Question

What is the concentration of OH− and pOH in a 0.00072 M solution of Ba(OH)2 at 25 ∘C? Assume complete dissociation.

Answer 1

Given :

0.00072 M solution of
`Ba(OH)_2` at
`25^oC` .

To Find :

The concentration of
`OH^-`and pOH .

Solution :

1 mole of
`Ba(OH)_2` gives 2 moles of
`OH^-` ions .

So , 0.00072 M mole of
`Ba(OH)_2` gives :

`[OH^-]=2 * 0.00072\ M`

`[OH^-]=0.00144\ M`

`[OH^-]=1.44* 10^(-3)\ M`

Now , pOH is given by :

`pOH=-log[OH^-]\\\\pOH=-log[1.44* 10^(-3)]\\\\pOH=2.84`

Hence , this is the required solution .