The pH of a 0.050 M solution of CaCl₂ is adjusted such that it maintains a value of 8.0. Which statement below is true regarding whether a precipitate of Ca(OH) ₂ will form?
K _sp[Ca(OH) ₂] = 6.5×10^{- 6}

Question

Quizplus · Accepted Answer

Since the pH of the solution is adjusted to 8.0, there will be OH- ions present in the solution. The reaction that occurs is:
Ca2+ + 2OH- → Ca(OH)2
The Ksp expression for Ca(OH)2 is:
Ksp[Ca(OH)2] = [Ca2+][OH-]2 = 6.5×10-6
Since the pH is adjusted to 8.0, the concentration of OH- can be calculated using the Kw expression:
Kw = [H+][OH-] = 1.0×10-14
[OH-] = 1.0×10-14/[H+] = 1.0×10-14/10-8 = 1.0×10-6 M
Substituting this value in the Ksp expression gives:
Ksp[Ca(OH)2] = [Ca2+][OH-]2 = (1.0×10-6)2[Ca2+] = 1.0×10-12[Ca2+]
Since the concentration of Ca2+ is 0.050 M, the Qsp can be calculated as follows:
Qsp = [Ca2+][OH-]2 = (0.050)(1.0×10-6)2 = 5.0×10-14
Comparing Qsp and Ksp, we can see that Qsp < Ksp. This means that the solution is not saturated and a precipitate of Ca(OH)2 will not form. Therefore, the correct answer is B.

The PH of a 0