A 0.205 g sample of impure NaOH requires 17.5 mL of 0.2180 M HCl for neutralization. What is the percent of NaOH in the sample, by weight?

Answer 1

moles acid=.2180*.0175

mass NaOH=formulamassNaOH*molesacid
mass NaOH=40*moles acid above

percent= massNaOH/.205

Answer 2

Thank you! That makes much more sense.