A saturated solution of NaOH is formed in a beaker. The change is written as NaOH(s) ⇌ Na+1 (aq) + OH-1 (aq) + 44.51 kJ,

Question

A saturated solution of NaOH is formed in a beaker. The change is written as

NaOH(s) ⇌ Na⁺¹ (aq) + OH^-1 (aq) + 44.51 kJ,

Explain, in terms of common ions, why adding NaCl (aq) to this change will cause NaOH (s) to precipitate out

Answer 1

In a saturated solution, there is an equilibrium between the solid and the soluble components.

NaOH (s) <======> Na¹⁺ + OH^1-. By adding NaCl which dissociates to Na¹⁺ + Cl^1-, the Na¹⁺ is an ion in common with the Na¹⁺ ions in equilibrium with solid NaOH. By LeChatelier's principle, the equilibrium will shift to the left to undue the stress to the equilibrium by the addition of additional Na¹⁺. In other words, more solid NaOH will be produced which will reduce the concentration of Na¹⁺.