# How do equilibrium shifts affect solids?

Say we have the following reversible reaction: $$\ce{NaOH{(s) }<=> Na^+{(aq) }+ OH^{-}{(aq)}} +10.6 \mathrm{kcal}$$

If we add $\ce{OH^-}$ and equilibrium shifts to the left, does that affect the amount of $\ce{NaOH{(s)}}$ present or does it remain constant?

I'm confused because I remember that pure solids and liquids don't affect equilibrium value, or does that only apply when we're talking about adding pure solids and liquids..not the effect on them by an equilibrium shift?

When a particular chemical process is at equilibrium, the respective rates of the forward and backward reactions are equal, meaning that no net change in the concentrations of products and reactants occurs and the composition of the reaction mixture remains stable. If $\ce{OH-}$ is added to a solution already at equilibrium, then there will be an excess of product relative to reactants and the rate of the reverse reaction will increase relative to the forward reaction until equilibrium is reestablished. This means that the ions will recombine into a crystal lattice and form a precipitate. So, to answer your first question, no, the amount of $\ce{NaOH_{(s)}}$ does not remain constant; more of it will be formed if additional ions are added to a solution already at equilibrium.