Question

A mass of 0.4145 g of an unknown acid, HA, is titrated with NaOH. If the acid reacts with 29.76 mL of 9.917×10-2 M NaOH(aq), what is the molar mass of the acid?

A. 0.1404 g/mol
B. 140.4 g/mol
C. 4.180 g/mol
D. 0.4145 g/mol
E. 1.223×10-3 g/mol

Answer 1

n(NaOH) = C*V = 9.917*10^-2mol/L*0.02976L = 0.29513*10^-2 mol

n(HA) = n(NaOH)

M(HA) = m(HA)/n(HA) = 0.4145g/0.29513*10^-2mol = 140.4 g/mol

B. 140.4 g/mol