Class 11th

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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 question as classified in NCERT Exemplar

NaH belongs to Saline hydrides and B₂H₆ and H₂O belong to molecular hydrides.

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 question as classified in NCERT Exemplar

These are formed by many d-block and/block elements. However, the metals of group 7, 8 and 9 do not form hydride. Unlike saline hydrides, they are almost always non- stoichiometric, being deficient in hydrogen. For example, LaH 2.87, YbH 2.55, TiH 1.5-1.8, ZrH 1.3-1.75. Dihydrogen forms molecular compounds with most of the jp-block elements.

Most familiar examples are CH₄, NH₃, H₂O and HF.

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.

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

Ans: At a given time the reaction quotient Q for the reaction will be given by the expression:

Q= $\frac{\left[\mathbf{H}2\right]\left[\mathbf{I}2\right]}{\left[\mathbf{H}\mathbf{I}\right]\mathrm{}}$  = $\frac{[1*10??*1*10??]}{{\left(2*\mathrm{}10^{-5}\right)}^2}$  = $\frac{1}{2}\mathrm{}$ = 0.25

 = 2.5*10⁻¹

As the value of the reaction quotient is greater than the value of K_c i.e. 1*10⁻⁴ the reaction will proceed in the reverse direction.