Question

A 9.15 g sample of an aqueous solution of hydroiodic acid contains an unknown amount of the acid. If 29.3 mL of 0.506 M sodium hydroxide are required to neutralize the hydroiodic acid, what is the percent by mass of hydroiodic acid in the mixture?

Answer 1

Answer:

Let

M1 = molarity of HI (hydroiodic acid)

M2 = Mlarity of NaOH = 0.506 M (given)

V1 = Volume of HI = 9.15 mL   (assuming aqueous solution have density near to water that is 1 g/mL so 9.15 g = 9.15 mL)

V2 = Volume of NaOH = 29.3 mL (given)

Reaction of neutralization is,

HI (aq) +   NaOH (aq) ---->    NaI (aq)    + H2O (l)

Since 1 mole of HI requires 1 mole of NaOH for neutralization, so equation for neutralization is,

M1V1 = M2V2

From above,

M1 = M2V2 / V1 = (0.506 M x 29.3 mL) / 9.15 mL = 1.62 M

So molarity of HI is 1.62 M

Also, Molarity of HI = Moles of HI x 1000 / volume of HI

1.62 M = Moles of HI x 1000 / 9.15

Above equation give,

moles of HI = 14.8 x 10^-3 mole

Molar mass of HI = 127.911 g/mol

So mass of 14.8 x 10^-3 mole HI = 14.8 x 10^-3 mole x 127.911 g/mol = 1.89 g

So mass percentage of HI = mass of HI x 100 / total mass = 1.89 g x 100 / 9.15 g = 20.6 %      Answer