A 0.0133mol sample of Ba(OH)2 is dissolved in water to make up 1.5L of solution. What is the pH of the solution at 25.0∘C? ?
- BobbyLv 72 months ago
Ba(OH)2 = Ba2+ + 2OH-
moles of Ba(OH)2 = 0.0133mol
moles of OH- = 2* 0.0133mol = 0.0266 moles
in 1.5 L [OH-] = 0.0266 moles / 1.5L = 0.0177 M
pOH = 1.75
pH = 12.2
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