Question

when a solution prepared by dissolving 4.00g of an unknown monoprotic acid in 1.00L of water is titrated with 0.600M NaOH, 38.7mL of the NaOH solution is needed to neutralize the acid. What was the molarity of the acid solution? what is the molecular weight of the unknown acid?

Answer 1

HA is the acid.
HA + NaOH ==> NaA + H2O

moles NaOH = M x L = 0.6000 x 0.0387 = 0.02322.
Therefore, moles HA = 0.02322
M of HA = moles/L.
moles = grams/molecular weight
m.w. = grams/moles = 4/0.02322 = ?? Then round to 3 significant figures (based on the 38.7)