# Question 231e7

Aug 3, 2017

$\left[{\text{H"_3"O}}^{+}\right] = 1.1 \times {10}^{-} 13$ $M$

#### Explanation:

We're asked to find the hydronium ion concentration $\left[{\text{H"_3"O}}^{+}\right]$, given the hydroxide ion concentration $\left[{\text{OH}}^{-}\right]$.

At $25$ $\text{^"o""C}$, we have the relationship

ul(K_w = ["OH"^(-)]["H"_3"O"^+] = 1.00xx10^-14color(white)(l)M^2

Since $\left[{\text{OH}}^{-}\right]$ is given as

$9.0 \times {10}^{-} 2$ $M$

we plug that into the equation and solve for $\left[{\text{H"_3"O}}^{+}\right]$:

["H"_3"O"^+] = (1.00xx10^-14color(white)(l)M^2)/(9.0xx10^-2color(white)(l)M) = color(red)(ulbar(|stackrel(" ")(" "1.1xx10^-13color(white)(l)M" ")|)#

rounded to $2$ significant figures.