0.100g of Cu(OH)₂ was added to a beaker which was filled to the 3.00L mark with water then the pOH of the solution is 3.2.
pOH of any solution will be calculated by using the below equation as:
pOH = -log[OH⁻]
Concentration will be calculated in terms of molarity as:
M = n/V, where
n = moles
V = volume = 3L
First we calculate moles of Cu(OH)₂ as:
Moles = 0.1g / 97.56g/mol = 0.001 mol
Dissocistaion reaction will be:
Cu(OH)₂ → Cu²⁺ + 2OH⁻
From the stoichiometry it is clear that:
1 moles of Cu(OH)₂ = produces 2 moles of OH⁻
0.001 moles of Cu(OH)₂ = produces 2×0.001=0.002 moles of OH⁻
Concentration of OH⁻ = 0.002 / 3 = 0.0006M
Now we put this value in pOH equation, we get
pOH = -log(0.0006)
pOH = -(-3.2) = 3.2
Hence required value of pOH is 3.2.
To know more about pOH, visit the below link:
https://brainly.com/question/27157023
