Question

What is the concentration of OH − and pOH in a 0.00066 M solution of Ba ( OH ) 2 at 25 ∘ C? Assume complete dissociation.

Answer 1

Answer: The hydroxide ion concentration and pOH of the solution is $1.32\times 10^{-3}M$  and 2.88 respectively

Explanation:

We are given:

Concentration of barium hydroxide = 0.00066 M

The chemical equation for the dissociation of barium hydroxide follows:

$Ba(OH)_2\rightarrow Ba^{2+}+2OH^-$

1 mole of barium hydroxide produces 1 mole of barium ions and 2 moles of hydroxide ions

pOH is defined as the negative logarithm of hydroxide ion concentration present in the solution

To calculate pOH of the solution, we use the equation:

$pOH=-\log[OH^-]$

We are given:

$[OH^-]=(2\times 0.00066)=1.32\times 10^{-3}M$

Putting values in above equation, we get:

$pOH=-\log(1.32\times 10^{-3})\\\\pOH=2.88$

Hence, the hydroxide ion concentration and pOH of the solution is $1.32\times 10^{-3}M$  and 2.88 respectively