Question

Assuming complete dissociation, what is the pH of a 3.06 mg/L Ba(OH)2 solution?

Answer 1

3.06 mg/L /1000 = 0.00306 g/L

0.00306 g/L / 171.34 g/mol = 1.78x10^-5 M

Ba(OH)2 -> Ba^2+ + 2OH-
Multiply the molarity by two (since when it dissociate, you get 2 OH- ions per Ba(OH)2)

1.78x10^-5 M x 2 = 3.56 x10^-5 M

pH= 14- pOH
pOH= -log [OH-]

pOH = -log[3.56x10^-5 M] = 4.44

pH = 14 - 4.44 = 9.72


Therefore the pH is 9.72