Final answer:
The pH of a 1L solution of 2.9 x 10⁻⁴ M Ca(OH)₂ is determined by first doubling the molarity to account for the two hydroxide ions per unit of Ca(OH)₂, then calculating the pOH by taking the negative logarithm of the hydroxide concentration, and finally subtracting this pOH from 14.
Step-by-step explanation:
To calculate the pH of a 1L solution of 2.9 x 10⁻⁴ M Ca(OH)₂, we first determine the hydroxide concentration [OH⁻].
Since Ca(OH)₂ is a strong base that dissociates completely in water, producing two OH⁻ ions for each formula unit, the actual [OH⁻] is twice the molarity of the Ca(OH)₂, which is 2 x 2.9 x 10⁻⁴ M = 5.8 x 10⁻⁴ M.
The pOH is calculated by taking the negative logarithm of this hydroxide concentration, thus pOH = -log(5.8 x 10⁻⁴). Finally, to find the pH, we use the expression pH = 14 - pOH. So, pH = 14 - (-log(5.8 x 10⁻⁴)).