Titrations

Question 1

1.50g of impure magnesium carbonate is added to 100cm³
of 0.400 moldm³ hydrochloric acid. The leftover acid was
made up to a total volume of 200cm³ in a volumetric flask.
A 25cm³ portion of this acid was placed into a conical
flask. During titration 26.25cm³ of 0.100moldm³ Sodium
hydroxide was needed to neutralise the sample.
Calculate the percentage purity of the carbonate sample.

Answer 1

① molesNaOH = 0.1 × 26.25/1000 = 2.625 × 10⁻3

② moles HCl in c.f. = 2.625 × 10⁻³

③ moles HCl in v.t. = 200/25=8 x 2.625 x10⁻³ = 0.021 moles

④ HCl start moles = 0.4 x 100/1000 = 0.04

⑤ HCl moles used = 0.04 - 0.021 = 0.019 =

6. Moles MgCO3 = 0.019 ÷ 2 = 0.0095 (pure)
7. Pure MgCO3 mass = Mr x moles = 84.3 x 0.0095 = 0.8019

8.%Purity = 0.801/ 1.50 x100 = 53.39% = 53.4% (3 sig. fig.)