I heard that if I saturate some water with NaCl and put it in a closed box, the relative humidity near the water surface should be around 75 % at room temperature.

I am using this to calibrate some electronic sensors measuring relative humidity.

If a saturated solution gives 75 % RH, how can i know which RH more diluted mixtures should result in?


My setup is a small 1l plastic container where in I put a mixture of table salt (iodized) and tap water (about 150 ml). The sensor is hanging a few cm above the water surface and measures for a day or so.

The solutions I use are:

  • Fully saturated water (Plenty of undissolved salt crystals)
  • 75 % saturated water (3 parts saturated, 1 part pure water)
  • 66 % saturated water (2 parts saturated, 1 part pure water)
  • 50 % saturated water (1 part saturated, 1 part pure water)

I am also monitoring the temperature in the container, and it is usually swinging a few degrees below 20 °C.

  • $\begingroup$ What do you mean by "part"? Mass, volume? $\endgroup$ – aventurin Oct 8 '17 at 14:43
  • $\begingroup$ I mean volume.. $\endgroup$ – user1361521 Oct 9 '17 at 9:01

Your information on relative humidity (RH) above saturated $\ce{NaCl}$ is valid within about 0.5% over the range of 0 °C to 80 °C. For other standard humidity solutions, such as $\ce{MgCl2}$ or $\ce{K2CO3}$ the U. S. National Institute of Standards & Technology provides a convenient reference.

However, though it is possible to adjust RH by varying concentration, there are, "special problems... because their concentrations must be measured and controlled." If you need only a few accurate RH points, it would be simpler to use saturated solutions of some alternate salts.

BTW, it may take some time for RH to equilibriate because of stratification. It would help to keep the solution on a slow magnetic stirrer, moving not only the liquid but causing eddies in the air above it.

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  • $\begingroup$ I don't need very accurate calibration. Most interesting for me is to be able to detect 90% RH, and 5% accuracy would be fine. $\endgroup$ – user1361521 Oct 8 '17 at 7:29

For ideal behaviour the vapour pressure above the sodium chloride solution is given by Raoult's law as

$$p_w = p_w^{\star}~x_w \tag{1}$$

where $p_w^{\star}$ is the vapour pressure of pure water, and $x_w$ the mole fraction of water in the solution.

Therefore the relative humidity

$$\mathrm{RH} = \frac{p_w}{p_w^{\star}} = x_w \tag{2}$$

equals the mole fraction of water in the solution. Note that this is independent of temperature.

The molar mass of water is $18.02~\pu{g mol^{-1}}$, that of sodium chloride is $58.44~\pu{g mol^{-1}}$.

The solubility of sodium chloride at $20\pu{°C}$ is $359.2~\pu{g l^{-1}}$.

The density of saturated sodium chloride solution at $20\pu{°C}$ is $1.200~\pu{g cm^{-1}}$.

From this it can be calculated that $1$ litre of saturated sodium chloride solution at $20\pu{°C}$ contains $n_w = 46.65~\pu{mol}$ water and $n_s = 6.146~\pu{mol}$ sodium chloride.

Thus the relative humidity above the solution is

$$\mathrm{RH} = x_w = \frac{n_w}{n_w + 2 \cdot n_s} = 0.791 \tag{3}$$

The factor $2$ in the denominator of equation $3$ comes from the assumption that $\ce{NaCl}$ as a strong electrolyte is completely dissociated in aqueous solution.

The actual behaviour of sodium chloride is not ideal, so the real value for the relative humidity may differ from the calculated one.

Since the mole fraction of water increases for more diluted solutions it follows that you can only prepare relative humidity in the range from $100$ to $79.1$ percent using sodium chloride solution.

The values for diluted solutions can be calculated using equation $3$. However, if you need more accurate values, you had to use tabulated ones.

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