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OpenStudy (anonymous):
calculate the poh of an aqueous solution of .014M NaOH at 25oC
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OpenStudy (anonymous):
@TheSmartOne
OpenStudy (anonymous):
@Cuanchi
OpenStudy (anonymous):
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
OpenStudy (anonymous):
I got 1.85
OpenStudy (anonymous):
@nmagnuson Is that what you get?
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thatonenerd:
Use Hvap of water = 40.7 kJ/mol, c of water = 4.184 J/g.u00b0C, and molar mass of HO = 18.
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