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

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

• Try to reformulate your question to make more clear, what exactly you are asking about. Jan 10 at 13:40

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.

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 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
• How do you obtain the volume of a solute, e.g., NaOH in this case? Jan 10 at 13:58
• Sorry, I am still thinking. You mean the density of solid NaOH? Jan 10 at 14:08
• @Achem Both solid and solution. V(NaOH(s))=m(NaOH(s)) /rho(NaOH(s)) // You wieght NaOH, obtain its density and calculate its volume. Jan 10 at 14:10
• Okay. Thanks for the clarification. Jan 10 at 14:14