# Relation between common ion effect and the temperature of the solution [closed]

Adding extra Na+ in a saturated NaCl/water solution decreases the solubility of NaCl. How does this exactly happen? AFAIK, the solubility of a solid only decreases with the temperature of the solution. Am I to assume that when I put Na+ in a saturated NaCl/water solution the temperature of the solution somehow drops? If that is what happens, how does it actually happen? If I add extra Na+, the extra Na+ will react with Cl- and will create precipitated NaCl. AFAIK, forming new bonds is exothermic. According to that, the temperature of the system should actually increase and the solubility should increase as well. What exactly is happening here?

• "solubility of a solid only decreases with the temperature" - for solubility in pure solvent that's a decent approximation. If you add another salt that's no longer pure. – Mithoron Jun 21 at 19:37
• You are mixing up different effects. The effect of temperature is determined by the enthalpy of solution of the salt. When you examine the effect of another salt (common ion effect) assume the temperature is constant unless told otherwise. – Buck Thorn Jun 22 at 10:21

For example, at $$25$$°C, $$\pu{1000 mL}$$ water can dissolve $$\ce{370 g NaCl (6.33 mol}$$). So $$1$$ Liter of such a saturated solution contains $$\ce{6.33 mol Na+ + 6.33 mol Cl-}$$. We'll see later that the so-called solubility product $$\ce{[Na+][Cl-]}$$ is important and has the numerical value $$\ce{K_s = [Na+][Cl-] = 6.33^2 = 40.1 mol^2 L^{-2}}$$ It should be stated that the dissolution of this amount of NaCl is supposed not to have modified the volume of the solution, which is of course a rough approximation.
If now $$\pu{1.00 mol}$$ of a more soluble salt like $$\ce{NaNO3}$$ is added to this solution, the concentration of the ion $$\ce{Na+}$$ per Liter increases up to $$\pu{6.33 + 1 = 7.33 mol/L}$$. In such a case, the thermodynamical laws of equilibrium require that the concentration of the other ion (here $$\ce{Cl-}$$) must decrease in such a way that the solubility product $$\ce{K_s = [Na+][Cl-]}$$ remains constant. So $$x$$ moles par liter of $$\ce{Na+}$$ and $$\ce{Cl-}$$ must appear as a deposit of $$\ce{NaCl}$$ in the container. The solubility of the ions $$\ce{Cl-}$$ must go down to : $$\ce{[Cl^-| = 6.33 - x mol/L}$$, and $$\ce{[Na+] = 7.33 - x}$$, in such a way that $$\ce{[Na+][Cl-] = K_s = 40.1 mol^2 L^{-2}}$$. This a $$2$$nd degree equation whose solution is : $$\ce{x = 0.48 mol (28 g)}$$.
Si the final concentration of the ions are : $$\ce{[Na+] = 7.33 - 0.48 = 6.85 mol L^{-1}}$$ $$\ce{[Cl-] = 6.33 - 0.48 = 5.85 mol L^{-1}}$$ $$\ce{[NO3^-] = 1.00 mol L^{-1}}$$
• @aNoNyMoUs. The solubility of $\ce{NaCl}$ is $\pu{5.85 mol/L ; 342 g/L}$). This is less than in pure water ($\pu{370 g/L}$) – Maurice Jun 22 at 8:13