Question

A 0.4126 g sample of primary-standard Na2CO3 was treated with 40.0 mL of diluted perchloric acid. The solution was boiled to remove CO2, following which the excess HClO4 was back titrated with 9.20 mL of dilute NaOH. In a separate experiment, it was stablished that 22.93 mL of the HClO4 neutralized the NaOH in a 25.00 mL portion. Calculate the molarity of the HClO4 and NaOH.

Please report correct Significant Figures and explain

Answer 1

Equations for back-titration:

n1(HClO4) = 2n(Na2CO3) + n1(NaOH) --- (eq 1)

Equation for acid-base titration:  

n2(NaOH) =n2(HClO4) ---(eq 2)

first calculate the moles for eq (1) :

(40/1000)[HClO4]=2*(0.4126g/ (105.99g/mol)) +(9.20/1000)[NaOH]

then for eq (2):

(22.93/1000)[HClO4]=(25/1000)[NaOH]

then substituting eq (2) in eq (1) you will get

molarity of HClO4 = [HCLO4] = 0.2467

and then putting this value in eq (2) you will get

molarity of NaOH = [NaOH] = 0.2263