A 709 g sample of an aqueous solution of hydrochloric acid c

A 7.09 g sample of an aqueous solution of hydrochloric acid contains an unknown amount of the acid.
If 11.5 mL of 0.915 M potassium hydroxide are required to neutralize the hydrochloric acid, what is the percent by mass of hydrochloric acid in the mixture?

Solution

The balanced chemical reaction equation for neutralization reaction is as follows

HCl + KOH --> KCl + H₂O

from above equation it is clear that 1 mol HCl requires 1 mol KOH

moles KOH = 11.5 mL x 10^-3 L x 0.915 M = 0.0105225 mol

same amount of moles of HCl required to neutralize so moles HCl = 0.0105225 mol

convert moles HCl to mass HCl by multiplying with molar mass of HCl

mass HCl = 0.0105225 mol x 36.46 g/mol = 0.38365035 g

percent by mass of hydrochloric acid in the mixture = [mass HCl / mass of mixture] x 100%

percent by mass of hydrochloric acid in the mixture = [0.38365035 g / 7.09 g] x 100% = 5.41114739069 %

Final Answer: percent by mass of hydrochloric acid in the mixture = 5.41 % (3 sig fig)

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