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) 0.003M HCl : H2O + HCl$\leftrightarrow$H3O+ + Cl-
Since HCl is completely ionized, [H3O] = [HCl] ⇒ [H3O+] = 0.003
Now, pH = - log [H3O+] = - log(.003) = 2.52
Hence, the pH of the solution is 2.52.

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(ii) 0.005MNaOH : NaOH(aq) ↔ Na+(aq) + HO-(aq) = [HO-] = [NaOH] ⇒ [HO-] = .005
pOH = -log [HO-] = - log(.005)
pOH = 2.30
∴ pH = 14 - 2.30 = 11.70
Hence, the pH of the solution is 11.70.

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(iii) 0.002 HBr : HBr + H2O $\leftrightarrow$ H3O+ + Br- = [H3O+] = [HBr] ⇒ [H3O+] = .002
∴ pH = - log (0.002) = 2.69
Hence, the pH of the solution is 2.69.
(iv) 0.002 M KOH : KOH(aq) $\leftrightarrow$ K+(aq) + OH-(aq) = [OH-] = [KOH] ⇒[OH-] = .002
Now, pOH = - log [OH-] = 2.69
∴ pH = 14 - 2.69 = 11.31
Hence, the pH of the solution is 11.31