The Total solution for NCERT class 6-12
Assumingcomplete 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
(i) 0.003MHCl:
SinceHCl is completely ionized,
Now,
Hence, the pH of the solution is 2.52.
(ii) 0.005MNaOH:
Hence,the pH of the solution is 11.70.
(iii) 0.002 HBr:
Hence,the pH of the solution is 2.69.
(iv) 0.002 M KOH:
Hence,the pH of the solution is 11.31.