Question

A solution of HClO4 was standardized by dissolving 0.4608g of primary standard grade HgO in a solution of KBr

Hgo (a) + 4Br- +H2O---->HgBr42- + 2OH-

the liberated OH- consumed 43.48 ml of the acid. calculate the molar concentration of HClO4?

Answer 1

Balanced equation:
HgO + 4 Br^- + H₂O ====> HgBr₄^2- + 2 OH^-

0.4608 g of HgO was required, which is equivalent = 0.4608 /  216.5894 = 2.1275 x10^-3 moles

From the balanced equation, we can calculate that = 2.1275 x10^-3 moles x 2 = 4.255 x10^-3 moles of OH- is produced.

Now, a solution of HClO4 acid consists of H⁺ ions and ClO₄^- ions. It is the H+ ions that react with the OH- ions according to the equation:

H+ + OH- --> H2O

The number of moles of OH- and H+ are therefore equal for neutralization. Since there are 4.255 x10^-3 moles of OH-, there must also be 4.255 x10^-3 of H+ in the 43.48 mL of acid solution,

Molarity = 4.255 x10^-3 / 0.04348 = 0.09786 mol/L = 0.09786 M