Answer:
[H3O (+)] = 7.1 x 10⁻¹⁰ M.
Explanation:
To solve this problem, we have to use the following equation:
[tex]\lbrack H_3O^+]\lbrack OH^-]=1\cdot10^{-14}.[/tex]As we have the value of [OH (-)], we just have to solve for [H3O (+)] and replace the data that we have, like this:
[tex]\begin{gathered} \lbrack H_3O^+]=\frac{1\cdot10^{-14}}{\lbrack OH^-]}=\frac{1\cdot10^{-14}}{1.4\cdot10^{-5}}, \\ \lbrack H_3O^+]=7.1\cdot10^{-10}\text{ M.} \end{gathered}[/tex]The answer would be that the concentration of [H3O (+)] = 7.1 x 10⁻¹⁰ M.