## Chemistry (7th Edition)

$[H_3O^+] = [OH^-] = 3.9 \times 10^{-6}M$ - The solution is neutral.
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.