Question

Suppose that 25.00 mL of a solution acid (H2 C2 O4) is titrated with 0.1 M NaOH and the stoichiometric point is reached when 38.0 mL of the solution of base is added. Products of the titration are sodium oxalate and water. Find the molarity of the Oxalic acid!

Answer 1

Due to the fact that in this task a solution of a weak acid is titrated with a solution of a strong base, we calculate the mass of oxalic acid according to the formula:

$$m(H_2C_2O_4) = {c(NaOH)V(NaOH)M(0.5H_2C_2O_4)100 \over V(H_2C_2O_4)1000}$$

$$m(H_2C_2O_4) = {0.1*38*45*100 \over 25*1000} = 0.45g$$

$$c(H_2C_2O_4) = n/v = m/(M*V) = 0.45/(90*0.025) = 0.2M$$

Answer: 0.2M