Question
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.
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.