3.7.4-8. If you were able to solve this problem, you are great, congratulations! If you could not solve it, do not be too upset, as this is a somewhat tricky one. I have met PhD chemists who were not able to find a solution to this problem. On mixing solutions of AlCl3 and NaOH, first insoluble Al(OH)3 is formed and precipitates out: AlCl3 + 3 NaOH = Al(OH)3↓ + 3 NaCl. However, if NaOH is used in excess, the insoluble amphoteric Al(OH)3 first formed will dissolve due to the reaction producing soluble NaAlO2 (see Figure 3-93, right). To tell which test tube contains which solution, a small amount (say a couple of drops) of either solution should be carefully added to the other solution without agitation. A white precipitate of Al(OH)3 will be formed at the top of the liquid within the time of mixing due to the formation of Al(OH)3. Now, swirl the contents of the tube. If the precipitated Al(OH)3 dissolves on agitation, the solution that was originally in that test tube is NaOH because Al(OH)3 dissolves in NaOH (see above). If the precipitate of Al(OH)3 does not dissolve on thorough mixing, the test tube with the mixed solution originally contained AlCl3 with which insoluble Al(OH)3 does not react.