Answer :
Answer : The pOH of pure water is, 6.68
Explanation :
As we are given that:
[tex]K_w=4.38=\times 10^{-14}[/tex]
First we have to calculate the concentration of hydroxide ion.
As, [tex]K_w=[H^+]\times [OH^-][/tex]
As we know that in pure water the hydrogen ion and hydroxide ion concentration are equal. That means,
[tex][H^+]=[OH^-][/tex]
So, [tex]K_w=[OH^-]\times [OH^-][/tex]
[tex]K_w=[OH^-]^2[/tex]
[tex]4.38\times 10^{-14}=[OH^-]^2[/tex]
[tex][OH^-]=2.09\times 10^{-7}M[/tex]
Now we have to calculate the pOH.
[tex]pOH=-\log [OH^-][/tex]
[tex]pOH=-\log (2.09\times 10^{-7})[/tex]
[tex]pOH=6.68[/tex]
Therefore, the pOH of pure water is, 6.68