Question

Assuming complete dissociation, calculate the pH of the given solution: 0.002 M HBr

Answer 1

On assuming complete dissociation, calculate the pH of the given solution:

0.002 M HBr0.002 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