Question

Equilibrium constants is given (in atm) for the following reaction $0^{\circ }C$:
$N{a}_{2}HP{O}_4.12H_{2}O\left(s\right)\rightleftharpoons N{a}_{2}HP{O}_4.7H_{2}O\left(s\right)+5H_{2}O\left(g\right);K_p=2.43\times {10}^{-13}$

The vapour pressure of water at $0^{\circ }C$ is $4.56torr$.
At what relative humidities will $N{a}_{2}HP{O}_4.12H_{2}O\left(s\right)$ be efflorescent when exposed to air at $0^{\circ }C$?

Answer 1

Efflorescence is the loss of water of crystallization of a salt to the atmosphere. If the water vapor pressure in the salt is more than the water vapor pressure in the atmosphere, it will effloresce.

Equillibrium constant for the given equillibrium at $0^{\circ }C$ is:

$N{a}_{2}HP{O}_4.12H_{2}O\left(s\right)\rightleftharpoons N{a}_{2}HP{O}_4.7H_{2}O\left(s\right)+5H_{2}O\left(g\right)$

and $K_p=2.43\times {10}^{-13}$

Using $K_p={P_{H_{2}O}}^5$ as solids are taken as unity and $P_{H_{2}O}$ = partial vapour pressure of water

$\therefore p_{H_{2}O}=3\times {10}^{-3}atm$

or $p_{H_{2}O}=3\times {10}^{-3}atm\times \left(760torr/atm\right)=2.28torr$

Given that the vapour pressure of water in air at $0^{\circ }C$ is $4.56torr$.

Since $p_{H_{2}O}$ is less than the water vapor pressure in the atmosphere at the same temperature. $N{a}_{2}HP{O}_4.12H_{2}O\left(s\right)$ will effloresce when the water vapor pressure in the air will be less than 2.28 torr.

Thus the relative humidity of the air should be more than (or atleast) $\frac{2.28}{4.56}=0.5$ or 50%.