Question

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

Answer 1

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:"`
   
   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`