Asked by susan
Calculate the pH of an aqueous solution
of 11.5 mg of KOH dissolved in 56 mL of
solution.
of 11.5 mg of KOH dissolved in 56 mL of
solution.
Answers
Answered by
DrBob222
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.
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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