Asked by Anonymous

Calculate the pOH of a 0.410 M Ba(OH) solution.
Answered by DrBob222
OH is 2x Ba(OH)2.
pOH = -log(OH^-)
Answered by Anonymous
8.62*10^-2
Answered by Thatoneguy
pOH = -log([OH])
[OH]=concentration*# of hydroxide ions
pOH=-log(2*0.41)
pOH=8.62*10^-2
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