Jee 2014 Booklet3 Hwt Solutions Ionic Equilibrium

Vidyamandir Classes

Ionic Equilibrium 1.(B)

HWT - 1

C1 12  C2 22 0.1  104  2.5  102  22 4  104  22

2.(A)

Acid is H+ donor.

3.(A)

[H  ]  K a C



2  2  102



K a1 C1  K a 2 C2

1.8  104  103  1.8  105  C2



C2  102

  A x By   xA  yB

4.(C) t=0 t = teq

C C(1  )

k eq 



0 xC

0 yC

 x c  x  y c   y c 1   

 K eq     cx  y  1 x x y y 

1   1 1

x  y   

5.(C)

K a  K a1  K a 2 = 5  1015

8.(B)

Lower the value of Kb ; weaker is the base.

9.(A)

 H    105  K C a  

K a  1010



Ionic Equilibrium 1.(D)

pH is nearest to 7 (will be slightly more than 7)

2.(B)

On increasing temp, pH decreases.

Buffer-1

3.(B)

3  pK a  log 

7.(D)

salt  acid 

Buffer-2

salt   acid  

HWT - 2

5  pK a  log

salt  acid 

1 10

salt  acid 

 10

Milli moles of [H+] = 2 Milli moles of [OH  ]  3 pH  14  log [OH  ]  14  log

1 500

= 14  log 500  11.3 8.(B)

pH   log102  2

VMC/Ionic Equilibrium

6

HWT-Solutions/Chemistry

Vidyamandir Classes

Ionic Equilibrium 1.(C) 5.(C)

HWT - 3

50  103  2 pH  14  pOH  14  log OH    14  log = 11   100 1 pOH   pK b  log C  2

4.(C)

pH   log  H     log 0.005  2  2  

Kb 2  105   0.14142   0.05 C 103 So, we can not use the standard formula. Solve for  . 

6.(A)

Kb 

C 2 and then use pOH   log    1 

pH 

1  pKa  log C  2

2

1  pKa  1 2

K a  103



7.(B)

Ka  0.1 C Combination of strong acid and salt is not a buffer

10.(C)

Since KSP of CuCl is very high compared to AgCl.



8.(B)

pK a   log K a

Cl    Cl    Cl       from AgCl  from CuCl = 103  Cl    from CuCl For AgCl 1.6  1010   Ag    103  



 Ag    1.6  107 Hence x = 7  

Ionic Equilibrium 1.(C)

HWT - 4

3    Cr(OH)3(s)   Cr(aq)  3OH (aq) s

3s

1/ 4

KSP  27s4 3.(B)

K  s   SP   27 



For precipitation IP > KSP [Zn+2] [S2] > KSP

[Cu 2 ] [S2 ]  KSP KSP



1.2  1010  6  1010 0.2

7.(D)

Solubility =

8.(C)

2    Hg 2 I 2 s    Hg 2aq  2Iaq

[Cl  ]

KSP  [Hg 22 ] [I]2 KSP BaCO3

5.1  109

9.(B)

[Ba 2 ] 

10.(D)

3    Fe(OH)3(s)   Fe(aq)  3OH (aq)

[CO32 ]



104

= 5.1  105 M

KSP  [Fe3 ] [OH  ]3 = 27X4

VMC/Ionic Equilibrium

7

HWT-Solutions/Chemistry

Vidyamandir Classes

Ionic Equilibrium 1.(B)

HWT - 5

 2   Ag 2S(s)   2Ag(aq)  S(aq)

KSP  [Ag  ]2 [S2 ] 1017  [Ag  ]2  101



[Ag  ]  108  2s



3.(C)

Adding CH3COONa to CH3COOH will suppress  and increase pH.

4.(A)

2    Mg  OH 2(s)   Mg(aq)  2OH (aq)

KSP  1.96  1011  4s3



 1.96  s  1011   4 

s  5  109 .

13

Now pH   14  log [OH  ]  14  log (2s) . 5.(C)

For precipitation

I.P. > KSP

Ionic Equilibrium

HWT - 7

1.(D)

Refer Illustration - 12 (Module)

2.(C)

pH  7  1/ 2  pK a  pK b   7  1/ 2  4.8  4.78  = 7.01

3.(D) 5.(B) 7.(C)

 [salt]  1   pH  14   pK b  log  = 14   5  log  =8 [base] 6.1     Na2CO3 on hydrolysis will give NaOH and H2CO3. Resultant solution will be basic. [salt] Since [salt] = [acid] pH  pK a  log [acid] pOH  14  pK a  9.5

Hence pH = pKa 10.(C)

K a  CH 2



h

Kh C

Ionic Equilibrium

HWT - 8

1.(C)

   CO32  2H 2 O   H 2 CO3  2OH

3.(A)

pH  p Kin  log

5.(B)

   Mg 2  2H 2 O   Ong(OH)2  2H

6.(C)

K Kb Mixture of CH3COOH & CH3COONa is a buffer.  [salt]  pH  14   pK b  log  = 14   3.3  log 4  = 10.1 [base]  

[In  ] = pKin + 1. [HIn]

[In  ] [HIn]

2.(D)

pH  pKin  log

4.(A)

Fe3 undergoes hydrolysis.

Kh 

7.(B) 9.(A)

Kh 

K  109  0.1 h 2 Ka

10.(A)

Kh 

K  109 Ka

109  103 h 2

VMC/Ionic Equilibrium







(H  )  8  1011

h  0.01%

h 2  106



h 2  106

8

HWT-Solutions/Chemistry