# The Henry's law constant for methylamine (CH_3NH_2) is 36M/bar. What is the equilibrium vapor pressure of methylamine over a solution that is 2.23 mM CH_3NH_2?

Dec 3, 2016

I think the units are off on the Henry's law constant. Shouldn't it be $\text{bar/M}$?

Henry's law is best applied for the solute in a dilute solution. Normally it's written like this, in its mol fraction form:

${P}_{j} = {\chi}_{j}^{l} {k}_{H , j}$

where:

• ${P}_{j}$ is the vapor pressure of solute $j$ in solution.
• ${\chi}_{j}^{l}$ is the mole fraction of $j$ in solution (in the liquid phase).
• ${k}_{H , j}$ is the Henry's law constant, and it is equal to ${P}_{j}^{\text{*}}$, the vapor pressure of pure solute $j$, when one extrapolates data to the limit of infinite dilution (i.e. ${\lim}_{{\chi}_{j}^{l} \to 0} {k}_{H , j} = {P}_{j}^{\text{*}}$).

The equilibrium vapor pressure of methylamine over the solution would be, then, ${P}_{j}$, since we are given ${k}_{H , j}$ and the solution is clearly dilute ($\text{mM}$ is millimolarity, a very small concentration scale).

In a dilute solution, we can assume that ${V}_{\text{solvent" ~~ V_"solution}}$.

Probably the most important assumption is that at such high dilution, we can actually use concentration instead of mol fraction if we wanted to, as long as we have a Henry's law constant in units of pressure over concentration.

Therefore, the vapor pressure of methylamine above the solution is:

$\textcolor{b l u e}{{P}_{\text{MA") = chi_"MA"^l k_(H,"MA") ~~ ["MA"]k_(H,"MA}}}$

$= \left(2.23 \times {10}^{- 3} \text{M")("36 bar/M}\right)$

$=$ $\textcolor{b l u e}{\text{0.08028 bar}}$

which makes sense because the concentration is so small.

When the concentration is small, it means there is little of it available in the solution in the first place, so when it vaporizes, little of it is present in the vapor phase. So, the vapor pressure should be small.