Asked by AMANDIP
What mass of KOH is necessary to prepare 799.9 mL of a solution having a pH = 11.53?
Answers
Answered by
DrBob22
KOH is a strong base in aqueous solution; therefore, the (KOH) = (OH%-).
For pH = 11.53, use pH + pOH = pKw = 14 to solve for pOH, then from
pOH = -log (OH^-), solve for OH^-. That will give you the molarity of the KOH you want for that pH.
Then M = moles/L. Substitute L and solve for moles.
Then moles = grams/molar mass, solve for grams.
For pH = 11.53, use pH + pOH = pKw = 14 to solve for pOH, then from
pOH = -log (OH^-), solve for OH^-. That will give you the molarity of the KOH you want for that pH.
Then M = moles/L. Substitute L and solve for moles.
Then moles = grams/molar mass, solve for grams.
Answered by
Yeki
112. 57
Answered by
Yeki
105. 84
Answered by
Yeki
4. 5×10^-13
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