Asked by Sara

Suppose that 0.483 g of an unknown monoprotic weak acid, HA, is dissolved in water. Titration of the solution with 0.250 M NaOH(aq) required 42.0 mL to reach the stoichiometric point. After the addition of 21.0 mL, the pH of the solution was found to be 3.75.

(a) What is the molar mass of the acid?
(b) What is the value of pKa for the acid?
Answered by DrBob222
mols acid = mols base
mols base = M x L = ?
Then mols acid = grams acid/molar mass. You know mols acid and grams, solve for molar mass.

Solve for mols base at 21.0 mL
Solve for mols acid remaining at 21.0 mL base
Substitute into H-H equation (see your post above) and solve for pKa.
Answered by Samson
Molar mass is 40.77g/mol
Pka=2.4×10M