Answer
$[H_3O^+] = [OH^-] = 3.9 \times 10^{-6}M$
- The solution is neutral.
Work Step by Step
For pure water: $[H_3O^+] = [OH^-]$, so we are going to create an unkown called "x", that has this value: $x = [H_3O^+] = [OH^-]$
Writing the Kw expression:
$[H_3O^+] * [OH^-] = K_w = 1.5 \times 10^{-11}$
$x * x = 1.5 \times 10^{-11}$
$x^2 = 1.5 \times 10^{-11}$
$x = \sqrt {1.5 \times 10^{-11}}$
$x = 3.9 \times 10^{-6}M$
So: $[H_3O^+] and [OH^-] = 3.9 \times 10^{-6}M$
- $Since [OH^-] = [H_3O^+]$, the solution is neutral.
