Calculate the pOH of a 0.410 M Ba(OH) solution.

Question

Anonymous · Accepted Answer

8.62*10^-2

DrBob222 · Answer

OH is 2x Ba(OH)2. pOH = -log(OH^-)

Thatoneguy · Answer

pOH = -log([OH]) [OH]=concentration*# of hydroxide ions pOH=-log(2*0.41) pOH=8.62*10^-2