Final answer:
The pH of a 0.00046 M Ca(OH)2 solution is determined by finding the [OH-], calculating the pOH, and then obtaining the pH from the relationship pH = 14 - pOH.
Step-by-step explanation:
The pH of an aqueous solution of 0.00046 M Ca(OH)₂ can be calculated by first determining the hydroxide ion concentration ([OH-]). Calcium hydroxide is a strong base that dissociates completely in water to yield two OH- ions per formula unit. Therefore, [OH-] is twice the concentration of Ca(OH)₂, which equals 2 × 0.00046 M or 0.00092 M. The next step is to calculate the pOH, which is the negative logarithm of the [OH-] concentration. Hence, pOH = -log(0.00092). After obtaining the pOH, we can find the pH by using the relationship pH = 14 - pOH. We solve for pH using the appropriate logarithmic calculations.
The pH of an aqueous solution of 0.00046 M Ca(OH)₂ can be calculated by finding the concentration of hydroxide ions ([OH-]) and then using the formula pH = -log[H+].
Since Ca(OH)₂ is a strong base, there are two OH- ions for every formula unit dissolved, so the concentration of hydroxide ions is 2 times the concentration of the solute: [OH-] = 2 × 0.00046 M = 0.00092 M.
Now, we can use the formula for pH to calculate it: pH = -log(0.00092) ≈ 3.04.