Answer: [tex][OH^{-}]= 0.01M or 1.0\times 10^{-2}M[/tex]
[tex][H^{+}]= 1.0\times 10^{-12}M[/tex]
pH = 12
pOH = 2
Explanation: Calcium hydroxide ([tex]Ca(OH)_{2}[/tex]) is a strong base that dissociates completely.
Dissociation equation of Calcium hydroxide is :
[tex]Ca(OH)_{2} \rightarrow Ca^{+2} + 2OH^{-}[/tex]
1. Concentration of [OH-]
1 mol [tex]Ca(OH)_{2}[/tex] produces 2 mol OH- ions.
The given solution is 0.005M [tex]Ca(OH)_{2}[/tex] , then  concentration of OH- would be twice the concentration of [tex]Ca(OH)_{2}[/tex]
[tex][OH^{-}] = 0.005\times 2 [/tex] = 0.01M or [tex]1.0\times 10^{-2}M[/tex]
2.Concentration of [H+]
Concentration of [H+] can be calculated by the formula: [tex][H^{+}] = \frac{Kw}{[OH^{-}]}[/tex]
kw = ionic product of water and its values is [tex](1\times 10^{-14})[/tex]
[OH-] = 0.01 M or [tex]1.0\times 10^{-2}M[/tex]
[tex][H^{+}] = \frac{(1\times 10^{-14})}{[OH^{-}]}[/tex]
[tex][H^{+}] = \frac{(1\times 10^{-14})}{[0.01]}[/tex]
[tex][H^{+}] = 1.0\times 10^{-12}M [/tex]
3. pH value
pH is calculated by the formula : [tex]pH = -log[H^{+}][/tex]
[tex]pH = -log[1.0\times 10^{-12}][/tex]
pH = 12
4. pOH value
pOH is calculated by the formula : pOH = 14 - pH
pOH = 14 - 12
pOH = 2
pOH can also be calculated by using a different formula which is :
[tex]pOH = -log(OH^{-})[/tex]
[tex]pOH = -log(0.01)[/tex]
pOH = 2.
