Question

If `"14 g"` of calcium nitrate is dissolved in `"200 g"` of water, and it experiences a `70%` dissociation, what is the solution's vapor pressure if the solvent began at the normal boiling point of water? Hint: the normal boiling point occurs at 760 torr.

Answer 1

`"748 torr"`.

(what is the answer in `"mm Hg"`?)

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Vapor pressure of the solution relative to the solvent is given by Raoult's law for ideal solutions:

`P_A = chi_(A(l)) P_A^"*"`

where:

• `P_A` is the vapor pressure of the solvent in the context of the solution.
• `"*"` indicates pure solvent.
• `chi_(A(l))` is the mol fraction of solvent in the solution phase.

When nonvolatile solute is added to a solvent, it decreases its vapor pressure. Thus, the answer will be less than `"760 torr"`.

In this case, we have a strong electrolyte. Normally, we assume 100% dissociation of strong electrolytes, but in this case there is ion pairing to form `"CaNO"_3^(+)`.

We ignore the `"CaNO"_3^(+)` species in solution for simplicity (which is supported by the fact the solution is dilute so that the ions have a hard time finding each other)

The salt dissociates into `"Ca"^(2+)` and `"NO"_3^(-)`, so:

`"Ca"("NO"_3)_2(aq) -> "Ca"^(2+)(aq) + 2"NO"_3^(-)(aq)`

To calculate the new vapor pressure, we'll need the mol fraction `chi` of the ions or of the water:

`chi_("ions") = n_"ions"/(n_"ions" + n_(H_2O))`

`chi_"ions" + chi_(H_2O) = 1`

We would have three mols of ions, `n_"ions"`, for every one mol of calcium nitrate, `n_(Ca(NO_3)_2)`.

For `"14 g"` of the salt...

`n_(Ca(NO_3)_2) = 14 cancel("g Ca"("NO"_3)_2) xx ("1 mol Ca"("NO"_3)_2)/(164.088 cancel("g Ca"("NO"_3)_2))`

`=` `"0.08532 mols Ca"("NO"_3)_2`

Again, three times the mols of ions result from 100% dissociation, so:

`n_"ions" = 3n_(Ca(NO_3)_2) = "0.2560 mols ions"`

With 70% dissociation, we simply have 70% of the mols, or

`0.7n_"ions" = n_"ions"' = "0.1792 mols"`.

Given `"200 g"` of water, we have:

`200 cancel("g H"_2"O") xx ("1 mol H"_2"O")/(18.015 cancel("g H"_2"O")) = "11.10 mols H"_2"O"`

This gives us a mol fraction of:

`chi_("ions")' = (n_"ions"')/(n_"ions"' + n_(H_2O))`

`= "0.1792 mols"/("0.1792 mols" + "11.10 mols")`

`= 0.01588`

or since all mol fractions for a solution when added together equal `1`,

`chi_(H_2O) = 1 - chi_("ions")' = 0.9841`

Therefore, the solution vapor pressure is:

`color(blue)(P_A) = 0.9841 cdot "760 torr"`

`=` `color(blue)("748 torr")`