Answer:
[tex][H_3O^+]=2.40x10^{-13}M[/tex]
[tex][OH^-]=0.0417M[/tex]
[tex]pOH=1.38[/tex]
Explanation:
Hello,
In this case, we are able to compute the concentration of hydroxyl and hydronium by using both he pH and the pOH as shown below:
[tex][OH^-]=10^{-pOH}[/tex]
[tex][H_3O^+]=10^{-pH}[/tex]
Moreover, we are able to compute the pOH by:
[tex]pH+pOH=14\\pOH=14-pH=14-12.62=1.38[/tex]
So we compute the concentrations:
[tex][OH^-]=10^{-1.38}=0.0417M[/tex]
[tex][H_3O^+]=10^{-12.62}=2.40x10^{-13}M[/tex]
Best regards.
