calculate the molarity of acetic acid soution if 25.2ml of an HC2H3O2 solution is titrated with 28.6 mL of a 0.164 M KOH solution

Question

equation is HC2H3)2(aq)+KOH(aq) arrow to rt H2O(l) +KC2H3O2(aq)

Can you please provide step by step... I need to learn how to do this....Thank you

Answer 1

HC₂H₃O₂(aq) + KOH(aq) ==> H2O(l) + HC₂H₃O₂(aq)

mols KOH used = 28.6 ml x 1 L / 1000 ml x 0.164 mol/L = 0.00469 mols KOH

mols HC₂H₃O₂ present = 0.00469 mol KOH x 1 mol HC₂H₃O₂ / mol KOH = 0.00469 mols HC₂H₃O₂

molarity of HC₂H₃O₂ = 0.00469 mols / 25.2 ml x 1000 ml / L = 0.186 M