Question

Determine the mean activity coefficient and mean activity of a 0.004 molal of $Ba{(HC{O_3})_2}$?

Answer 1

We can solve this problem by using the Debye-Huckel Limiting Law, which relates the mean activity coefficient with the ionic strength (I). The ionic strength is the summation of the product of the concentration (molal) and the charge on the respective disintegrated ion.

Complete answer:
The Debye-Huckel Theory gives us an equation for calculating the mean activity coefficient denoted by $\gamma \pm$ for solutions having concentrations less than or equal to 0.01 M. The equation can be given as:
$\log \gamma \pm = \dfrac{{1.824 \times {{10}^6}}}{{{{(\varepsilon T)}^{3/2}}}} \times |{Z^ + }{Z^ - }|\sqrt I$
Where, $\varepsilon$ is the dielectric constant, ${Z^ + },{Z^ - }$ is the charge on the cation and anion respectively T is the temperature and I is the ionic strength.
To find the mean activity coefficient we’ll first have to find out the ionic strength of the 0.004 molal $Ba{(HC{O_3})_2}$ solution. The ionic strength can be given by the formula: $I = \dfrac{1}{2}\sum {{m_i}{z_i}^2}$
Where m is the concentration of the respective ion in molal and z is the charge on that ion.
The disintegration of the compound given to us can hereby happen like this:
$Ba(HC{O_3}) \to B{a^{ + 2}} + 2HCO_3^ -$
The charge on Ba is +2 and that on $HC{O_3}$ is -1. The ionic strength therefore can be calculated as:
$I = \dfrac{1}{2}[(0.004){( + 2)^2} + (2 \times 0.004){( - 1)^2}]$
$I = \dfrac{1}{2}[(0.004 \times 4) + (2 \times 0.004)]$
$I = \dfrac{1}{2}[0.016 + 0.008]$
$I = \dfrac{1}{2}[0.024] = 0.012m$
Now that we know the ionic strength I we can calculate the mean activity coefficient $\gamma \pm$ by substituting the respective values. Since the temperature isn’t given we’ll consider it as 298 K.
The information given to us is:
$\varepsilon = 78.54,T = 298K,{Z^ + } = + 2,{Z^ - } = - 1,I = 0.012m$
Substituting the values in the equation above:
$\log \gamma \pm = \dfrac{{1.824 \times {{10}^6}}}{{{{(78.54 \times 298)}^{3/2}}}} \times |2 \times ( - 1)|\sqrt {0.01}$
$\log \gamma \pm = 0.10188$
To find the value of $\gamma \pm$ we’ll take the antilog on both sides.
$\gamma \pm = AL(0.10188) = 0.7909$
Next we’ll have to determine the activity coefficient denoted by $a \pm$ . The relationship between activity and mean activity coefficient can be given as: $\gamma = \dfrac{a}{m}$
Therefore, $a \pm = \gamma \pm .m \pm$
To find $a \pm$ we’ll have to know the value of mean ionic molality (m). The formula for finding the same is: $m \pm = {(m_ + ^{v + } + m_ - ^{v - })^{1/v}}$
Where ${v^ + },{v^ - }$ are the stoichiometries of the respective ions and ${m^ + },{m^ - }$ are the molalities of the respective cations and anions. $v$ is the sum of the stoichiometries of both cations and anions.
The values of: ${v^ + } = 1,{v^ - } = 2,{m^ + } = 0.02m,{m^ - } = 0.02.2 = 0.04m,\gamma \pm = 0.7909$
$m \pm = {[{(0.02)^1} + {(0.04)^2}]^{1/(1 + 2)}}$
$m \pm = {[(0.02) + 0.0016]^{1/3}} = 0.2785m$
Now that we know the value of m, we can find the value of a: $a \pm = 0.2785 \times 0.7909 = 0.2203m$.

Note:
Since the charges do not have any units, the units of activity and mean activity coefficients are the same as that of concentrations i.e. molal. If we are told to find only the mean activity coefficient we can stop right after finding the values for $\gamma \pm$ .