# Chemistry Class 11 NCERT Solutions: Chapter 7 Equilibrium Part 19

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# Chemistry Class 11 NCERT Solutions Chapter 7 Equilibrium Part 19

Q: 51. The pH of 0.005M codeine solution is 9.95. Calculate its ionization constant and .

Answer:

C = 0.005

pH = 9.95

pOH = 4.05

pH = -log (4.105)

Thus,

= 5.80

Q: 52. What is the pH of 0.001 M aniline solution? The ionization constant of aniline can be taken from Table 7.7. Calculate the degree of ionization of aniline in the solution. Also calculate the ionization constant of the conjugate acid of aniline

Answer:

Then,

Now,

∴

Thus, the ionization constant of the conjugate acid of aniline is

Q: 53. Calculate the degree of ionization of 0.05M acetic acid if its value is 4.74. How is the degree of dissociation affected when its solution also contains (A) (B) ?

Answer:

When HCl is added to the solution, the concentration of ions will increase. Therefore, the equilibrium will shift in the backward direction i.e., dissociation of acetic acid will decrease.

Case I: When 0.01 M HCl is taken

Let x be the amount of acetic acid dissociated after the addition of HCl.

As the dissociation of a very small amount of acetic acid will take place, the value i.e., 0.05-x and 0.01+ x can be taken as 0.05 and 0.01 respectively.

Now,

Case II: When 0.1 M HCl is taken.

Let the amount of acetic acid dissociated in this case be X. As we have done in the first case, the concentrations of various species involved in the reaction are:

∴

Now,