(a) 0.003 M HCl

HCl + H₂O ↔ H⁺ + Cl^-

Since HCl is completely  dissociate

[H⁺] = HCl ,

 Then, [H⁺] = 0.003

pH = -log [H⁺] = -log [0.003] = 2.52

(b) 0.005 M NaOH

NaOH ↔ Na⁺ + OH^-     [OH^-] = NaOH,

[OH^-] = 0.005

pOH = -log [OH^-] = -log [0.005] = 2.30

Therefore, pH = 14 - 2.30 = 11.70

(c) 0.002 M HBr

Hbr ↔ H⁺ + Br^-

[H⁺] = HBr,

[H⁺] = 0.002

pH = -log [H⁺] = -log[0.002] = 2.69

(d) 0.002 M KOH

KOH ↔ K⁺ + OH^-

[OH^-] = [KOH]

[OH^-] = 0.002

pOH = - log [OH^-] =-log[0.002] = 2.69

pH= 14-2.69 = 11.70