Question
If 33.7 ml of 0.210 M KOH is required to completely neutralize 27.0 ml of a HC2H3O2 solution, what is the molarity of the acetic acid solution?
HC2H3O2(aq)+ KOH(aq)-->KC2H3O2(aq)+H2O(l)
HC2H3O2(aq)+ KOH(aq)-->KC2H3O2(aq)+H2O(l)
Answers
mols KOH = M x L = ?
mols acetic acid = mols KOH (from the coefficients in the balanced equation.)
M acid = mols acid/L acid. You know mols and L, solve for M.
mols acetic acid = mols KOH (from the coefficients in the balanced equation.)
M acid = mols acid/L acid. You know mols and L, solve for M.
Related Questions
How many ml of a .25M HCL would be required to neutralize completely 2.5g of NaOH
how many mL of a .25 M HCL would be required to neutralize completely 30mL of a .2 M Ca(OH)2
How much 1.0 M NaOH would be required to completely neutralize 40.0 mL of 0.60 M HCl?
How much 0.200 M KOH is required to completely neutralize 25.0 mL of 0.150 M HClO4?