Is the process of volume decrease in NaOH solutions reversible during evaporation of water from the solution? [closed]








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While preparing w/w 50% NaOH solution I noticed a decrease in solution volume (5-10mL) in a 2L volumetric flask, so I am wondering if there is an increase in water volume during the water evaporation or concentration of NaOH? Is it the same amount and is it measurable?


[QuantumYitian question regarding the decrease of volume during the dissolution of NaOH](https://chemistry.stackexchange.com/questions/167434/why-is-there-a-decrease-in-the-total-volume-when-naoh-dissolves-in-water)




There is no water volume in solution. $\ce{Na+}$ and $\ce{OH-}$ ions and $\ce{H2O}$ molecules are packed together in solution in average tighter than is the average packing tightness for $\ce{NaOH(s)}$ and $\ce{H2O(l)}$ alone.


At water evaporation, the mass concentration of $\ce{NaOH}$ obviously increases until solution saturation. After that, extra $\ce{NaOH}$ crystalizes.


If all water evaporates and then condenses, the initial volumes are obtained, if we neglect experimental troubles of removing all water.




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The difference of volumes during dissolution (assuming the same initial and final temperature) can be determined from experimental or tabulated values of volume, mass or density:


* Measured mass/volume and measured/known density of solute
* Measured volume/mass and measured/known density of solvent
* Measured volume or measured/tabulated density of solution at give concentration


Many solutions, including $\ce{NaOH}$ have [tabulated densities](https://www.handymath.com/cgi-bin/naohtble3.cgi?submit=Entry) that can be interpolated and the volume difference calculated.


Here is the simple formula to calculate the difference:


$$\Delta V = V\_\mathrm{solution} - V\_\mathrm{solute} - V\_\mathrm{solvent} = \\
\frac{m\_\mathrm{solute} + m\_\mathrm{solvent}}{\rho\_\mathrm{solution}} - \frac{m\_\mathrm{solute}}{\rho\_\mathrm{solute}} - \frac{m\_\mathrm{solvent}}{\rho\_\mathrm{solvent}}=
\\
\frac{m\_{\ce{NaOH}} + m\_{\ce{H2O}}}{\rho\_\mathrm{solution}} - \frac{m\_{\ce{NaOH}}}{\rho\_{\ce{NaOH(s)}}} - \frac{m\_{\ce{H2O}}}{\rho\_{\ce{H2O}}}=
\\
\frac{m\_{\ce{NaOH}} + V\_{\ce{H2O}}\cdot \rho\_{\ce{H2O}} }{\rho\_\mathrm{solution}} - \frac{m\_{\ce{NaOH}}}{\rho\_{\ce{NaOH(s)}}} - V\_{\ce{H2O}}$$


* $V$ is volume
* $\rho$ is density
* $m$ is mass


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