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

Jul 30, 2016

The solution has a $\left[O {H}^{-}\right]$ of $1.0 \times {10}^{- 8} M$

#### 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 $\left[O {H}^{-}\right]$ by dividing both sides of the equation by $\left[{H}_{3} {O}^{+}\right] :$

$\left[O {H}^{-}\right] = \frac{1.0 \times {10}^{- 14}}{\left[{H}_{3} {O}^{+}\right]}$

Plug in the known concentration of ${H}_{3} {O}^{+}$ ions:
$\left[O {H}^{-}\right] = \frac{1.0 \times {10}^{- 14}}{1.0 \times {10}^{- 6}}$

$\left[O {H}^{-}\right] = 1.0 \times {10}^{- 8} M$

2.
Determine the pH of the solution by taking the negative logarithm
(-log) of the concentration of ${H}_{3} {O}^{+}$ ions.

$p H = - \log \left(1.0 \times {10}^{- 6}\right) = 6$

Then obtain the pOH using the equation: $p H + p O H = 14$. Rearrange to solve for pOH:

$p O H = 14 - 6 = 8$

Finally, you take the anti log (inverse of the natural logarithm), ${10}^{\text{negative number}}$ of the pOH to obtain the concentration of $O {H}^{-}$ ions.

${10}^{- p O H}$
${10}^{- 8} = 1.0 \times {10}^{- 8} M$