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.
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?
1 answer