Equilibrium

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans:  Option(iv)

The relationship between Kp and Kc is

Kp=Kc(RT)Δn

Where ?n = (number of moles of gaseous products) – (number of moles of gaseous reactants)

For the reaction,

NH4C1(s)?NH3(g)+HCl(g)

Δn=2–0=2

This is a Short Answer Type Questions as classified in NCERT Exemplar

Ans:  Δ_rH^?= ?_fH? [CaO(s)]+ ?_fH?[CO2(g)]− ?_fH?[ [CaCO₃(s)]

∴ ?_fH? =178.3kJ mol⁻¹.

The reaction is endothermic. Hence, according to Le Chatelier's principle, reaction will proceed in forward direction on increasing temperature.

This is a Short Answer Type Questions as classified in NCERT Exemplar

Ans: As, NH₃ is Lewis base while BF₃  is Lewis acid. Lewis electronic theory of acids and bases explains it. Hybridisation state: N in NH₃ is sp³ hybridised and Boron in BF₃ is sp² hybridised.

This is a Short Answer Type Questions as classified in NCERT Exemplar

Ans:

K_sp=3.2 * 10^-8

PbCl₂? Pb₂⁺ +2Cl^-

K_sp= (S) (2S)²

K_sp=4S³

s³=43.2 * 10^-8

s³=8 * 10^-9

s=2 * 10^-3

Molar   mass   of   PbCl₂=278g/mol   

By using Molarity formula,      

2 * 10^-3= 278 *  x0.1

x= $\frac{0.1}{278\mathrm{ }*\mathrm{ }2\mathrm{ }*\mathrm{ }{10}^{-3}}$

= $\frac{100}{278\mathrm{ }*\mathrm{ }2\mathrm{ }}$

x=0.18L   

This is a Short Answer Type Questions as classified in NCERT Exemplar

Ans: Let us assume S be the solubility of Al (OH)³.

K_sp= [Al³⁺] [OH^-]3= (S) (3S)³=27S⁴

S⁴=27K_sp=27 * 10²⁷ * 10^-11=1 * 10^-12

S=1*10⁻³ mol L^-1.

(i) Solubility of Al (OH)³ ? : Molar mass of Al (OH)³  is 78g. Hence, solubility of Al (OH)³ ? in gL^-1=1 * 10³ * 78gL^-1=78 * 10-3gL¹=7.8 * 10^-2gL^-1
(ii) pH of the solution: S=1* 10⁻³mol L^-1

OH=3S= 3*1* 10⁻³ = 3* 10⁻³

pOH=3−log³

pH=14−pOH= 11+log³= 11.4771.

This is a Short Answer Type Questions as classified in NCERT Exemplar

Ans: pH of HOCL = 2.85

But, - pH = log [H⁺]

 -2.85=log [H⁺]

= -3.15 = log [H⁺]

  [H⁺]=anti log3.15

  [H⁺]= 1.413* 10^-3 

or weak mono basic acid

[H⁺]= $\sqrt[]{K_a*}$  C

= $K_a$  = $\frac{{\left. [{\mathrm{H}}^{+}\right]}^2}{\mathrm{C}}$  = $\frac{{\left(\mathrm{}1.43*{10}^{-3}\right)}^2\mathrm{ }\mathrm{ }}{0.08}$

=24.957 $*\mathrm{}$ 10^-6

=2.4957 $*\mathrm{}$ 10^-5

This is a Short Answer Type Questions as classified in NCERT Exemplar

Ans: Given with the pH = 5 as, pH= - log  [ H⁺ ] 

  [ H⁺ ] =10^-5M

On 100 times dilution-

  [ H⁺ ] =10^-7M.

This is a Short Answer Type Questions as classified in NCERT Exemplar
Ans: Here concentration of water cannot be neglected since the solution is very dilute. pH will be less than 7.0. Hence, the total concentration is given as -

  [ H₃O⁺ ] =10^-8+10^-7M.