Acids, bases and titration — Question 28

28a)

28b)

• determines moles of base in aliquot equals moles of HCl at equivalence point [1 mark]

$\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{s}\,\mathrm{o}\mathrm{f}\,\mathrm{B}\mathrm{O}\mathrm{H}=0.1\,\times 0.020=2.0\,\times 10^{-3}\,\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{s}$

$[\mathrm{d}\mathrm{i}\mathrm{l}\mathrm{u}\mathrm{t}\mathrm{e}\mathrm{d}\,\mathrm{B}\mathrm{O}\mathrm{H}]=\frac{2.0\,\times 10^{-3}}{0.025}=0.08\,\mathrm{m}\mathrm{o}\mathrm{l}\,\mathrm{L}^{-1}$

• calculates concentration of BOH in aliquot [1 mark]

$[\mathrm{B}\mathrm{O}\mathrm{H}]=0.08\,\times 5\,\,$

$=0.40\,\mathrm{m}\mathrm{o}\mathrm{l}\,\mathrm{L}^{-1}$

• calculates concentration of BOH [1 mark]

28c)

The pH of the equivalence point: would increase to pH 7 because NaOH is a strong base. The volume of HCl required to reach the equivalence point: would remain the same, as the moles of base being neutralised has not changed.

• determines that the pH at the equivalence point is 7 [1 mark]

• explains that the pH at the equivalence point would increase when a strong base is titrated with a strong acid [1 mark]

• determines that the volume of HCl required to reach the equivalence point would not change [1 mark]

• explains that the moles of base to be neutralised remains the same [1 mark]