initially.(iii) Finally, equilibrium will be restored when the rates of the forward and backward processes become equal. However, the vapour pressure will remain unchanged because it depends upon the temperature and not upon the volume of the container.

Question 19

What is Kc​ for the following equilibrium when the equilibrium concentration of each substance is: [SO2​]=0.60M,∣O2​∣=0.82M and [SO3​]=1.90M ?2SO2​(g)+O2​(g)⇌2SO3​(g)

Accepted Answer

Kc​=[SO2​]2[O2​][SO3​]2​Kc​=(0.60)2×0.82(1.90)2​=12.229 L/mol

Question 20

At a certain temperature and total pressure of 105 Pa, iodine vapour contains 40% by volume of I atoms.I2​(g)⇌2I(g) Calculate Kp​ for the equilibrium.

Accepted Answer

The partial pressure of iodine atoms is P=100105×40​=0.4×106 Pa.The parital ressure of I2​ molecules is P′=100105×60​=0.6×105 Pa.The value of the equilibium constant is KP​=P′P2​=0.6×105(0.4×105)2​=2.67×104.

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Solution for Class 11, Chemistry, Equilibrium

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