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 :

$[OH^-]=2 \times 0.00072\ M$

$[OH^-]=0.00144\ M$

$[OH^-]=1.44\times 10^{-3}\ M$

Now , pOH is given by :

$pOH=-log[OH^-]\\\\pOH=-log[1.44\times 10^{-3}]\\\\pOH=2.84$

Hence , this is the required solution .