When a-lactose is added in excess to water at 20°C, about 7 g per 100 g dissolves immediately (this is the true solubility of a-lactose). Some a-lactose then mutarotates to b-lactose until an equilibrium of 62.7b:37.3a is achieved. The solution is now unsaturated with respect to a-lactose and since the solubility of b-lactose is higher than that of a-, as b-lactose is produced by mutarotation, more lactose goes into solution until, at equilibrium, the final solubility is 18.2 g lactose per 100 g water (7 g a-lactose plus 11.2 g b-lactose).
If excess b-lactose is added to water, its initial solubility is ~50 g per 100 g. Some b-lactose then mutarotates to a-lactose to establish an equilibrium of 62.7b:37.3a. However, at this ratio and starting with 50 g b-lactose, the solution would contain 30.8 g b- and 19.2 g a-lactose, and thus would be supersaturated with respect to a-lactose. Some a-Lactose then crystallizes out of solution, upsetting the equilibrium and leading to further mutarotation from b- to a-lactose. These two events, i.e., crystallization and mutarotation continue until two criteria are met: 7 g a-lactose per 100 g in solution and a ratio of 62.7b:37.3a. The final solubility is the same, 18.2 g lactose (a + b), whether one starts with a- or b-lactose. However, since b-lactose is more soluble than a-lactose and mutarotation is slow, it is possible to form more highly saturated solutions by dissolving b- rather than a-lactose. However, the final equilibrium concentration and ratio is the same.