Question

A 0.4188-g sample of primary-standard Na2Co3 was treated with 39.50 mL of dilute perchloric acid. The solution was boiled to remove Co2 , following which the excess HclO4 was back-titrated with 9.35 mL of dilute NaOH . In a separate experiment, it was established that 26.75 mL of HCLO4 the neutralized the NaOH in a 25.00-mL portion. Calculate the molarities of HCLO4 the  and NaOH .

Answer 1

NaOH + HClO_4 rightarrow NaClO_4 + H_2O Na_2 CO_3 + 2 HClO_4 rightarrow 2NaClO_4 + H_2 O + CO_2 No. of moles of Na_2 CO_3 = 0.4188 g/105.988 gmol^-1 = 3.951 times 10-3 mol Molarity of NaOH = C_NaOH Molarity of HClO_4 = C_HClO_4 In the back titration, Moles of NaOH reacted = 9.35 cm^3 times 10^-3 times C_NaOH therefore Excess moles of HClO_4 = 9.35 times 10^-3 times C_NaOH Total added moles of HClO_4 = 39.50 times 10^-3 dm^3 times C_HClO_4 Reacted moles of HClO_4 with Na_2 CO_3} = (39.50 times 10^-3 times C_HClO_4 - 9.35 times 10^-3 times C_NaOH) Reacted moles of HClO_4 with Na_2 CO_3 = 3.951 times 10^-3 times 2 (1)- therefore 3.951 times 10^-3 times 2 = 39.50 times 10^-3 C_HClO_4 - 9.35 times 10^-3 times C_NaOH Separate experiment. Moles of HClO_4 = 26.75 times 10^-3 dm^3 times C_HClO_4 Moles of NaOH = 25.00 ml times 10^-3 times C_NaOH