what is pOH of 0.02M of Ca(OH)2
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I will answer assuming you mean :
What is the pOH of 0.02 M Ca(OH)2
Ca(OH)2 is a strong base that dissociates completely
Ca(OH)2(aq) → Ca2+ (aq) + 2 OH- (aq)
Therefore [OH-] = 0.04M
pOH = - log [OH-]
pOH = - log 0.04
pOH = 1.40
And for completion:
pH = 14.00 - 1.40 = 12.60
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