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:
\(\mathbf{m}_{\pm}=\left(\mathbf{m}_{+}^{\nu+}+\mathbf{m}_{-}^{\nu-}\right)^{1 / \nu}\)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:
\(\begin{aligned}&a_{\pm}=0.7909 \cdot 0.2785 \mathrm{~m} \\
&=0.2203 \mathrm{~m}
\end{aligned}\)