Question

A 7.36 g sample of an aqueous solution of hydrochloric acid contains an unknown amount of the acid.

If 23.3 mL of 0.104 M potassium hydroxide are required to neutralize the hydrochloric acid, what is the percent by mass of hydrochloric acid in the mixture?

_______% by mass

Answer 1

Balanced chemical equation is:

KOH + HCl ---> KCl + H2O

lets calculate the mol of KOH

volume , V = 23.3 mL

= 2.33*10^-2 L

use:

number of mol,

n = Molarity * Volume

= 0.104*2.33*10^-2

= 2.423*10^-3 mol

According to balanced equation

mol of HCl reacted = (1/1)* moles of KOH

= (1/1)*2.423*10^-3

= 2.423*10^-3 mol

This is number of moles of HCl

Molar mass of HCl,

MM = 1*MM(H) + 1*MM(Cl)

= 1*1.008 + 1*35.45

= 36.458 g/mol

use:

mass of HCl,

m = number of mol * molar mass

= 2.423*10^-3 mol * 36.46 g/mol

= 8.835*10^-2 g

mass % of HCl = mass of HCl * 100 / mass of sample

= (8.835*10^-2)*100/7.36

= 1.20 %

Answer: 1.20 %