Question

If 10.90 mL of a standard 0.1090 M Ba(OH)2 solution reacts with 45.00 mL of CH3COOH solution, what is the molarity of the acid solution? I got the chem equation as Ba(OH)2 + CH3COOH = 2H2O + Ba(C2H3O2)2, but where do i go from there?

Answer 1

Since you seem to have already balanced the equation, at this point you want to find out how many moles of Ba(OH)2 you have by multiplying the volume x M of Ba(OH)2. Make sure the volume is in Liters, of course. That will give you moles of Ba(OH)2, and from there you multiply it times whatever the stoichiometric ratio is between Ba(OH)2 and CH3COOH. That will give you number of moles of CH3OOH. Since you are given the volume of the CH3COOH, and you know the moles of CH3COOH now, you can plug it into the equation [Molarity = moles/liter]