What is the #[OH^-]# in a solution that has a #[H_3O]=1.0*10^-6# #M#?

1 Answer
Jul 30, 2016

Answer:

The solution has a #[OH^(-)]# of #1.0xx10^(-8)M#

Explanation:

The answer can be obtained one of two ways:

  1. #color(red)("Use the auto-ionization constant of water equation, Kw:"#
    study.com
    All you would have to do is rearrange the equation to solve for #[OH^(-)]# by dividing both sides of the equation by #[H_3O^(+)]:#

#[OH^(-)] =(1.0xx10^(-14))/[[H_3O^(+)]]#

Plug in the known concentration of #H_3O^(+)# ions:
#[OH^(-)] =(1.0xx10^(-14))/(1.0xx10^(-6))#

#[OH^(-)] = 1.0xx10^(-8)M#

2.
Determine the pH of the solution by taking the negative logarithm
(-log) of the concentration of #H_3O^(+)# ions.

#pH = -log(1.0xx10^(-6)) = 6#

Then obtain the pOH using the equation: #pH + pOH = 14#. Rearrange to solve for pOH:

#pOH = 14 - 6 = 8#

Finally, you take the anti log (inverse of the natural logarithm), #10^("negative number")# of the pOH to obtain the concentration of #OH^(-)# ions.

#10^(-pOH)#
#10^(-8) = 1.0xx10^(-8)M#