Acids, bases and titration — Question 3

3a)

( ) ( )

( ) ( )

−

−

CH COOH aq OH aq CH COO aq H O I

+

+

⇌

3

3

2

The conjugate base formed is CH₃COO⁻.

• provides correct balanced chemical equation [1 mark]

• identifies CH₃OO⁻ as conjugate base [1 mark]

3b)

3c)

$\mathrm{A}\mathrm{t}\,\mathrm{e}\mathrm{q}\mathrm{u}\mathrm{i}\mathrm{v}\mathrm{a}\mathrm{l}\mathrm{e}\mathrm{n}\mathrm{c}\mathrm{e}\,\mathrm{p}\mathrm{o}\mathrm{i}\mathrm{n}\mathrm{t},\,\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{s}\,\mathrm{O}\mathrm{H}^{-}\,=\,\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{s}\,\mathrm{H}^{+}\,\,$

$=\,\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{s}\,\mathrm{C}\mathrm{H}_{3}\mathrm{C}\mathrm{O}\mathrm{O}^{-}\,\,$

$\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{s}\,\mathrm{C}\mathrm{H}_{3}\mathrm{C}\mathrm{O}\mathrm{O}^{-}\,=\,\mathrm{n}\,\times \,\mathrm{V}\,=\,0.10\,\times \,0.015\,=\,1.50\,\times \,10^{-3}\,$

• determines moles CH₃COO⁻ is 1.50 × 10⁻³ [1 mark]

• calculates [CH₃COO⁻] is 2.31 × 10⁻² mol L⁻¹ [1 mark]

3d)

• provides correct substitution [1 mark]

• calculates [OH⁻] is 3.64 × 10⁻⁶ M [1 mark]

• determines pOH is 5.4 [1 mark]

• calculates pH is 8.6 [1 mark]