a domestic cleaner contains concentrated hydrochloric acid as the active ingredient.

Question

a titration experiment was carried out to determine the concentration of hydrochloric acid in the domestic cleanser.
20.0mL of the cleanser were diluted to 250.0mL with distilled water.
25.0mL of the diluted cleanser were than titrated against 0.5mol L-1 sodium hydroxide solution.
15.00mL of sodium hydroxide solution were required for complete neutralization.

Q1.) Calculate the molarity of hydrochloric acid in the cleanser.

Q2.) Calculate the concentration, in g dm⁻³, of hydrochloric acid in the cleanser.

thx.

Answer 1

For the diluted acid,
C1,V1=diluted acid
C2,V2=sodium hydroxide

NaOH+HCl=NaCl+H2O

Use CiV1=C2V2
C1*25.0 mL=0.5M * 15.00 mL

C1=0.5M*15.00/25.0= 0.3 M

Concentration of original hydrochloric acid
= 0.3 M * (250.0/25.0)
= 3 M

(note only one significant figure is retained as a result of the precision of NaOH being 0.5 M.)

For Q2, use the molar mass of HCl multiplied by the molarity to get concentration in g/L=g/dm³ (approx.)