Why can solubility be used instead of K in van't Hoff equation?

[_Greg_]

Just done an experiment where you dissolve benzoic acid in water at different temperatures and different volumes of water.

From the van't hoff equation you then plot a graph of ln(solubility) against 1/Temp which produces a straight line with gradient -ΔH/RT from which you get the enthalpy change.

(vant hoff equation): lnk = -ΔH/RT + constant

The first question was:

why isn't it necessary to express solubility in mol/l despite ΔH beight evaluated in J/mol?

I answered with:

It isn’t necessary to convert solubility into mol/l because it’s the difference in solubility that’s important. Therefore plotting solubility with either units will always produce a line of the same gradient but with different Y intercept (the intercept being insignificant)

Is that ok?

The second question I am stuck on is:

Why is it that solubility, S, may be used instead of the equilibrium constant, K, in the van't hoff equation?

I would guess they are proportional to each other or something?

would really appreciate any help

 

[ShawnD]

(vant hoff equation): lnk = -ΔH/RT + constant

why isn't it necessary to express solubility in mol/l despite ΔH beight evaluated in J/mol?

I answered with:

It isn’t necessary to convert solubility into mol/l because it’s the difference in solubility that’s important. Therefore plotting solubility with either units will always produce a line of the same gradient but with different Y intercept (the intercept being insignificant)

Is that ok?

I believe you are right. As long as -H/RT is using the correct units, the slope should be the same. Remember that this is only true if the units being used have a direct linear correlation to the other units being used.

Why is it that solubility, S, may be used instead of the equilibrium constant, K, in the van't hoff equation?

I would guess they are proportional to each other or something?

Correct, but you should probably say why that is true. Wikipedia has an explanation of solubility equilibrium. Under nonionic it says this:

If the activity of the substance in solution is constant (i.e. not affected by any other solutes that may be present) it may be replaced by the concentration.

$$K_s = \left[\mathrm{{C}_{12}{H}_{22}{O}_{11}}(aq)\right]\$$

The square brackets mean molar concentration in mol dm-3 (sometimes called molarity with symbol M).

This statement says that water at equilibrium with solid sugar contains a concentration equal to K. For table sugar (sucrose) at 25 °C, K = 1.971 mol/L.

Remember that this is only true because benzoic acid is a weak acid meaning the majority of it is COOH (which applies to the above quote). Only a tiny fraction of it is the ionic COO- form, to which Ka applies, but it's so small you can probably just ignore it.

edit: Firefox actually copies tex code from wikipedia. Fascinating.