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Assuming complete dissocaition, what is the pH of 4.93 mg/L BaOH2 solution

  • Chem -

    There is no BaOH2. You must mean Ba(OH)2
    Ba(OH)2 ==> Ba^2+ + 2OH^-
    4.93 mg/L = 0.00493 g/L.
    mols/L = M = 0.00493/molar mass Ba(OH)2
    1 mol Ba(OH)2 gives 2 mols OH^-
    Therefore, M OH^- must be twice M Ba(OH)2.
    Then pOH = -log(OH^-) and
    pH + pOH = pKw = 14. You know pOH and pKw, solve for pH.

  • Chem -

    Ba(OH)2 is a strong electrolyte. Determine the concentration of each of the individual ions in a 0.800 M Ba(OH)2 solution.

    (OH-): ??? M

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