Question

Question #0bd3b

Answer 1

The titration requires 2 cm³ of 0.1 mol•dm⁻³ HCl. Both bases react.

Both NaOH and Ba(OH)₂ are strong bases. They ionize completely in solution:

NaOH(s) → Na⁺(aq) + OH⁻(aq)

Ba(OH)₂(s) → Ba²⁺(aq) + 2OH⁻(aq)

We must figure out how many moles of OH⁻ there are and then calculate the volume of HCl required to neutralize the OH⁻.

Mass of NaOH = 2.0 g mixture × `(60 g NaOH)/(100 g mixture)` = 1.2 g NaOH

Moles of NaOH = 1.2 g NaOH × `(1 mol NaOH)/(40.00 g NaOH)` = 0.030 mol NaOH

Moles of OH⁻ from NaOH = 0.030 mol NaOH × `(1 mol OH^-)/(1 mol NaOH)` = 0.030 mol OH⁻

Mass of Ba(OH)₂ = 2.0 g mixture × `(40 g Ba(OH)_2)/(100 g mixture)` = 0.80 g Ba(OH)₂

Moles of Ba(OH)₂ = 0.80 g Ba(OH)₂ × `(1 mol Ba(OH)_2)/(171.3 g Ba(OH)_2)` =
4.7 × 10⁻³ mol Ba(OH)₂

Moles of OH⁻ from Ba(OH)₂ =

4.7 × 10⁻³ mol Ba(OH)₂ × `(2 mol OH^-)/(1 mol NaOH)` = 9.3 × 10⁻³ mol OH⁻

Total moles of OH⁻ = 0.030 mol + 9.3 × 10⁻³ mol = 0.039 mol OH⁻

Moles of OH⁻ in the aliquot = 0.039 mol OH⁻ × `(25 cm^3)/(500 cm^3)` =
2.0 × 10⁻³ mol OH⁻

Volume of HCl required = 2.0 × 10⁻³ mol OH⁻ × `(1 mol H^+)/(1 mol OH^-) × (1 mol HCl)/(1 mol H^+) × (1000 cm³)/(1 mol HCl)` = 2 cm³ HCl