A) 0.11 M NaOH:
Since NaOH is a strong base, it dissociates completely in water:
[OH-] = 0.11 M
[H₃O+] = 1 x10⁻¹⁴ / [OH-] = 1 x 10⁻¹⁴ 0.11 = 9.09 x 10⁻¹⁴M
pOH = -log[OH-] = -log(0.11) ≈ 0.96
pH = 14 - pOH = 14 - 0.96 ≈ 13.04
B) 1.5 x 10⁻³ M Ca(OH)₂:
Ca(OH)₂ dissociates to form two OH- ions per formula unit:
[OH-] = 2 x 1.5 x 10⁻³ = 3 x 10⁻³ M
[H3O+] = 1 x 10⁻¹⁴ / [OH-] = 1 x 10⁻¹⁴/ 3 x 10⁻³ = 3.33 x 10⁻¹²M
pOH = -log[OH-] = -log(3 x 10⁻³) ≈ 2.52
pH = 14 - pOH = 14 - 2.52 ≈ 11.48
C) 4.8 x 10⁻⁴ M Sr(OH)₂:
Sr(OH)₂ dissociates to form two OH- ions per formula unit:
[OH-] = 2 x 4.8 x 10⁻⁴ = 9.6 x 10⁻⁴ M
[H3O+] = 1 x 10⁻¹⁴/ [OH-] = 1 x 10⁻¹⁴ / 9.6 x 10⁻¹⁴ = 1.04 x 10⁻¹¹M
pOH = -log[OH-] = -log(9.6 x 10⁻⁴) ≈ 3.02
pH = 14 - pOH = 14 - 3.02 ≈ 10.98
D) 8.7 x 10⁻⁵ M KOH:
Since KOH is a strong base, it dissociates completely in water:
[OH-] = 8.7 x 10⁻⁵ M
[H3O+] = 1 x 10⁻¹⁴ / [OH-] = 1 x 10⁻¹⁴ / 8.7 x 10⁻⁵ = 1.15 x 10⁻¹⁰ M
pOH = -log[OH-] = -log(8.7 x 10⁻⁵) ≈ 4.06
pH = 14 - pOH = 14 - 4.06 ≈ 9.94
To determine the concentrations of hydroxide ions ([OH-]), hydronium ions ([H3O+]), pH, and pOH for the given strong base solutions, we can use the fact that strong bases dissociate completely in water. Here are the calculations for each solution:
A) 0.11 M NaOH:
Since NaOH is a strong base, it dissociates into Na+ and OH- ions. Therefore, [OH-] is equal to the concentration of NaOH, which is 0.11 M. In water, the concentration of H₃O+ is negligible because NaOH does not provide H+ ions. As a result, the pH can be calculated by taking the negative logarithm of the [OH-] concentration, which is approximately 13.04. The pOH is the negative logarithm of the [H₃O+] concentration, which is negligible.
B) 1.5 x 10⁻³ M Ca(OH)₂:
Calcium hydroxide ( Ca(OH)₂) dissociates into Ca₂+ and two OH- ions. Since the concentration of Ca(OH)₂ is 1.5 x 10⁻³ M, the concentration of OH- ions is twice that, or 3 x 10⁻³ M. The concentration of H₃O+ is negligible in this case. Therefore, the pOH can be calculated by taking the negative logarithm of the [OH-] concentration, resulting in approximately 2.52. The pH is 14 minus the pOH, which is approximately 11.48.
C) 4.8 x 10⁻⁴ M Sr(OH)₂:
Strontium hydroxide (Sr(OH)2) dissociates into Sr₂+ and two OH- ions. Thus, the concentration of OH- ions is twice the concentration of Sr(OH)2, which is 9.6 x 10⁻⁴ M. Since the concentration of H₃O+ is negligible, the pOH can be calculated as approximately 3.02. The pH is 14 minus the pOH, which is approximately 10.98.
D)8.7 x 10⁻⁵ M KOH:
As KOH is a strong base, it dissociates into K+ and OH- ions. Consequently, the concentration of OH- ions is equal to the concentration of KOH, which is 8.7 x 10⁻⁵ M. Since there are no H₃O+ ions provided by KOH, the pH is calculated by taking the negative logarithm of the [OH-] concentration, resulting in approximately 9.94. The pOH is negligible in this case.
These calculations provide the values for [OH-], [H₃O+], pH, and pOH for each of the given strong base solutions.