Friday
July 25, 2014

Homework Help: chem

Posted by chris on Thursday, March 22, 2007 at 11:09pm.

hey guys I posted this awhile ago im not sure if anyone has read it or whatever but here it is again find the PH of a .00580 M solution of the strong base KOH.

is it -log(.00580)= 2.24

pOH = -log[OH-]
pH + pOH = 14.00

Therefore,
pOH = -log(.00580)
pH + 2.24 = 14
pH = 11.76

Remember, a pH above 7 is a base, and below is an acid.

Hope this helps.






right. Your 2.24 is ok. But you must remember that is pOH.
Then pH + pOH = pKw = 14.

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