# 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"?

## 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 ${\text{0.787 g/cm}}^{3}$, while its dielectric constant is ${\epsilon}_{r} = 32.6$. Main Equation: $\ln {\gamma}_{\pm} = - | {z}_{+} {z}_{-} | \left(A {I}_{m}^{\text{1/2")/(1 + BaI_m^"1/2}}\right)$ Sub-Equations: $A = {\left(2 \pi {N}_{A} {\rho}_{A}\right)}^{\text{1/2"(e^2/(4piepsilon_0 epsilon_(r,A)k_BT))^"3/2}}$ $B = e {\left(\frac{2 {N}_{a} {\rho}_{A}}{{\epsilon}_{0} {\epsilon}_{r , A} {k}_{B} T}\right)}^{\text{1/2}}$ Molal-scale Ionic Strength: ${I}_{m} = {\sum}_{i} {z}_{i}^{2} {m}_{i} = \stackrel{\text{If single strong electrolyte}}{\overbrace{\frac{1}{2} | {z}_{+} {z}_{-} | \left({\nu}_{+} + {\nu}_{-}\right) {m}_{i}}}$ Assume $a \approx 3 \times {10}^{- 10} \text{m}$, 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 $\text{mol/kg}$. ${N}_{A}$ is Avogadro's number, $6.022 \times {10}^{23} {\text{mol}}^{- 1}$. ${\rho}_{A}$ is the density of the solvent in ${\text{kg/m}}^{3}$. ${\epsilon}_{0} = 8.854 \times {10}^{- 12}$ ${\text{C"^2"/N"cdot"m}}^{2}$ is the vacuum permittivity. ${\epsilon}_{r , A}$ is the dielectric constant of the solvent, with no units. ${k}_{B} = 1.3807 \times {10}^{- 23}$ $\text{J/K}$ is the Boltzmann constant. $T$ is the temperature in $\text{K}$. $e$ is the proton charge, $1.602 \times {10}^{- 19}$ $\text{C}$

Nov 21, 2016

From the definitions above, I will evaluate $A$ and $B$, followed by ${I}_{m}$, and then $\ln {\gamma}_{\pm}$, and then ${\gamma}_{\pm}$.

Sub-Equation Evaluations

$\textcolor{g r e e n}{A} = {\left(2 \pi {N}_{A} {\rho}_{A}\right)}^{\text{1/2"(e^2/(4piepsilon_0 epsilon_(r,A)k_BT))^"3/2}}$

= (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"

$= \textcolor{g r e e n}{3.8885}$ color(green)(("kg"/"mol")^"1/2")

$\textcolor{g r e e n}{B} = e {\left(\frac{2 {N}_{A} {\rho}_{A}}{{\epsilon}_{0} {\epsilon}_{r , A} {k}_{B} T}\right)}^{\text{1/2}}$

$= \left(1.602 \times {10}^{- 19} \text{C}\right)$
[(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"

$= \textcolor{g r e e n}{4.5245 \times {10}^{9}}$ 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:

$\textcolor{g r e e n}{{I}_{m}} = \frac{1}{2} {\sum}_{i} {z}_{i}^{2} {m}_{i}$

$= \frac{1}{2} \left[{z}_{+}^{2} {m}_{+} + {z}_{-}^{2} {m}_{-}\right]$

$= \frac{1}{2} \left[{z}_{+}^{2} {\nu}_{+} {m}_{i} + {z}_{-}^{2} {\nu}_{-} {m}_{i}\right]$

$= \frac{1}{2} \left[| {z}_{+} {z}_{-} | {\nu}_{+} {m}_{i} + | {z}_{+} {z}_{-} | {\nu}_{-} {m}_{i}\right]$

$= \frac{1}{2} | {z}_{+} {z}_{-} | \left({\nu}_{+} + {\nu}_{-}\right) {m}_{i}$

$= \frac{1}{2} | 1 \cdot - 1 | \left(1 + 1\right) \left(\text{0.0200 mol HCl"/"kg methanol") = color(green)("0.0200 mol HCl"/"kg methanol}\right)$

Main Equation Evaluation

Now that those are calculated, we can use them in the main equation to get ${\gamma}_{\pm}$. Again, assuming $a \approx 3 \times {10}^{- 10} \text{m}$.

$\ln {\gamma}_{\pm} = - | {z}_{+} {z}_{-} | \left(A {I}_{m}^{\text{1/2")/(1 + BaI_m^"1/2}}\right)$

$= - | 1 \cdot 1 |$
[(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:

$\textcolor{b l u e}{{\gamma}_{\pm}} = {e}^{\ln {\gamma}_{\pm}} = \textcolor{b l u e}{0.6304}$