Answer :
Final answer:
Based on the analysis, a precipitate will form in Beaker 3, where the Ca(OH)₂ solution is added to the 1.0 M CaCl₂ solution.
So, the correct answer is c) Beaker 3
Explanation:
To determine in which beakers a precipitate will form, we need to consider the solubility product constant (Ksp) and the reaction that occurs between the solutes in each beaker.
The Ksp of Ca(OH)₂ is given as 6.5 x 10⁻⁶. The dissociation reaction of Ca(OH)₂ is:
Ca(OH)₂(s) ⇌ Ca²⁺(aq) + 2OH⁻(aq)
The Ksp expression for this reaction is:
Ksp = [Ca²⁺][OH⁻]²
Given the Ksp value, we can determine the solubility of Ca(OH)₂ in water.
Addition of Ca(OH)₂ Solution to Beakers
Now, let’s consider the addition of 50 ml of the 1.0 M Ca(OH)₂ solution to each of the beakers:
Beaker 1 contains 50 ml of 1.0 M HCl. When Ca(OH)₂ is added to this beaker, the following reaction occurs:
Ca²⁺(aq) + 2OH⁻aq) + 2H+(aq) ⇌ CaCl₂(aq) + H₂O(l)
Since HCl is a strong acid, it will completely dissociate, and the H+ ions will react with the OH- ions from Ca(OH)₂. This will result in the formation of CaCl₂ and water, and no precipitate will form.
Beaker 2 contains 50 ml of 1.0 M NaCl. Adding Ca(OH)₂ to this beaker will not result in a reaction, as NaCl is a neutral salt and will not affect the concentration of H+ or OH- ions in the solution.
Beaker 3 contains 50 ml of 1.0 M CaCl₂. Adding Ca(OH)₂2 to this beaker will result in the following reaction:
Ca²⁺(aq) + 2OH-⁻aq) ⇌ Ca(OH)₂(s)
This reaction is the reverse of the dissociation reaction of Ca(OH)₂, and it will shift to the left to counteract the increase in the concentration of Ca²⁺ ions. As a result, a precipitate of Ca(OH)2 will form.
Complete question:
Ksp of Ca(OH)2 = 6.5 x 10⁻⁶
- Breker 1) 50 ml 1.0 M HCl
- Breker 2) 50 ml 1.0 M NaCl
- Breker 3) 50 ml 1.0 M CaCl₂
If 50 ml of the solution from part a is added to each of the beakers shown here, in which beakers, if any, will a precipitate form?
a) Beaker 1
b) Beaker 2
c) Beaker 3
d) No precipitate will form in any beaker
