find the PH of a .00580 M solution of the strong base KOH.
is it -log(.00580)=2.24

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asked by chris

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    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,

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