Answer : The correct option is, (A) [tex]5.0\times 10^{-12}M[/tex]
Explanation:
First we have to calculate the pH.
[tex]pH=-\log [H^+][/tex]
[tex]pH=-\log (2.0\times 10^{-3})[/tex]
[tex]pH=2.69[/tex]
Now we have to calculate the pOH.
[tex]pH+pOH=14\\\\pOH=14-pH\\\\pOH=14-2.69=11.31[/tex]
Now we have to calculate the [tex]OH^-[/tex] concentration.
[tex]pOH=-\log [OH^-][/tex]
[tex]11.31=-\log [OH^-][/tex]
[tex][OH^-]=5.0\times 10^{-12}M[/tex]
Therefore, the [tex]OH^-[/tex] concentration is, [tex]5.0\times 10^{-12}M[/tex]
