Question 48
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
(i) 0.003MHCl:
H2O + HCl ↔ H3O+ + Cl-
Since HCl is completely ionized,
[H3O+] = [ HCl]
⇒ [H3O+] = = 0.003
Now
pH = -log [H3O+] = -log (0.003)
= 2.52
Hence, the pH of the solution is 2.52.
(b) 0.005 M NaOH
NaOH(aq) ↔ Na+(aq) + HO-(aq)
[NaOH] = [ HO-]
⇒ [ HO-] = 0.05
pOH = -log[ HO-] = -log (0.05)
= 2.30
∴ pH = 14 - 2.30 = 11.70
Hence, the pH of the solution is 11.70.
(c) 0.002 M HBr
HBr + H2O ↔ H3O+ + Br-
[HBr] = [H3O+]
⇒ [H3O+] = 0.002
∴ pH = -log [H3O+] = -log (0.002)
= 2.69
Hence, the pH of the solution is 2.69.
(d) 0.002 M KOH
KOH(aq) ↔ K+(aq) + OH-(aq)
[OH-] = [KOH]
⇒ [OH-] = 0.002
Now pOH = -log[OH-] = -log (0.002)
= 2.69
∴ pH = 14-2.69 = 11.31
Hence, the pH of the solution is 11.31.
