Assuming complete dissociation, what is the pH of a 3.68 mg/L Ba(OH)2 solution?

Question

          Assuming complete dissociation, what is the pH of a 3.68 mg/L Ba(OH)2 solution?
The prior answer of 11.87  was incorect.
          Assuming complete dissociation, what is the pH of a 3.68 mg/L Ba(OH)2 solution?
The prior answer of 11.87  was incorect.
     Assuming complete dissociation, what is the pH of a 3.68 mg/L Ba(OH)2 solution?
The prior answer of 11.87  was incorect.
Assuming complete dissociation, what is the pH of a 3.68 mg/L Ba(OH)2 solution?
The prior answer of 11.87  was incorect.

Explanation / Answer

convert mg/L to g/L (multiplying by "1g/1000mg")
then convert g/L to mol/L (by dividing by the molas mass of Ba(OH)2)

So : 3.68 mg/L /1000 = 0.00368 g/L

0.00368g/L / 171.34 g/mol = 2.147 x10^-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)

2.147x10^-5 M x 2 = 4.3x10^-5 M

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

pOH = -log4.3 10^-5 M] = 4.36

pH = 14 - 4.36 = 9.63


Therefore the pH is 9.63

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