# Question 0bd3b

Mar 17, 2014

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 × $\frac{60 g N a O H}{100 g m i x t u r e}$ = 1.2 g NaOH

Moles of NaOH = 1.2 g NaOH × $\frac{1 m o l N a O H}{40.00 g N a O H}$ = 0.030 mol NaOH

Moles of OH⁻ from NaOH = 0.030 mol NaOH × $\frac{1 m o l O {H}^{-}}{1 m o l N a O H}$ = 0.030 mol OH⁻

Mass of Ba(OH)₂ = 2.0 g mixture × $\frac{40 g B a {\left(O H\right)}_{2}}{100 g m i x t u r e}$ = 0.80 g Ba(OH)₂

Moles of Ba(OH)₂ = 0.80 g Ba(OH)₂ × $\frac{1 m o l B a {\left(O H\right)}_{2}}{171.3 g B a {\left(O H\right)}_{2}}$ =
4.7 × 10⁻³ mol Ba(OH)₂

Moles of OH⁻ from Ba(OH)₂ =

4.7 × 10⁻³ mol Ba(OH)₂ × $\frac{2 m o l O {H}^{-}}{1 m o l N a O H}$ = 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⁻ × $\frac{25 c {m}^{3}}{500 c {m}^{3}}$ =
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