chemistry

What is the pOH of a 0.032 M Ca(OH)2 (strong base) solution?

i solved it by doing

14=-log (0.064)
14-1.19= 12.81

but the right answer is 1.19, why? i thought the OH in the problem meant i had to use 14?

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1. You calculated the pH not the pOH

pOH=-log([OH]/mole l^-1)

pH=-log([H]/mole l^-1)

Note that pOH=14-pH is only valid at 25C.

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posted by Dr Russ

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