What are the concentrations of OH Solution Ba(OH)2 -----> Ba2+ + 2OH- since in t

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Ba(OH)2 -----> Ba2+ + 2OH- since in the question there is no mention about the volume of solution so i am assuming it as 1 L solution now molarity = number of moles of solute/volume of solution in litres so number of moles of solute = molarity X volume of solution in litres so number of moles of Ba(OH)2 = 0.00049 X 1 = 0.00049 moles now 1 mole of Ba(OH)2 gives 2 moles of OH- so 0.00075 moles of Ba(OH)2 will give 0.00049 X 2 = .98 X 10^-3 moles of OH- molarity of OH- = .98 X 10^-3 / 1 = .98 X 10^-3 M also as Kw = [H+] [OH-] so 10^-14 = [H+] X 1.5 X 10^-3 [H+] = 10^-14/.98 X 10^-3 = 10.204 X 10^-12

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