Question

A 7.55 g sample of an aqueous solution of hydrochloric acid contains an unknown amount of the acid.  
If 29.1 mL of 0.321 M sodium hydroxide are required to neutralize the hydrochloric acid, what is the percent by mass of hydrochloric acid in the mixture?

% by mass

Answer 1

Given: 29.1ml of 0.321M Sodium Hydroxide neutralizes 7.55g sample of Hydrochoric acid.

To Calculate : Percentage of Hydrochloric acid solution.

No of moles of Sodium Hydroxide present in 29.1ml of 0.321M solution = 0.321 x 29.1/1000 = 0.00934moles.

( No. of moles = Molarity x Volume expressed in litre.)

1 mole of sodium hydroxide neutralizes 1 mole of hydrochloric acid.

This means that 0.00934 moles of hydrocholric acid are present in the given solution.

Mass of 0.00934 moles of Hydrochloric acid = 0.00934 x 36.5 (Molar mass of HCl)

= 0.34091g of HCl.

Calculation of percent by mass of HCl:

   

= 4.545%

Ans : The percentage by mass of Hydrochloric acid in the geven solution is 4.545%.