Question

What is the poh of pure water that has a kw value of 4.38 x 10-14?

Answer 1

Answer : The pOH of pure water is, 6.68

Explanation :

As we are given that:

`K_w=4.38=* 10^(-14)`

First we have to calculate the concentration of hydroxide ion.

As,
`K_w=[H^+]* [OH^-]`

As we know that in pure water the hydrogen ion and hydroxide ion concentration are equal. That means,

`[H^+]=[OH^-]`

So,
`K_w=[OH^-]* [OH^-]`

`K_w=[OH^-]^2`

`4.38* 10^(-14)=[OH^-]^2`

`[OH^-]=2.09* 10^(-7)M`

Now we have to calculate the pOH.

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

`pOH=-\log (2.09* 10^(-7))`

`pOH=6.68`

Therefore, the pOH of pure water is, 6.68