Answer:
pH = 11,17
Explanation:
[tex][OH^{-} ] = \frac{n_{OH^{-} } }{V} = \frac{2n_{Ca(OH)_{2} } }{V} = 2[ Ca(OH)_{2} ]\\[/tex]
[tex]Ca(OH)_{2}[/tex] → [tex]Ca^ {2+}[/tex] + [tex]2OH ^ -[/tex]
[tex]7,53.10^-4[/tex] → [tex]7,53.10^-4[/tex] → [tex]1,506^-3[/tex]
pOH = -log [[tex]OH^-[/tex]] = [tex]-log 1,506^{-3}[/tex] ≈ 2,822
pH = 14- pOH = 14-2,822 = 11,178
