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.
                                                    There are no AI answers yet. The ability to request AI answers is coming soon!
                                            
                Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.