Answer: The hydroxide ion concentration and pOH of the solution is [tex]1.32\times 10^{-3}M[/tex] and 2.88 respectively
Explanation:
We are given:
Concentration of barium hydroxide = 0.00066 M
The chemical equation for the dissociation of barium hydroxide follows:
[tex]Ba(OH)_2\rightarrow Ba^{2+}+2OH^-[/tex]
1 mole of barium hydroxide produces 1 mole of barium ions and 2 moles of hydroxide ions
pOH is defined as the negative logarithm of hydroxide ion concentration present in the solution
To calculate pOH of the solution, we use the equation:
[tex]pOH=-\log[OH^-][/tex]
We are given:
[tex][OH^-]=(2\times 0.00066)=1.32\times 10^{-3}M[/tex]
Putting values in above equation, we get:
[tex]pOH=-\log(1.32\times 10^{-3})\\\\pOH=2.88[/tex]
Hence, the hydroxide ion concentration and pOH of the solution is [tex]1.32\times 10^{-3}M[/tex] and 2.88 respectively
