stoichiometry help find molarity of naoh

you are reacting a 1m hcl solution with a sodium hydroxide solution of unknown concentration. if it takes 20.0 ml of the acid solution to react with 35 ml of the naoh solution, what is the molarity of naoh?

hi mariah,
 
let's start this problem by looking at the equation of the reaction:
 
                                      hcl + naoh --> nacl + h2o
 
the first thing we need to find is the number of moles of acid that were used to react with the base.  we find this using the known concentration of the acid as a conversion between volume and moles:
 
20.0ml * (1l/1000ml)*(1mol hcl/l) = 0.0200 mol hcl        * remember to convert ml to l first (m = mol/l)
 
next, using mole-to-mole stoichiometry, we convert moles of hcl to moles of naoh:
 
0.0200 mol hcl * (1 mol naoh/1 mol hcl) = 0.0200 mol naoh
 
now we use the volume of the naoh and the moles naoh to find the molarity:
 
0.0200 mol naoh/0.035l naoh = 0.571m naoh