## Determine the mean activity coefficient and mean activity of a 0.004 molal of \(\mathrm{Ba}\left(\mathrm{HCO}_{3}\right)_{2}\)?

Answer:

MEAN ACTIVITY COEFFICIENT

According to Debye-Huckel Theory, the equation for calculating the mean activity coefficient γ±, for solutions with concentrations on the order of 0.01 M or less, is:

where ε is the dielectric constant, z± is the charge of each ion, and of course, \(T\) is temperature in \(K\) \(I\) is the ionic strength, defined as:

\(\mathbf{I}=\frac{\mathbf{1}}{\mathbf{2}} \sum_{\mathbf{i}} \mathbf{m}_{\mathbf{i}} \mathbf{z}_{\mathbf{i}}^{\mathbf{2}}\)where m is the molality and z is the charge (here, the sign doesn’t matter).

In water, ε = 78.54. I don’t know what temperature you are referring to, but I will assume \(T\) = \(298K\)

In \(\mathrm{Ba}\left(\mathrm{HCO}_{3}\right)_{2}\) you have the ions \(\mathrm{Ba}^{2+} \text { and } \mathrm{HCO}_{3}^{-}\) Thus, z_{+}= +2 and z_{–}=-1.

Next, we need I:

\(\begin{aligned}&I=\frac{1}{2} \sum_{i} m_{i} z_{i}^{2}=\frac{1}{2}\left[(0.004)(+2)^{2}+(0.004)(-1)^{2}\right] \\

&=0.01 \mathrm{~m}

\end{aligned}\)

Overall, we have:

- ε = 78.54
- T = 298 k
- z
_{+}= +2 - z
_{–}= -1 - I = 0.01 m

Now that we have all this, we can write it all out to be:

If you get a number higher than 1, you know you’ve made a mistake.

γ±≤ 1 because at best, a = m in γ = \(\frac{a}{m}\)

MEAN ACTIVITY

Next, the mean activity a± can be determined using the following equations:

\(\gamma=\frac{a}{m}\) therefore

\(\mathbf{a}_{\pm}=\gamma_{\pm} \mathbf{m}_{\pm}\)where γ_{±} is the mean activity coefficient, a_{±} is the mean activity in m, and m_{±} is the mean ionic molality:

where ν is the sum of the stoichiometric coefficients for both ions, v_{+,} and v_{–} are the stoichiometric coefficients of each ion, and m_{+} and m_{–} are the molalities of each ion (incorporating stoichiometric coefficients!!!).

For this case:

- v
_{+}= 1 - v
_{–}= 2 - m
_{+}= 0.02 . 1 = 0.02 m - m
_{–}= 0.02 . 2 = 0.04 m - γ
_{±}= 0.7909

Now we have everything we need!

\(\begin{aligned}

&m_{\pm}=\left[(0.02)^{1}+(0.04)^{2}\right]^{1 /(1+2)} \\

&=[0.02+0.0016]^{1 / 3} \\

&=0.2785 \mathrm{~m}

\end{aligned}\).

We’re almost done. Now that we have m_{±} and γ_{±}, we can determine a_{±} to be:

&a_{\pm}=0.7909 \cdot 0.2785 \mathrm{~m} \\

&=0.2203 \mathrm{~m}

\end{aligned}\)