Homework Help: chemistry

Posted by Anonymous on Monday, July 30, 2012 at 5:36am.

If 35.7{\rm mL} of 0.205{\rm M} {\rm KOH} is required to completely neutralize 20.0{\rm mL} of a {\rm HC_2H_3O_2} solution, what is the molarity of the acetic acid solution?
{\rm HC_2H_3O_2}(aq) + {\rm KOH}(aq) \to {\rm KC_2H_3O_2}(aq) + {\rm H}_{2}{\rm O}(l)

chemistry - Steve, Monday, July 30, 2012 at 9:35am
35.7 mL of 0.205M KOH contains 0.00732 moles of KOH.

Since one mole of KOH reacts with one mole of acid, you neutralized 0.00732 moles of acid.

.00732moles/.02L = 0.366M

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