Given :
0.00072 M solution of [tex]Ba(OH)_2[/tex] at [tex]25^oC[/tex] .
To Find :
The concentration of [tex]OH^-[/tex]and pOH .
Solution :
1 mole of [tex]Ba(OH)_2[/tex] gives 2 moles of [tex]OH^-[/tex] ions .
So , 0.00072 M mole of [tex]Ba(OH)_2[/tex] gives :
[tex][OH^-]=2 \times 0.00072\ M[/tex]
[tex][OH^-]=0.00144\ M[/tex]
[tex][OH^-]=1.44\times 10^{-3}\ M[/tex]
Now , pOH is given by :
[tex]pOH=-log[OH^-]\\\\pOH=-log[1.44\times 10^{-3}]\\\\pOH=2.84[/tex]
Hence , this is the required solution .
