Final answer:
The pOH of a 0.169 M Ca(OH)2 solution is approximately 0.47, calculated by doubling the molarity for the dissociation into two hydroxide ions and then taking the negative logarithm of that concentration.
Step-by-step explanation:
To determine the pOH of a 0.169 M Ca(OH)2 solution, we begin by calculating the hydroxide ion concentration, [OH-]. Since Ca(OH)2 is a strong base that dissociates completely in solution, each molecule yields two OH- ions. Thus, the [OH-] is double the concentration of the Ca(OH)2 solution.
[OH-] = 2 × [Ca(OH)2] = 2 × 0.169 M = 0.338 M
The next step is to find the pOH:
pOH = - log [OH-] = - log (0.338) ≈ 0.47
Therefore, the pOH of a 0.169 M Ca(OH)2 solution is approximately 0.47.