${H}_2O(s)\leftrightharpoons{H}_2O(l)$

A) the mass of ice must equal to the mass of water B) the mass water and the mass of ice each remain constant

At equilibrium, the amount of reactant being converted to product is equal to the amount of product being converted to reactant. Therefor, the concentration of both the reactant and product are no longer changing and the ratio of these no longer changing concentrations can be represented as a fixed number or constant - the equilibrium constant. $$\mathrm{K_{eq}=\frac{[Product]}{[Reactant]}}$$ where the brackets denote concentrations. In the case at hand, a liquid-solid equilibrium, the expression becomes (thanks to DavePhD for pointing this out, see his UC Davis link in the comments if you'd like more info.) $$\ce{K_{eq}=[H2O~ (l)]}$$ This equation tells us that at equilibrium, the concentration of liquid water is a constant, it is not changing. If the concentration of liquid water is not changing, then the concentration of ice cannot be changing. This does not say that the concentration (or mass, since concentration is related to mass divided by molecular weight per unit volume) of liquid water must equal the concentration or mass of ice; it just says that the concentrations or masses of liquid water and ice must be constant. Hence "B" is the correct choice.

• Does that definition of equillibrium constant apply to a two phase system? Whether I have a ton of ice and a kg of liquid water or a ton of liquid water and a kg of ice, they are still in equillibrium as long as they are at the freezing point. – DavePhD Jun 5 '14 at 19:39
• Yes the definition still applies, but in a two-phase system not all parts of the system are in equilibrium. In your example, only the ice that is in contact with the water could eventually be in equilibrium with the water. – ron Jun 5 '14 at 19:50
• Isn't that the formula for density? – most venerable sir Jun 5 '14 at 19:57
• you mean where I said, "mass divided by molecular weight per unit volume"? density is just mass per unit volume – ron Jun 5 '14 at 20:07
• ron, "A heterogeneous equilibrium is more complicated than homogeneous examples discussed above and has species present in more than one phase. Writing an expression for K for a heterogeneous equilibrium is similar to homogeneous reactions, with the important difference that you don't include any term for a solid in the equilibrium expression." chemwiki.ucdavis.edu/Physical_Chemistry/Equilibria/… – DavePhD Jun 5 '14 at 20:23

If the ice and liquid water are in equillibrium, and no heat is added to or removed from the system, B) the mass of water and the mass of ice each remain constant