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Calculate the pH of a [tex]$2.71 \times 10^{-5} \, M \, \text{Ba(OH)}_2$[/tex] solution.

[tex]pH = \square[/tex]


Sagot :

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].