# Determine the mean activity coefficient and mean activity of a "0.004 molal" of "Ba"("HCO"_3)_2?

Oct 14, 2015

Note: This will be long. Bear with me!

MEAN ACTIVITY COEFFICIENT

According to Debye-Huckel Theory, the equation for calculating the mean activity coefficient ${\gamma}_{\pm}$, for solutions with concentrations on the order of $0.01 M$ or less, is:

$\setminus m a t h b f \left(\log {\gamma}_{\pm} = \frac{1.824 \times {10}^{6}}{\epsilon T} ^ \text{3/2} | {z}_{+} {z}_{-} | \sqrt{I}\right)$

where $\epsilon$ is the dielectric constant, ${z}_{\pm}$ is the charge of each ion, and of course, $T$ is temperature in $K$. $I$ is the ionic strength, defined as:

$\setminus m a t h b f \left(I = \frac{1}{2} {\sum}_{i} {m}_{i} {z}_{i}^{2}\right)$

where $m$ is the molality and $z$ is the charge (here, the sign doesn't matter).

In water, $\epsilon = 78.54$. I don't know what temperature you are referring to, but I will assume $T = \text{298 K}$.

In $B a {\left(H C {O}_{3}\right)}_{2}$, you have the ions $B {a}^{2 +}$ and $H C {O}_{3}^{-}$. Thus, ${z}_{+} = \text{+2}$ and ${z}_{-} = \text{-1}$.

Next, we need $I$:

$I = \frac{1}{2} {\sum}_{i} {m}_{i} {z}_{i}^{2} = \frac{1}{2} \left[\left(0.004\right) {\left(+ 2\right)}^{2} + \left(0.004\right) {\left(- 1\right)}^{2}\right]$

$\text{= 0.01 m}$

Overall, we have:

• $\textcolor{g r e e n}{\epsilon = 78.54}$
• $\textcolor{g r e e n}{T = \text{298 K}}$
• $\textcolor{g r e e n}{{z}_{+} = \text{+2}}$
• $\textcolor{g r e e n}{{z}_{-} = \text{-1}}$
• $\textcolor{g r e e n}{I = \text{0.01 m}}$

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

$\log {\gamma}_{\pm} = \frac{1.824 \times {10}^{6}}{\epsilon T} ^ \text{3/2} | {z}_{+} {z}_{-} | \sqrt{I}$

$= \frac{1.824 \times {10}^{6}}{78.54 \cdot 298} ^ \text{3/2} | \left(+ 2\right) \left(- 1\right) | \sqrt{0.01}$

$= 0.10188$

$\textcolor{b l u e}{{\gamma}_{\pm}} = {10}^{- 0.10188} \approx \textcolor{b l u e}{0.7909}$

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

${\gamma}_{\pm} \le 1$ because at best, $a = m$ in $\gamma = \frac{a}{m}$.

MEAN ACTIVITY

Next, the mean activity ${a}_{\pm}$ can be determined using the following equations:

$\gamma = \frac{a}{m}$, therefore

$\setminus m a t h b f \left({a}_{\pm} = {\gamma}_{\pm} {m}_{\pm}\right)$

where ${\gamma}_{\pm}$ is the mean activity coefficient, ${a}_{\pm}$ is the mean activity in $\text{m}$, and ${m}_{\pm}$ is the mean ionic molality:

$\setminus m a t h b f \left({m}_{\pm} = {\left({m}_{+}^{\nu +} + {m}_{-}^{\nu -}\right)}^{\text{1/} \nu}\right)$

where $\nu$ is the sum of the stoichiometric coefficients for both ions, ${\nu}_{+}$ and ${\nu}_{-}$ are the stoichiometric coefficients of each ion, and ${m}_{+}$ and ${m}_{-}$ are the molalities of each ion (incorporating stoichiometric coefficients!!!).

For this case:

• $\textcolor{g r e e n}{{\nu}_{+} = 1}$
• $\textcolor{g r e e n}{{\nu}_{-} = 2}$
• $\textcolor{g r e e n}{{m}_{+} = 0.02 \cdot 1 = \text{0.02 m}}$
• $\textcolor{g r e e n}{{m}_{-} = 0.02 \cdot 2 = \text{0.04 m}}$
• $\textcolor{g r e e n}{{\gamma}_{\pm} = 0.7909}$

Now we have everything we need!

${m}_{\pm} = {\left[{\left(0.02\right)}^{1} + {\left(0.04\right)}^{2}\right]}^{\text{1/} \left(1 + 2\right)}$

$= {\left[0.02 + 0.0016\right]}^{\text{1/3}}$

$\text{= "color(green)"0.2785 m}$

We're almost done. Now that we have ${m}_{\pm}$ and ${\gamma}_{\pm}$, we can determine ${a}_{\pm}$ to be:

$\textcolor{b l u e}{{a}_{\pm}} = 0.7909 \cdot \text{0.2785 m}$

"= "color(blue)("0.2203 m")