To determine the pH of a [tex]$2.71 \times 10^{-5} \, M \, Ba(OH)_2$[/tex] solution, we need to follow these steps:
1. Determine the concentration of hydroxide ions ([tex]$OH^-$[/tex]):
Barium hydroxide, [tex]$Ba(OH)_2$[/tex], dissociates fully in water according to the equation:
[tex]$
Ba(OH)_2 \rightarrow Ba^{2+} + 2OH^-
$[/tex]
For each mole of [tex]$Ba(OH)_2$[/tex] that dissociates, we get one mole of [tex]$Ba^{2+}$[/tex] ions and two moles of [tex]$OH^-$[/tex] ions. Therefore, the concentration of hydroxide ions can be calculated as:
[tex]$
[OH^-] = 2 \times (2.71 \times 10^{-5} \, M) = 5.42 \times 10^{-5} \, M
$[/tex]
2. Calculate the pOH of the solution:
The pOH is calculated using the hydroxide ion concentration:
[tex]$
pOH = -\log [OH^-]
$[/tex]
Substituting the value of [tex]$[OH^-]$[/tex]:
[tex]$
pOH = -\log (5.42 \times 10^{-5}) \approx 4.266
$[/tex]
3. Determine the pH of the solution:
The pH and pOH are related by the equation:
[tex]$
pH + pOH = 14
$[/tex]
Rearranging to find pH:
[tex]$
pH = 14 - pOH
$[/tex]
Substituting the value of pOH:
[tex]$
pH = 14 - 4.266 \approx 9.734
$[/tex]
So, the pH of a [tex]$2.71 \times 10^{-5} \, M \, Ba(OH)_2$[/tex] solution is approximately [tex]$9.734$[/tex].
