Answer:
[OH⁻] = 0.0627M
pOH = 1.20
pH = 12.8
[H⁺] = 1.59x10⁻¹³M
Explanation:
To solve this question we must, as first, find the molarity of Al(OH)₃ in the solution -Molar mass Al(OH)₃: 78.00g/mol-:
0.300g * (1mol/ 78.00g) = 3.846x10⁻³ moles
In 184mL = 0.184L:
3.846x10⁻³ moles / 0.184L = 0.0209M Al(OH)₃. Three times this molarity = [OH⁻]:
[OH⁻] = 0.0209M * 3
[OH⁻] = 0.0627M
pOH = -log [OH⁻] =
pOH = 1.20
pH = 14 - pOH
pH = 12.8
And [H⁺] = 10^-pH
[H⁺] = 1.59x10⁻¹³M
