Does the solid Zn(OH)2 dissolve when concentrated lye, NaOH solution, is added to a beaker containing a saturated solution of Zn (OH)2 and a small amount of Zn(OH)2?
Yes. The Zn(OH)2 is amphoteric.
To determine if the solid Zn(OH)2 will dissolve when concentrated lye (NaOH) solution is added to a beaker containing a saturated solution of Zn(OH)2 and a small amount of Zn(OH)2, we need to consider the solubility of Zn(OH)2 and the effect of adding a concentrated NaOH solution.
The solubility of a compound refers to its ability to dissolve in a solvent. For Zn(OH)2, it is a sparingly soluble compound, meaning it only dissolves to a limited extent in water. When a saturated solution of Zn(OH)2 is already present in the beaker, it indicates that the maximum amount of Zn(OH)2 that can dissolve in the given amount of water has already been dissolved.
Now, when concentrated NaOH solution is added to the beaker, it introduces a high concentration of hydroxide ions (OH-) into the solution. These hydroxide ions can react with the dissolved Zn2+ ions from the Zn(OH)2, forming Zn(OH)4^2- ions through the following reaction:
Zn(OH)2 + 4OH- ⇌ Zn(OH)4^2-
Since the formation of Zn(OH)4^2- ions increases the solubility of Zn, it allows more Zn(OH)2 to dissolve in the solution. Therefore, the solid Zn(OH)2 would indeed dissolve when concentrated NaOH solution is added.
In summary, based on the solubility of Zn(OH)2 and the introduction of hydroxide ions from the concentrated NaOH solution, the solid Zn(OH)2 will dissolve in the presence of the added solution.