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:

$K_w=4.38=\times 10^{-14}$

First we have to calculate the concentration of hydroxide ion.

As, $K_w=[H^+]\times [OH^-]$

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

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

So, $K_w=[OH^-]\times [OH^-]$

$K_w=[OH^-]^2$

$4.38\times 10^{-14}=[OH^-]^2$

$[OH^-]=2.09\times 10^{-7}M$

Now we have to calculate the pOH.

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

$pOH=-\log (2.09\times 10^{-7})$

$pOH=6.68$

Therefore, the pOH of pure water is, 6.68