1. Imagine you have a saturated solution of Ca(OH)2, at equilibrium, that has some undissolved
Ca(OH)2 precipitated on the bottom of the test tube. If you add an additional scoop of Ca(OH)2 to
the solution will concentration of [OH-] in the solution (increase, decrease, or be unchanged)?
Explain. Assume you let the solution reach equilibrium again.
Unchanged because the solution is already in equilibrium of ions.
2. Imagine you have a saturated solution of Ca(OH)2, at equilibrium, that has some undissolved
Ca(OH)2 precipitated on the bottom of the test tube. If you add pure water to the test tube will
the amount of solid Ca(OH)2 at the bottom (increase, decrease, or be unchanged)? Explain.
Assume you let the solution reach equilibrium again.
Shift to the right because the increased volume of solution with reduce the concentration of the products so the reactant solid with decrease and the product ions will increase until it reaches equilibrium.
3. Imagine you have a saturated solution of Ca(OH)2, at equilibrium, that has some undissolved
Ca(OH)2 precipitated on the bottom of the test tube. If you add pure water will the concentration
of [OH-] in the solution (increase, decrease, or be unchanged)? Explain. Assume you let the
solution reach equilibrium again and the solution remains saturated.
Unchanged because while the mass of [OH-] ions will increase the concentration in the solution will be the same because that is the equilibrium.
So I answered these they way I thought was correct but I wanted to make sure. I have trouble understanding number 2 in particular.
I thought you did a super job.
Here are step-by-step explanations for each question:
1. Imagine you have a saturated solution of Ca(OH)2, at equilibrium, that has some undissolved Ca(OH)2 precipitated on the bottom of the test tube. If you add an additional scoop of Ca(OH)2 to the solution, will the concentration of [OH-] in the solution (increase, decrease, or be unchanged)? Explain. Assume you let the solution reach equilibrium again.
- When you add more Ca(OH)2 to the solution, it will dissolve and contribute more hydroxide ions (OH-) to the solution.
- Since the added compound will dissociate to form OH- ions, the concentration of [OH-] in the solution will increase.
- The solution will reach a new equilibrium point with a higher concentration of hydroxide ions.
2. Imagine you have a saturated solution of Ca(OH)2, at equilibrium, that has some undissolved Ca(OH)2 precipitated on the bottom of the test tube. If you add pure water to the test tube, will the amount of solid Ca(OH)2 at the bottom (increase, decrease, or be unchanged)? Explain. Assume you let the solution reach equilibrium again.
- Adding pure water to the test tube will dilute the solution.
- The increased volume of the solution will decrease the concentration of the dissolved Ca(OH)2 ions.
- As a result, according to Le Chatelier's principle, the equilibrium will shift to the right to restore the balance.
- More solid Ca(OH)2 will dissolve to increase the concentration of dissolved ions, eventually reaching a new equilibrium point with a reduced amount of solid Ca(OH)2 at the bottom.
3. Imagine you have a saturated solution of Ca(OH)2, at equilibrium, that has some undissolved Ca(OH)2 precipitated on the bottom of the test tube. If you add pure water, will the concentration of [OH-] in the solution (increase, decrease, or be unchanged)? Explain. Assume you let the solution reach equilibrium again, and the solution remains saturated.
- Adding pure water to the saturated solution does not change the overall concentration of hydroxide ions in the solution.
- Although the mass of OH- ions will increase due to the added water, the volume of the solution will also increase proportionally.
- As a result, the concentration of [OH-] in the solution will remain unchanged.
- The system will reach a new equilibrium that maintains a constant concentration of [OH-].
Let's break down question 2 to better understand it.
The question asks about adding pure water to a saturated solution of Ca(OH)2. In a saturated solution, there is a dynamic equilibrium between the dissolved ions and the undissolved solid Ca(OH)2.
When pure water is added to the test tube, the total volume of the solution increases. This means that the concentration of the dissolved ions, including Ca^2+ and OH-, will decrease as the same amount of ions are now spread over a larger volume.
According to Le Chatelier's principle, when the concentration of reactants or products is changed, the system will try to shift in a direction that will counteract the change. In this case, the system will shift towards the dissolution of more solid Ca(OH)2 (increasing the concentration of dissolved ions) in order to restore equilibrium.
Therefore, by adding pure water, the amount of solid Ca(OH)2 at the bottom of the test tube will decrease as more of it dissolves. The solution will reach a new equilibrium with a higher concentration of dissolved ions but a smaller amount of undissolved solid.
It's important to note that this explanation assumes the solution remains saturated. If the additional water added exceeds the solubility limit of Ca(OH)2, the solution will become unsaturated, and the concentration of dissolved ions will increase even further.