Question

Find the activity (not the activity coefficient) of the ${\mathbf{\left(}{\mathbf{C}}_{\mathbf{3}}{\mathbf{H}}_{\mathbf{7}}\mathbf{\right)}}_{\mathbf{4}}\mathbf{}{\mathbf{N}}^{\mathbf{+}}$(tetra propylammonium) ion in a solution containing${\mathbf{\left(}{\mathbf{C}}_{\mathbf{3}}{\mathbf{H}}_{\mathbf{7}}\mathbf{\right)}}_{\mathbf{4}}\mathbf{}{\mathbf{N}}^{\mathbf{+}}{\mathbf{Br}}^{\mathbf{-}}\mathbf{plus}\mathbf{0}\mathbf{.}\mathbf{0050}\mathbf{M}{\mathbf{\left(}{\mathbf{CH}}_{\mathbf{3}}\mathbf{\right)}}_{\mathbf{4}}\mathbf{}{\mathbf{N}}^{\mathbf{+}}{\mathbf{CI}}^{\mathbf{-}}$

Short answer

The activity of${\left({\mathrm{C}}_3{\mathrm{H}}_7\right)}_4\hspace{0.17em}{\mathrm{N}}^{+}$ ion is 0.0046.

Step-by-step solution

Define the formula for activity species.

Any species' activity value is calculated by multiplying its concentration by its activity coefficient.

For a species C, the activity is given by

${\mathbf{A}}_{\mathbf{c}}\mathbf{=}\left[C\right]{\mathbf{\gamma }}_{\mathbf{c}}$

where,

A - Activity

[c] - Concentration

- Activity coefficient.

Calculate the activity species.

Given

$0.0050\mathrm{M}\left({\mathrm{C}}_3{\mathrm{H}}_7\right){\mathrm{N}}^{+}{\mathrm{Br}}^{-}$plus$0.0050\mathrm{M}{\left({\mathrm{CH}}_3\right)}_4{\mathrm{N}}^{+}{\mathrm{CI}}^{-}$

The ionic strength of this solution is$\mathrm{\mu }=0.010\mathrm{M}$

The size of the ion is${\left({\mathrm{C}}_3{\mathrm{H}}_7\right)}_4\hspace{0.17em}{\mathrm{N}}^{+}$is 800pm

$\mathrm{\mu }=0.010\mathrm{M},$

At$\gamma =0.912$ for an ion of charge${}_{-}{}^{+}1$ with$\mathrm{\alpha }=800\mathrm{pm}$

The activity ofrole="math" localid="1654762931695" ${\left({\mathrm{C}}_3{\mathrm{H}}_7\right)}_4{\mathrm{N}}^{-}$ by using activity coefficient and concentration values

$$\begin{gathered} {\mathrm{A}}_{{\left({\mathrm{C}}_3{\mathrm{H}}_7\right)}_4}{\mathrm{N}}^{+}=\left[{\left({\mathrm{C}}_3{\mathrm{H}}_7\right)}_4{\mathrm{N}}^{+}\right]{\mathrm{\gamma }}_{{\left({\mathrm{C}}_3{\mathrm{H}}_7\right)}_4{\mathrm{N}}^{-}} \\ =\left(0.0050\right)\left(0.912\right) \\ =0.0046 \end{gathered}$$

The activity of ${\left({\mathrm{C}}_3{\mathrm{H}}_7\right)}_4{\mathrm{N}}^{+}$ion is 0.0046.