Question

A 8.72 g sample of an aqueous solution of hydroiodic acid contains an unknown amount of the acid.
If 21.6 mL of 0.567 M sodium hydroxide are required to neutralize the hydroiodic acid, what is the percent by mass of hydroiodic acid in the mixture?

% by mass

Answer 1

moles of NaOH = 21.6 x 0.567 / 1000

                         = 0.01225

moles of HI = 8.72 / 127.9 = 0.0682

NaOH   + HI   --------------> NaI + H2O

     1          1

moles of hydroiodic acid reacts = 0.01225

mass of HI = 0.01225 x 127.9 = 1.566

mass % of HI =( 1.566 / 8.72 ) x 100

mass % of HI = 17.96 %