You titrate 0.500 g of an unknown monoprotic acid with 0.0990 M NaOH. You discover that it takes 23.56 ml of base to completely neutralize the acid. what is the formula weight of the unknown acid?
If the unknown were a triprotic acid, what would its formula weight be?
moles of NaOH reacted = 0.099 x 23.56 / 1000 = 0.00233
if acid is monoprotic
moles of acid = 0.00233
molar mass = mass / moles = 0.500 / 0.00233
molar mass = 214.6 g/mol
if acid is triprotic acid
moles of acid reacted = 0.00233 / 3 = 0.00078 moles
molar mass = mass / moles = 0.500 / 0.00078
molar mass = 641 g/mol
You titrate 0.500 g of an unknown monoprotic acid with 0.0990 M NaOH. You discover that...
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