# Increasing the solubility of sodium hydroxide

$\ce{NaOH -> Na+ + OH-} \quad \Delta H^\circ = -44.51\ \mathrm{kJ/mol}$

The dissolution of sodium hydroxide in water is an exothermic process, and so, according to Le Chatelier’s principle, cooling the container should shift the reaction to the right. Shouldn’t this mean that cooling the container will increase the solubility?

According to a solubility chart for sodium hydroxide, heating the container will increase the solubility; why is this the case?

This site explains it.

What I understood from the explanation on that site:

Solubility is defined as the concentration of the solute in a saturated solution. So when considering the increase in solubility with temperature, you have to check the enthalpy of solution of $\ce{NaOH}$ in a saturated solution of $\ce{NaOH}$.

When you dissolve $\ce{NaOH}$ in pure water, the process is highly exothermic, but as the concentration of $\ce{NaOH}$ increases, the process becomes less exothermic and eventually near the saturation point it becomes endothermic (it becomes less favourable for more $\ce{NaOH}$ to dissolve).

So according to Le Chatelier’s Principle, although the dissolution process is overall exothermic, since solubility is determined only at the saturation point, the solubility increases with increase in temperature.

• The website link now points to a generic "Educational Programs" website, and does not make any reference to $\ce{NaOH}$ dissolution. If you found the reasons on this site indeed, then please use a specific link to point to the exact webpage. Otherwise, simply remove the link as your explanation is self-sufficient. Thanks! – Gaurang Tandon Jan 17 '18 at 2:37

This reaction has a positive change of entropy (ΔS > 0). Hence, from the equation ΔG = ΔH - TΔS, we find that its Gibbs energy will decrease with an increase of the temperature. As a results, the solubility of NaOH increases when we increase the temperature (heating the container), since we have ΔG = -RTlnK.

According to

$$\Delta G = -RT \ln K$$

K is dependent on Temperature.

Le Chatelier's principle is explicitly a mass balance issue, and thus is not affected by temperature, although that distinction may be pedantic.

• No. The Free Energy is a description of Le Chaetleier's principle. However temperature is not a 'product' or 'reactant' and so cannot 'shift' the equilibrium in any way traditionally related to the application of Le Chateliers's principle. – Lighthart Mar 21 '13 at 14:34
• No, raising temp does affect the Keq (see Haber process, or NO2 -> N2O4). Plus, change in entropy in this reaction is negative not positive (-11 for OH + 59 for Na - 64 for solid NaOH = -16 J/K*mol). – LanguagesNamedAfterCofee Mar 21 '13 at 19:43
• T does affect Keq, per DG=-RT ln K above. However, this is not leChatelier's principle. Also, the entropies you've quoted are not relevant; you need the solvated ions. – Lighthart Mar 21 '13 at 19:47
• It is part of Le Chatelier's principle... (see here) The entropies above are of the solvated ions Na+ and OH- (see here) Thus, the change in entropy of the reaction is negative. – LanguagesNamedAfterCofee Mar 21 '13 at 23:15
• Le Chateliers principle is explicitly mass balance. Temperature is irrelevant in that context. Using products minus reactant for entropy, 59 - 11 (products) - (-64) reactants = 112, a positive entropy gain, based on your numbers. – Lighthart Mar 21 '13 at 23:28