Question

Ba(OH)2 is a strong electrolyte. Determine the concentration of each of the individual ions in a 0.300 M Ba(OH)2 solution.?

Answer 1

`Ba^+^2` = 0.3 M

`OH^-` = 0.6 M

Step-by-step explanation:

We can start wit the ioniation reaction:

`Ba(OH)_2~->~Ba^+^2~+~2OH^-`

We have the ions
`Ba^+^2` and
`OH^-`. Now, we can calculate the concentration of each ion:

Concentration of
`Ba^+^2`

In balanced reaction we have 1 mol of
`Ba(OH)_2` and 1 mol of
`Ba^+^2`, so, we have a 1:1 mol ratio, with this in mind:

`0.300~M~Ba(OH)_2(1~Ba^+^2)/(1~Ba(OH)_2)=0.300~M~Ba^+^2`

Concentration of
`OH^-`

In balanced reaction we have 1 mol of
`Ba(OH)_2` and 2 mol of
`OH^-`, so, we have a 1:2 mol ratio, with this in mind:

`0.300~M~Ba(OH)_2(2~OH^-)/(1~Ba(OH)_2)=0.600~M~OH^-`

I hope it helps!