Question 48

Assuming complete dissociation, calculate the pH of the following solutions:

(a) 0.003 M HCl

(b) 0.005 M NaOH

(c) 0.002 M HBr

(d) 0.002 M KOH

Answer

(i) 0.003MHCl:

H2O   +   HCl  ↔  H3O+   +   Cl-

Since HCl is completely ionized,

[H3O+] = [ HCl]

⇒ [H3O+] = = 0.003

Now

pH = -log [H3O+]  = -log (0.003)

= 2.52

Hence, the pH of the solution is 2.52.

 

(b) 0.005 M NaOH

NaOH(aq)  Na+(aq) + HO-(aq)

[NaOH] =  [ HO-]

⇒  [ HO-]  = 0.05

pOH =  -log[ HO-]  = -log (0.05)

= 2.30

∴ pH  = 14 - 2.30 = 11.70

Hence, the pH of the solution is 11.70.

 

(c) 0.002 M HBr

HBr  + H2↔  H3O+ + Br-

[HBr] = [H3O+]

⇒  [H3O+] = 0.002

∴ pH = -log [H3O+]  = -log (0.002)

= 2.69

Hence, the pH of the solution is 2.69.

 

(d) 0.002 M KOH

KOH(aq)  ↔  K+(aq)  + OH-(aq)

[OH-]  =  [KOH]

⇒ [OH-] = 0.002

Now pOH  = -log[OH-]  = -log (0.002)

= 2.69

∴ pH = 14-2.69  =  11.31

Hence, the pH of the solution is 11.31.

Popular Questions of Class 11 Chemistry

Write a Comment: