0.0784 M C₆H₅NH₂, aniline, a weak base”

We are told to assume the ionization of water can be neglected.

Make assumptions to simplify the equilibrium concentration expressions.

You will need to verify this assumption.

[latex]\begin{gathered} \mathrm{C}_6\mathrm{H}_5\mathrm{NH}_2\mathrm{(aq)}+\mathrm{H}_2\mathrm{O}\,\mathrm{(l)}\rightleftharpoons\mathrm{C}_6\mathrm{H}_5\mathrm{NH}_3^+\mathrm{(aq)}+\mathrm{OH}^-\mathrm{(aq)} \end{gathered}[/latex]

Create an ICE Table (initial, change, equilibrium) with all species represented. ‘x’ will represent your change.

We are told to neglect the ionization of water, so the initial [OH^–] is 0.

Assume the change in concentration (x) is negligible compared to the initial concentration of the weak base.

[latex]\begin{gathered}\mathrm{K}_\mathrm{b}=4.0\times10^{-10}\end{gathered}[/latex]
(from Table E2. Base Dissociation Constants at 25°C (2))

Assumptions:

Aniline is a weak base, with a small K_b value, so any change in concentration will be small as the reactants are favoured at equilibrium. This means that the change in concentration (x) can be assumed to be negligible compared to the initial concentration of aniline.

0.0784 M >> x, therefore x is negligible and the equilibrium concentration of aniline,  [C₆H₅NH₂], is 0.0784M.

[latex]\begin{aligned} \mathrm{K_b}&=\frac{[\text {Products}]}{[\text {Reactants}]}\\ \mathrm{K_b}&=\frac{\left[\mathrm{C}_6\mathrm{H_5}\mathrm{N}\mathrm{H_3}^{+}\right]\left[\mathrm{OH}^{-}\right]}{\mathrm{C}_6\mathrm{H_5}\mathrm{N}\mathrm{H_2}} \end{aligned}[/latex]

Plug in your equilibrium concentrations from the last line of the ICE table:

• [C₆H₅NH₂] = 0.0784 M
• [C₆H₅NH₃⁺] = x
• [OH^–] = x

[latex]\begin{aligned} 4.0\times 10^{-10}&=\frac{(x)(x)}{0.0784} \\ 4.0\times 10^{-10}&=\frac{x^2}{0.0784} \\ \left(4.0\times 10^{-10}\right)(0.0784)&=x^2 \\ 3.14\times 10^{-11}&=x^2 \\ \sqrt{3.14\times 10^{-11}}&=\sqrt{x^2} \\ 5.60\times 10^{-6}\mathrm{M}&=x \end{aligned}[/latex]

[latex]\begin{aligned} \,[\mathrm{C}_6\mathrm{H}_5\mathrm{NH}_2]&=0.0784\mathrm{~M}-\mathrm{x}\\ &=0.0784\mathrm{~M}-5.60\times10^{-6}\mathrm{~M}\\ &=0.0784\mathrm{~M}\\\\ \,[\mathrm{C}_6\mathrm{H}_5\mathrm{NH}_3]&=\mathrm{x}\\ &=-5.60\times10^{-6}\mathrm{~M}\\\\ \,[\mathrm{OH}^-]&=\mathrm{x}\\ &=-5.60\times10^{-6}\mathrm{~M}\\\\ \mathrm{K}_{\mathrm{w}}&=[\mathrm{H}_3\mathrm{O}^+][\mathrm{OH}^{-}]\\ \,[\mathrm{H}_3\mathrm{O}^+]&=\frac{\left[\mathrm{K}_{\mathrm{w}}\right]}{\left[\mathrm{OH}^{-}\right]}\\ \left[\mathrm{H}_3\mathrm{O}^{+}\right]&=\frac{1.00\times10^{-14}\mathrm{M}}{5.60\times10^{-6}\mathrm{M}}\\ \left[\mathrm{H}_3\mathrm{O}^{+}\right]&=1.79\times10^{-9}\mathrm{M} \end{aligned}[/latex]

Check your assumptions:

Use the 5% rule to prove that the change is negligible for the initial concentration of the reactant. Recall that if the change is truly negligible the calculated value from this test will be less than 5%.

Valid if:

[latex]\begin{gathered} \frac{\text { Change }}{[\text { initial }]}\times100\text{ is }<5\% \end{gathered}[/latex]

[latex]\begin{gathered} \frac{\mathrm{X}}{\left[\mathrm{C}_6\mathrm{H}_5\mathrm{NH}_2\right]}=\frac{5.60\times 10^{-6}}{0.0784}\times100=7.14\times10^{-3}=<5\% \end{gathered}[/latex]

So, the assumptions were valid.

We have shown that the change in the initial weak base concentration can be neglected, since when we subtracted ‘x’ from the initial concentration, the value did not change.

Answer:

• [C₆H₅NH₂] = 0.0784 M
• [C₆H₅NH₃⁺] = 5.60 × 10^-6 M
• [OH^–] = 5.60 × 10^-6 M
• [H₃O⁺] = 1.79 × 10^-9