Titration answer 1

10.0 cm³ of hydrochloric acid was titrated with 0.100 M sodium hydroxide solution . The titres were 18.00 cm³, 17.30 cm³ and 17.40 cm³. What is the concentration of the hydrochloric acid in mol dm^-3?

Average titre, ignoring 18.00 cm³ = (17.30 + 17.40)/2 = 17.35 cm³

Number of moles of NaOH = 17.35/1000 × 0.100

NaOH(aq) + HCl(aq) ⇒ NaCl(aq) + H₂O(l)

NaOH º HCl

\number of moles of HCl in 10.0 cm³ = 17.35/1000 × 0.100

\concentration of HCl = (17.35/1000 × 0.100)/0.01 = 0.174 mol dm^-3