1.6g of an unknown monoprotic acid (HA) required 5.79 mL of a 0.35 M NaOH solution to reach the equivalence point. Determine the molar mass of the acid and given that the pH at the half way point to the equivalent point is 3.86. Calculate the Ka of the unknown acid.

Answer 1

HA + NaOH ==> NaA + H2O

mols NaOH = M x L = ?
mols HA = mols NaOH (look at the coefficient in the balanced equation).
mols HA = grams HA/molar mass HA. You know mols and grams, solve for molar mass.

If the pH at the half way point is 3.86, then the pKa is 3.86.
pKa = -log Ka. Solve for Ka.

Answer 2

molar mass= 90.01 g/mol and Ka= 1.38 x 10^-4

Answer 3

Na(oh)3