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Calculate the pH of an aqueous solution
of 11.5 mg of KOH dissolved in 56 mL of

  • chemistry -

    mols KOH = grams/molar mass
    Then (KOH) = mols KOH/L solution
    Then (OH^-) = (KOH) since KOH is a strong electrolyte and dissociates completely.
    I would then do pOH = -log(OH^-) and follow with pH + pOH = pKw = 14 and solve for pH.

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