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calculate the poh of an aqueous solution of .014M NaOH at 25oC
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@TheSmartOne
@Cuanchi
NaOH is a strong base so all you need to do is \[pOH=-\log[Base]\] Please note that this does NOT apply to weak bases, only strong bases. For weak bases there are a few extra steps involved. Anyways, just take the -log of the concentration of NaOH to find the pOH
I got 1.85
@nmagnuson Is that what you get?
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