# Use the Debye-Huckel equation to evaluate #gamma_(pm)# for a #"0.0200 mol/kg"# #"HCl"# solution, with methanol as the solvent, at #25^@ "C"# and #"1 atm"#?

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The answer is #gamma_(pm) = 0.6304# . I am writing this answer because this equation is rather difficult to evaluate correctly.

The density of methanol at this #T# and #P# is #"0.787 g/cm"^3# , while its dielectric constant is #epsilon_r = 32.6# .

**Main Equation:**

#ln gamma_(pm) = -|z_(+)z_(-)|(A I_m^"1/2")/(1 + BaI_m^"1/2")#

**Sub-Equations:**

#A = (2piN_A rho_A)^"1/2"(e^2/(4piepsilon_0 epsilon_(r,A)k_BT))^"3/2"#

#B = e((2N_arho_A)/(epsilon_0 epsilon_(r,A) k_B T))^"1/2"#

Molal-scale Ionic Strength:

#I_m = sum_i z_i^2 m_i = stackrel("If single strong electrolyte")overbrace(1/2 |z_(+)z_(-)|(nu_(+) + nu_(-)) m_i)#

Assume #a ~~ 3xx10^(-10) "m"# , or 3 angstroms, the mean ionic radius (including the hydration sphere).

**Definitions:**

#gamma_(pm)# is the activity coefficient of a solution containing some set of strong electrolytes, ignoring ion pairing interactions.
#nu_(+)# (#nu_(-)# ) is the stoichiometric coefficient of the cation (anion).
#z_(-)# (#z_(+)# ) is the charge of the anion (cation).
#m_i# is the molality of the solute in #"mol/kg"# .
#N_A# is Avogadro's number, #6.022xx10^(23) "mol"^(-1)# .
#rho_A# is the density of the solvent in #"kg/m"^3# .
#epsilon_0 = 8.854xx10^(-12)# #"C"^2"/N"cdot"m"^2# is the vacuum permittivity.
#epsilon_(r,A)# is the dielectric constant of the solvent, with no units.
#k_B = 1.3807xx10^(-23)# #"J/K"# is the Boltzmann constant.
#T# is the temperature in #"K"# .
#e# is the proton charge, #1.602xx10^(-19)# #"C"#

The answer is

The density of methanol at this

**Main Equation:**

**Sub-Equations:**

Molal-scale Ionic Strength:

Assume

**Definitions:**

#gamma_(pm)# is the activity coefficient of a solution containing some set of strong electrolytes, ignoring ion pairing interactions.#nu_(+)# (#nu_(-)# ) is the stoichiometric coefficient of the cation (anion).#z_(-)# (#z_(+)# ) is the charge of the anion (cation).#m_i# is the molality of the solute in#"mol/kg"# .#N_A# is Avogadro's number,#6.022xx10^(23) "mol"^(-1)# .#rho_A# is the density of the solvent in#"kg/m"^3# .#epsilon_0 = 8.854xx10^(-12)# #"C"^2"/N"cdot"m"^2# is the vacuum permittivity.#epsilon_(r,A)# is the dielectric constant of the solvent, with no units.#k_B = 1.3807xx10^(-23)# #"J/K"# is the Boltzmann constant.#T# is the temperature in#"K"# .#e# is the proton charge,#1.602xx10^(-19)# #"C"#

##### 1 Answer

From the definitions above, I will evaluate

**Sub-Equation Evaluations**

#color(green)(A) = (2piN_A rho_A)^"1/2"(e^2/(4piepsilon_0 epsilon_(r,A)k_BT))^"3/2"#

#= (2pi(6.022xx10^(23) "mol"^(-1))("787 kg/m"^3))^"1/2"#

#[(1.602xx10^(-19) "C")^2/(4pi(8.854xx10^(-12) "C"^2"/N"cdot"m"^2)(32.6)(1.3807xx10^(-23) "J/K")("298.15 K"))]^"3/2"#

#= color(green)(3.8885)# #color(green)(("kg"/"mol")^"1/2")#

#color(green)(B) = e((2N_Arho_A)/(epsilon_0 epsilon_(r,A) k_BT))^"1/2"#

#= (1.602xx10^(-19) "C")#

#[(2(6.022xx10^(23) "mol"^(-1))("787 kg/m"^3))/((8.854xx10^(-12) "C"^2"/N"cdot"m"^2)(32.6)(1.3807xx10^(-23) "J/K")("298.15 K"))]^"1/2"#

#= color(green)(4.5245xx10^9)# #color(green)(("kg"/"mol")^"1/2" "m"^(-1))#

Now that those crazy ones are out of the way, here's the fairly easy one. The **molal-scale ionic strength** is the measure of total ion concentration, basically. Here's how you could use the equation:

#color(green)(I_m) = 1/2sum_i z_i^2 m_i#

#= 1/2[z_(+)^2m_(+) + z_(-)^2 m_(-)]#

#= 1/2[z_(+)^2 nu_(+)m_i + z_(-)^2nu_(-)m_i]#

#= 1/2[|z_(+)z_(-)|nu_(+)m_i + |z_(+)z_(-)|nu_(-)m_i]#

#= 1/2|z_(+)z_(-)|(nu_(+) + nu_(-))m_i#

#= 1/2|1*-1|(1+1)("0.0200 mol HCl"/"kg methanol") = color(green)("0.0200 mol HCl"/"kg methanol")#

**Main Equation Evaluation**

Now that those are calculated, we can use them in the main equation to get

#ln gamma_(pm) = -|z_(+)z_(-)|(A I_m^"1/2")/(1 + BaI_m^"1/2")#

#= -|1*1|#

#[(3.8885 ("kg"/"mol")^"1/2"("0.0200 mol"/"kg")^"1/2")/(1 + (4.5245xx10^9 ("kg"/"mol")^"1/2" "m"^(-1))(3xx10^(-10) "m")("0.0200 mol"/"kg")^"1/2")]#

#= -0.4614#

Therefore, we finally have that:

#color(blue)(gamma_(pm)) = e^(ln gamma_(pm)) = color(blue)(0.6304)#