Question

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

Answer 1

Note: This will be long. Bear with me!

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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:

`\mathbf(log gamma_pm = (1.824xx10^6)/(epsilonT)^"3/2"|z_+z_-|sqrtI)`

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:

`\mathbf(I = 1/2 sum_i m_i z_i^2)`

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 = "298 K"`.

In `Ba(HCO_3)_2`, you have the ions `Ba^(2+)` and `HCO_3^(-)`. Thus, `z_(+) = "+2"` and `z_(-) = "-1"`.

Next, we need `I`:

`I = 1/2 sum_i m_iz_i^2 = 1/2[(0.004)(+2)^2 + (0.004)(-1)^2]`

`"= 0.01 m"`

Overall, we have:

• `color(green)(epsilon = 78.54)`
• `color(green)(T = "298 K")`
• `color(green)(z_(+) = "+2")`
• `color(green)(z_(-) = "-1")`
• `color(green)(I = "0.01 m")`

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

`log gamma_pm = (1.824xx10^6)/(epsilonT)^"3/2"|z_+z_-|sqrtI`

`= (1.824xx10^6)/(78.54*298)^"3/2"|(+2)(-1)|sqrt(0.01)`

`= 0.10188`

`color(blue)(gamma_pm) = 10^(-0.10188) ~~ color(blue)(0.7909)`

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

`gamma_pm <= 1` because at best, `a = m` in `gamma = a/m`.

MEAN ACTIVITY

Next, the mean activity `a_pm` can be determined using the following equations:

`gamma = (a)/(m)`, therefore

`\mathbf(a_pm = gamma_pm m_pm)`

where `gamma_pm` is the mean activity coefficient, `a_pm` is the mean activity in `"m"`, and `m_pm` is the mean ionic molality:

`\mathbf(m_pm = (m_(+)^(nu+) + m_(-)^(nu-))^("1/"nu))`

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:

• `color(green)(nu_(+) = 1)`
• `color(green)(nu_(-) = 2)`
• `color(green)(m_(+) = 0.02*1 = "0.02 m")`
• `color(green)(m_(-) = 0.02*2 = "0.04 m")`
• `color(green)(gamma_pm = 0.7909)`

Now we have everything we need!

`m_pm = [(0.02)^1 + (0.04)^2]^("1/"(1 + 2))`

`= [0.02 + 0.0016]^"1/3"`

`"= "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:

`color(blue)(a_pm) = 0.7909 * "0.2785 m"`

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