Determine the mean activity coefficient and mean activity of a 0.004 molal of Ba(HCO_3)_2?

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:

\(\log \gamma_{\pm}=\frac{1.824 \times 10^{6}}{(\varepsilon T)^{3 / 2}}\left|z_{+} z_{-}\right| \sqrt{\mathrm{I}}\)

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:

\(\begin{aligned}
&a_{\pm}=0.7909 \cdot 0.2785 \mathrm{~m} \\
&=0.2203 \mathrm{~m}
\end{aligned}\)