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

1 Answer
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:

#\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")#