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

##### 1 Answer

*Note: This will be long. Bear with me!*

**MEAN ACTIVITY COEFFICIENT**

According to **Debye-Huckel Theory**, the equation for calculating the **mean activity coefficient**

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

where

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

where *molality* and

In water,

In

Next, we need

#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.*

*best*,

**MEAN ACTIVITY**

Next, the mean activity

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

#\mathbf(a_pm = gamma_pm m_pm)#

where

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

where *both* ions,

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

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

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