Question

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 1

(i) 0.003MHCl:

Since HCl 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.