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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

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Solution

(i) 0.003MHCl:

Since HCl is completely ionized,

Now,

Hence, the pH of the solution is

(ii) 0.005MNaOH:

Hence, the pH of the solution is

(iii) 0.002 HBr:

Hence, the pH of the solution is

(iv) .002 M KOH:

Hence, the pH of the solution is


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