The pH of this solution is 13.529.
Given:
Number of moles Ba(OH)₂ = 0.0550 moles
Volume = 325 mL
The calculation for moles:
[tex]Ba (OH)_{2}[/tex] → [tex]Ba^{2}[/tex]⁺ + [tex]2OH[/tex]
In 1 moles Ba(OH)₂ we have 2 moles OH-
In 0.0550 moles Ba(OH)₂ we have 2*0.0550 = 0.11 moles OH-
The calculation for concentration:
[OH-] = moles / volume
[OH-] = 0.11 moles / 0.325 L
[OH-] = 0.338 M
Calculation of pOH:
pOH = -log[OH-]
pOH = -log(0.338)
pOH = 0.471
Calculation of pH:
pH + pOH = 14
pH = 14 - pOH
pH = 14 - 0.471
pH = 13.529
Thus, the pH of this solution is 13.529.
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