Final answer:
The pOH of a 0.08 M aqueous solution of Ca(OH)2 is 0.80.
Step-by-step explanation:
The pOH of a 0.08 M aqueous solution of Ca(OH)2 can be calculated using the concentration of hydroxide ions ([OH-]). First, calculate the concentration of hydroxide ions by multiplying the concentration of Ca(OH)2 by 2, since there are two OH ions for every formula unit of Ca(OH)2. In this case, [OH-] = 2 * 0.08 M = 0.16 M.
Next, use the pOH formula: pOH = -log([OH-]). Plug in the value of [OH-] to calculate the pOH: pOH = -log(0.16) = 0.80.
Therefore, the pOH of the solution is 0.80. The correct answer is option A) 10.30.