Question

0.25g of acid is titrated with 108M NaOH, if it takes 27.22ml NaOH to reach the endpoint, what is the equivalent weight of the acid

Answer 1

C_M = n / V

n = C_M x V

n = m / M

M = m / n

n (NaOH) = 0.108 x 0.02722 = 0.003 mol

M_acid = 0.25 / 0.003 = 83.3 g/mol