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* 10^(-3)M` and 2.88 respectively

Step-by-step 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* 0.00066)=1.32* 10^(-3)M`

Putting values in above equation, we get:

`pOH=-\log(1.32* 10^(-3))\\\\pOH=2.88`

Hence, the hydroxide ion concentration and pOH of the solution is
`1.32* 10^(-3)M` and 2.88 respectively