Final answer:
Calculating the [OH-], [H3O+], pH, and pOH for solutions of the strong bases Ba(OH)2, KOH, and Ca(OH)2 involves using their molarities, understanding the stoichiometry of hydroxide production for each compound, and applying the relationships between [OH-] and [H3O+], as well as pH and pOH calculations.
Step-by-step explanation:
The subject matter involves calculating the concentration of hydroxide ions ([OH-]), hydronium ions ([H3O+]), pH, and pOH in solutions of strong bases, which are Ba(OH)2, KOH, and Ca(OH)2. These calculations are essential for understanding the chemical properties of these solutions. For Ba(OH)2 and Ca(OH)2, remember that these bases provide two moles of OH- per mole of compound as they are both dihydroxide compounds.
- Ba(OH)2 1.13×10−2 M: Since Ba(OH)2 produces two moles of OH- for every mole of compound dissolved, the [OH-] = 2 × (1.13×10−2 M) and we proceed with further calculations from there.
- KOH 1.8×10−4 M: KOH produces one mole of OH- per mole of compound, so [OH-] is the same as the molarity of the solution.
- Ca(OH)2 5.3×10−4 M: Similar to Ba(OH)2, two moles of OH- are produced per mole of Ca(OH)2, so the [OH-] = 2 × (5.3×10−4 M).
Once [OH-] is known, we use Kw = [H3O+] × [OH-] to find [H3O+], where Kw = 1×10−14 at 25°C. We then calculate the pOH by taking the negative logarithm of [OH-] and determine the pH by subtracting the pOH from 14.