# Active masses of solids and liquids [duplicate]

In the calculation for equilibrium constant, we take the active mass of solids and liquids as unity. Intuition tells that more amount of a solid reactant should result in more product being formed, but according to the formula for the equilibrium constant it has no role in determining that. Is it unity in all possible cases? How can we assume it to be one and have it not affect the equilibrium calculation?

• Chemists don't call it the active mass, but rather the activity. The activity of a solid is taken as unity if the reaction is happening in a liquid or gas. The activity of a liquid is taken as unity if the reaction is happening in a gas. // Consider putting some calcium carbonate in pure water. A tiny amount will dissolve. But once the solution is saturated it doesn't matter if there is one microgram or 1 kilogram of calcium carbonate excess, no more calcium carbonate will dissolve. – MaxW Mar 28 '20 at 6:50

A good example of the relevant principle is a mixture of two reactive solids or two immiscible liquids of constant activity that react at the interface between them. In either case, if the reaction is favorable ($$Q\gt K_{eq}$$) then it will proceed until one or both of the substances has been completely consumed (both being completely consumed if present in stoichiometric proportion). On the other hand, when a solid dissolves into a solvent, the activity of the solid remains constant and it will continue to dissolve provided the activity of the dissolved species does not increase above that of the solid. If it doesn't, then the solid is completely dissolved (exhausted).