# What is the [OH^-] for a water solution if the [H_3O^+] is 6.0 times 10^-11 M?

May 28, 2017

$1.7 \times {10}^{-} 4 M$

#### Explanation:

For solutions at ${25}^{o} \text{C}$, the total concentration of $\left[{\text{OH}}^{-}\right]$ and $\left[{\text{H"_3"O}}^{+}\right]$ is given by the equation

K_"w" = ["OH"^-]["H"_3"O"^+] = 1.00 xx 10^-14 M^2

This equation is applicable to both pure water and to any aqueous solution. Although this equilibrium is somewhat affected by the presence of other ions in solution, we generally disregard this unless we're dealing with calculations involving great accuracy.

Since the product of the concentrations of the hydroxide and hydronium ions equals a constant, the two concentrations are inversely proportional. That is, if one increases, the other must decrease.

We can plug the given $\left[{\text{H"_3"O}}^{+}\right]$ into the above equation and solve for $\left[{\text{OH}}^{-}\right]$:

["OH"^-] = (1.00 xx 10^-14 M^cancel(2))/(6.0 xx 10^-11 cancel(M)) = color(red)(1.7 xx 10^-4 M