Asked by Kaleen
What mass of KOH is necessary to prepare 799.2 mL of a solution having a pH = 11.65?
Answers
Answered by
DrBob222
pH = 11.65. Convert to pOH from
pH + pOH = pKw = 14, then convert (OH^-) in moles/L = approximately 4E-3M but you should do it more accurately. You want only 799.2 mL; therefore,
M x L = moles = 4E-3 x 0.7992 = ?moles
Then grams = moles/molar mass. Solve for grams.
pH + pOH = pKw = 14, then convert (OH^-) in moles/L = approximately 4E-3M but you should do it more accurately. You want only 799.2 mL; therefore,
M x L = moles = 4E-3 x 0.7992 = ?moles
Then grams = moles/molar mass. Solve for grams.
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