Asked by Anonymous
Calculate the Ph of an aqueous solution that is 3.00% KOH, by mass, and has a density of 1.0242 g/ml.
Answers
Answered by
bobpursley
Assume the mass of the solution is 1024.2 grams, or one liter.
Then three percent (by mass) is KOH. How many moles is that?
Assuming total dissociation, then
[OH}=MolarityKOH= moles/1.0000liter
Then [H]=14-[OH]
and pH= -log[H]
Then three percent (by mass) is KOH. How many moles is that?
Assuming total dissociation, then
[OH}=MolarityKOH= moles/1.0000liter
Then [H]=14-[OH]
and pH= -log[H]
Answered by
GK
I think bobpursley meant that:
pOH = -log[OH-]
and
pH = 14 - pOH
Use these relationships to get the pH after finding the molarity of KOH = [OH-] as bobpursley suggested.
pOH = -log[OH-]
and
pH = 14 - pOH
Use these relationships to get the pH after finding the molarity of KOH = [OH-] as bobpursley suggested.
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