Answer:
[tex]Ba^+^2[/tex] = 0.3 M
[tex]OH^-[/tex] = 0.6 M
Explanation:
We can start wit the ioniation reaction:
[tex]Ba(OH)_2~->~Ba^+^2~+~2OH^-[/tex]
We have the ions [tex]Ba^+^2[/tex] and [tex]OH^-[/tex]. Now, we can calculate the concentration of each ion:
Concentration of [tex]Ba^+^2[/tex]
In balanced reaction we have 1 mol of [tex]Ba(OH)_2[/tex] and 1 mol of [tex]Ba^+^2[/tex], so, we have a 1:1 mol ratio, with this in mind:
[tex]0.300~M~Ba(OH)_2\frac{1~Ba^+^2}{1~Ba(OH)_2}=0.300~M~Ba^+^2[/tex]
Concentration of [tex]OH^-[/tex]
In balanced reaction we have 1 mol of [tex]Ba(OH)_2[/tex] and 2 mol of [tex]OH^-[/tex], so, we have a 1:2 mol ratio, with this in mind:
[tex]0.300~M~Ba(OH)_2\frac{2~OH^-}{1~Ba(OH)_2}=0.600~M~OH^-[/tex]
I hope it helps!
