Question

15.Activity coefficient of a neutral molecule. We use the approximation

that the activity coefficient $\left(\gamma \right)$of neutral molecules is 1.00.A more accurate relation is $\mathbf{log\gamma }\mathbf{=}\mathbf{k\mu }$, where$\mathbf{\mu }$is ionic strengthand $\mathbf{k}\mathbf{\approx }\mathbf{0}\mathbf{.}\mathbf{11}$for ${\mathbf{NH}}_{\mathbf{3}}$and ${\mathbf{CO}}_{\mathbf{2}}$and $\mathbf{k}\mathbf{\approx }\mathbf{0}\mathbf{.}\mathbf{2}$for organic molecules.With activity coefficients for $\mathbf{HA}\mathbf{,}\mathbf{}{\mathbf{A}}^{\mathbf{-}}$,and${\mathbf{H}}^{\mathbf{+}}$, predict the value ofthe quotient below for benzoic acid${}^{(\mathrm{HA}=C_6{H}_5{\mathrm{CO}}_{2}H)}$The observed quotient is${}^{\mathbf{0}\mathbf{.}\mathbf{63}\mathbf{\pm }\mathbf{0}\mathbf{.}\mathbf{03}}$.

Concentration quotient$\mathbf{=}\frac{{\displaystyle \frac{\mathbf{\left[}{\mathbf{H}}^{\mathbf{+}}\mathbf{\right]}\mathbf{\left[}{\mathbf{A}}^{\mathbf{-}}\mathbf{\right]}}{\left[\mathrm{HA}\right]}}\mathbf{(}\mathbf{at}\mathbf{}\mathbf{\mu }\mathbf{=}\mathbf{0}\mathbf{)}}{{\displaystyle \frac{\mathbf{\left[}{\mathbf{H}}^{\mathbf{+}}\mathbf{\right]}\mathbf{\left[}{\mathbf{A}}^{\mathbf{-}}\mathbf{\right]}}{\left[\mathrm{HA}\right]}}\mathbf{(}\mathbf{at}\mathbf{}\mathbf{\mu }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{1}\mathbf{M}\mathbf{)}}$

Short answer

The predicted value of the concentration quotient is 0.63

Step-by-step solution

Definition of activity coefficients

The activity coefficient $\mathit{\gamma }$measures the deviation of behavior from ideality,rapidly decreases as ionic strength increases.

:Obtaining the quotient value

With activity coefficients for $\mathrm{HA},{\mathrm{A}}^{-}\mathrm{and}{\mathrm{H}}^{+}$we will predict the value of the quotient forbenzoic acid.

The observed quotient from task is $0.63\pm 0.03$

The equilibrium constant is given as:

$$\begin{gathered} \mathrm{HA֏}{\mathrm{H}}^{+}+{\mathrm{A}}^{-} \\ {\mathrm{k}}_{\mathrm{a}}=\frac{\left[{\mathrm{H}}^{+}\right]\mathrm{\gamma }\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{A}}^{-}\right]\mathrm{y}\left({\mathrm{A}}^{-}\right)}{\left[\mathrm{HA}\right]\mathrm{\gamma }\left(\mathrm{HA}\right)} \end{gathered}$$

From Pg166 Table 8-1 we can see that $\mathrm{\gamma }\left({\mathrm{H}}^{+}\right)=0.83\mathrm{and}\mathrm{\gamma }\left({\mathrm{A}}^{-}\right)=0.80\mathrm{when}0.1\mathrm{M}$

For HA we estimate the following:

$$\begin{gathered} \log \mathrm{\gamma }\left(\mathrm{HA}\right)=\mathrm{k\mu }=0.2\times 0.1=0.02 \\ \mathrm{\gamma }\left(\mathrm{HA}\right)={10}^{0.02}=1.05 \end{gathered}$$

First we will consider the (ionic strength $\mathrm{\mu }=0.1\mathrm{M}$)

role="math" localid="1654918428253" $$\begin{gathered} {\mathrm{k}}_{\mathrm{a}}=\frac{\left[{\mathrm{H}}^{+}\right]\left(0.83\right)\left[{\mathrm{A}}^{-}\right]\left(0.80\right)}{\left[\mathrm{HA}\right]\left(1.05\right)}\Rightarrow 0.63\times \frac{\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{A}}^{-}\right]}{\left[\mathrm{HA}\right]} \\ =\frac{{\mathrm{k}}_{\mathrm{a}}}{0.63} \end{gathered}$$

Next we will consider the (ionic strength $\mathrm{\mu }=0\mathrm{M}$)

$$\begin{gathered} {\mathrm{k}}_{\mathrm{a}}=\frac{\left[{\mathrm{H}}^{+}\right]\left(1\right)\left[{\mathrm{A}}^{-}\right]\left(1\right)}{\left[\mathrm{HA}\right]\left(1\right)}\Rightarrow \frac{\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{A}}^{-}\right]}{\left[\mathrm{HA}\right]} \\ ={\mathrm{k}}_{\mathrm{a}} \end{gathered}$$

consider that when ionic strength$\mathrm{\mu }=0\mathrm{M}$ the activities are all 1

Finally, we will take the two values and put them into theconcentration quotient equationgiven in the task.

$$\begin{gathered} =\frac{{\mathrm{k}}_{\mathrm{a}}}{{\mathrm{k}}_{\mathrm{a}}/0.63} \\ =0.63 \end{gathered}$$

Therefore, the predicted value of the concentration quotient is 0.63